Systems of Linear Equations in Two Variables

Form and Graphical Representation of Linear Systems

In many real-world situations, two different conditions must be true at the same time. When we write each condition as a linear equation, we get a system of linear equations.

A linear system in two variables is a pair of equations with the same variables, usually \(x\) and \(y\). The solution to the system is the ordered pair \((x,y)\) that makes both equations true at the same time.

In this lesson, you will learn how to:

• recognize the form of a linear system,
• write equations from given conditions,
• graph both lines on the same coordinate plane, and
• interpret what the point of intersection means.

1. What is a linear equation?

A linear equation in two variables can be written in forms such as:

• Slope-intercept form: \(y=mx+b\)
• Standard form: \(Ax+By=C\)

These equations graph as straight lines.

Examples of linear equations are:

• \(y=2x+1\)
• \(3x+y=7\)
• \(x-4y=12\)

2. What is a system of linear equations?

A system of linear equations is two or more linear equations considered together. In 10th Grade, we usually work with two equations in two variables.

For example:

$$ \begin{cases} y=2x+1\\ y=-x+7 \end{cases} $$

Each equation makes a line. When both lines are graphed on the same coordinate plane, the solution is where the lines intersect.

3. What does the graph tell us?

When graphing a system, there are three possible results:

• One solution: the lines cross at one point.
• No solution: the lines are parallel and never meet.
• Infinitely many solutions: the lines are the same line.

These cases can be understood visually:

• If two different lines have different slopes, they usually intersect once.
• If they have the same slope but different intercepts, they are parallel.
• If they have the same slope and the same intercept, they overlap completely.

4. Writing a linear system from conditions

To form a system, translate each condition into an equation.

For example, suppose:

• The sum of two numbers is 10.
• Their difference is 2.

Let the numbers be \(x\) and \(y\).

Then the system is:

$$ \begin{cases} x+y=10\\ x-y=2 \end{cases} $$

Each sentence becomes one equation. Together, they form a system.

5. How to graph a linear system

To graph a system:

1. Graph the first equation as a line.
2. Graph the second equation on the same axes.
3. Find the point where the lines intersect.
4. Check that the point satisfies both equations.

You can graph a line in different ways:

• by using slope and \(y\)-intercept if the equation is in \(y=mx+b\) form,
• or by finding two points if the equation is in standard form.

6. Worked Example 1: Graphing from slope-intercept form

Graph the system:

$$ \begin{cases} y=x+1\\ y=-x+5 \end{cases} $$

Step 1: Graph \(y=x+1\).

This line has slope \(1\) and \(y\)-intercept \(1\).

• Start at \((0,1)\).
• Use slope \(1=\frac{1}{1}\): go up 1 and right 1.
• Another point is \((1,2)\).

Step 2: Graph \(y=-x+5\).

This line has slope \(-1\) and \(y\)-intercept \(5\).

• Start at \((0,5)\).
• Use slope \(-1\): go down 1 and right 1.
• Another point is \((1,4)\).

Step 3: Find the intersection.

The lines meet at \((2,3)\).

Step 4: Check the solution.

Substitute \((2,3)\) into both equations:

• For \(y=x+1\): \(3=2+1\), true.
• For \(y=-x+5\): \(3=-2+5\), true.

So the solution is:

$$ (2,3) $$

7. Worked Example 2: Graphing from standard form

Graph the system:

$$ \begin{cases} 2x+y=4\\ x-y=2 \end{cases} $$

Step 1: Graph \(2x+y=4\).

Find two points that satisfy the equation.

• If \(x=0\), then \(y=4\). So one point is \((0,4)\).
• If \(x=2\), then \(2(2)+y=4\), so \(y=0\). Another point is \((2,0)\).

Draw the line through \((0,4)\) and \((2,0)\).

Step 2: Graph \(x-y=2\).

Again, find two points.

• If \(x=2\), then \(2-y=2\), so \(y=0\). One point is \((2,0)\).
• If \(x=3\), then \(3-y=2\), so \(y=1\). Another point is \((3,1)\).

Draw the line through those points.

Step 3: Find the intersection.

The lines intersect at \((2,0)\).

Step 4: Check.

• \(2(2)+0=4\), true.
• \(2-0=2\), true.

So the solution is:

$$ (2,0) $$

8. Worked Example 3: No solution

Graph the system:

$$ \begin{cases} y=2x+1\\ y=2x-3 \end{cases} $$

Both lines have slope \(2\), so they rise at the same rate.

But their \(y\)-intercepts are different: one crosses the \(y\)-axis at \(1\), and the other crosses at \(-3\).

That means the lines are parallel. Parallel lines never intersect.

So this system has:

$$ \text{no solution} $$

9. Worked Example 4: Writing and graphing a real-world system

A school snack stand sells apples for \$1 each and juice boxes for \$2 each. A student buys a total of 5 items and spends \$8.

Let:

• \(x\) = number of apples
• \(y\) = number of juice boxes

Step 1: Write the equations.

The total number of items is 5:

$$ x+y=5 $$

The total cost is 8 dollars:

$$ x+2y=8 $$

So the system is:

$$ \begin{cases} x+y=5\\ x+2y=8 \end{cases} $$

Step 2: Graph the first equation.

For \(x+y=5\):

• If \(x=0\), then \(y=5\), giving \((0,5)\).
• If \(x=5\), then \(y=0\), giving \((5,0)\).

Step 3: Graph the second equation.

For \(x+2y=8\):

• If \(x=0\), then \(2y=8\), so \(y=4\), giving \((0,4)\).
• If \(x=4\), then \(4+2y=8\), so \(y=2\), giving \((4,2)\).

Step 4: Find the intersection.

The lines intersect at \((2,3)\).

Step 5: Interpret the answer.

The student bought 2 apples and 3 juice boxes.

This is why graphing matters: the intersection point is not just a math answer. It represents a real situation that satisfies both conditions.

10. Important ideas when reading a graph

• The solution to a linear system is an ordered pair.
• That ordered pair must lie on both lines.
• If the lines do not meet, there is no ordered pair that works for both equations.
• In real-world problems, the solution should make sense in the context.

For example, if \(x\) and \(y\) represent numbers of items, negative values may not make sense.

11. Tips for success

• Make sure both equations are graphed on the same coordinate plane.
• Use at least two accurate points for each line.
• Label your lines or points if needed.
• Check the intersection point in both equations.
• If the equations are in standard form, finding intercepts can help you graph quickly.

12. Common mistakes to avoid

• Graphing only one equation instead of both.
• Reading the wrong intersection point from the graph.
• Using points that do not actually satisfy the equation.
• Forgetting that the solution must work in both equations.
• Assuming every system has one solution.

Brief Summary

A linear system in two variables is made of two linear equations that are considered together. When you graph both lines on the same coordinate plane, the solution is the point where they intersect. Some systems have one solution, some have no solution, and some have infinitely many solutions. In real-world problems, the intersection represents the values that satisfy all given conditions at the same time.

Conditions for Consistency and Inconsistency help us decide whether a pair of linear equations has one solution, no solution, or infinitely many solutions.

When we solve a system of linear equations in two variables, we are trying to find values of the variables that satisfy both equations at the same time.

A general pair of linear equations in two variables can be written as:

$$a_1x + b_1y + c_1 = 0$$ $$a_2x + b_2y + c_2 = 0$$

To check whether this system is consistent or inconsistent, we compare the ratios:

$$\frac{a_1}{a_2}, \quad \frac{b_1}{b_2}, \quad \frac{c_1}{c_2}$$

These ratios tell us how the two equations are related.

Important words:

• Consistent system: A system that has at least one solution.
• Inconsistent system: A system that has no solution.
• Dependent equations: Two equations that represent the same line.
• Independent equations: Two equations that represent different lines and meet at one point.

There are three possible cases for a pair of linear equations.

Case 1: Unique Solution

If

$$\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$$

then the system has one unique solution.

This means the two lines have different slopes, so they intersect at exactly one point.

So this system is consistent.

Case 2: Infinite Solutions

If

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

then the system has infinitely many solutions.

This means both equations represent the same line. Every point on that line satisfies both equations.

This system is also consistent, because it has solutions.

Case 3: No Solution

If

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$$

then the system has no solution.

This means the two lines are parallel. They never meet, so there is no point that satisfies both equations.

This system is inconsistent.

Quick table of conditions

• If $$\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$$ → one unique solution → consistent
• If $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$ → infinitely many solutions → consistent
• If $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$$ → no solution → inconsistent

Why do these conditions work?

Each linear equation represents a straight line on a graph.

• If the lines cross once, there is one solution.
• If the lines are the same, there are infinitely many solutions.
• If the lines are parallel, there is no solution.

The ratios help us recognize these situations without drawing the graph.

How to test a system step by step

1. Write both equations in the form $$a_1x + b_1y + c_1 = 0$$ and $$a_2x + b_2y + c_2 = 0$$.
2. Identify the values of $$a_1, b_1, c_1, a_2, b_2, c_2$$.
3. Find the three ratios $$\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$$.
4. Compare the ratios carefully.
5. Use the condition table to decide whether the system is consistent or inconsistent.

Worked Example 1: A system with a unique solution

Determine whether the following system is consistent or inconsistent:

$$2x + 3y - 5 = 0$$ $$4x - y + 1 = 0$$

Here,

$$a_1 = 2, \quad b_1 = 3, \quad c_1 = -5$$ $$a_2 = 4, \quad b_2 = -1, \quad c_2 = 1$$

Now find the ratios:

$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$$ $$\frac{b_1}{b_2} = \frac{3}{-1} = -3$$

Since

$$\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$$

the system has a unique solution.

So the system is consistent.

Worked Example 2: A system with no solution

Check the system:

$$2x + 4y - 6 = 0$$ $$x + 2y - 1 = 0$$

Identify the coefficients:

$$a_1 = 2, \quad b_1 = 4, \quad c_1 = -6$$ $$a_2 = 1, \quad b_2 = 2, \quad c_2 = -1$$

Now calculate the ratios:

$$\frac{a_1}{a_2} = \frac{2}{1} = 2$$ $$\frac{b_1}{b_2} = \frac{4}{2} = 2$$ $$\frac{c_1}{c_2} = \frac{-6}{-1} = 6$$

We see that

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$$

So the system has no solution.

This means the lines are parallel.

Therefore, the system is inconsistent.

Worked Example 3: A system with infinitely many solutions

Check the system:

$$3x + 6y - 9 = 0$$ $$x + 2y - 3 = 0$$

Identify the coefficients:

$$a_1 = 3, \quad b_1 = 6, \quad c_1 = -9$$ $$a_2 = 1, \quad b_2 = 2, \quad c_2 = -3$$

Now calculate the ratios:

$$\frac{a_1}{a_2} = \frac{3}{1} = 3$$ $$\frac{b_1}{b_2} = \frac{6}{2} = 3$$ $$\frac{c_1}{c_2} = \frac{-9}{-3} = 3$$

Since

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

the system has infinitely many solutions.

The two equations represent the same line.

So the system is consistent.

Worked Example 4: First rewrite in standard form

Determine the nature of the system:

$$x - 2y = 7$$ $$2x - 4y = 10$$

First write both equations in the form $$ax + by + c = 0$$:

$$x - 2y - 7 = 0$$ $$2x - 4y - 10 = 0$$

Now identify the coefficients:

$$a_1 = 1, \quad b_1 = -2, \quad c_1 = -7$$ $$a_2 = 2, \quad b_2 = -4, \quad c_2 = -10$$

Find the ratios:

$$\frac{a_1}{a_2} = \frac{1}{2}$$ $$\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$$ $$\frac{c_1}{c_2} = \frac{-7}{-10} = \frac{7}{10}$$

So,

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$$

Therefore, the system has no solution.

Hence, it is inconsistent.

Common mistakes to avoid

• Not writing equations in standard form first. Make sure both equations are written as $$ax + by + c = 0$$ before comparing coefficients.
• Forgetting signs. A negative sign in a coefficient changes the ratio.
• Checking only two ratios when all three are needed. To decide between no solution and infinitely many solutions, you must also compare $$\frac{c_1}{c_2}$$.
• Thinking “same ratio” always means same line. That is true only when all three ratios are equal.

Connecting to graphs

You can also understand consistency by imagining the graph of the two equations.

• Intersecting lines → one point of intersection → one unique solution
• Coincident lines (same line) → every point common → infinitely many solutions
• Parallel lines → no intersection → no solution

This is why the ratio test is so useful: it gives the same conclusion without actually drawing the graph.

Mini practice check

Try to classify each system:

1. $$x + y - 2 = 0$$ and $$2x - y + 3 = 0$$
2. $$2x + 6y - 8 = 0$$ and $$x + 3y - 4 = 0$$
3. $$3x + 9y - 12 = 0$$ and $$x + 3y - 5 = 0$$

Answers:

• 1. Unique solution, so consistent
• 2. Infinitely many solutions, so consistent
• 3. No solution, so inconsistent

Summary

To decide whether a pair of linear equations is consistent or inconsistent, compare the ratios $$\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$$.

• If $$\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$$, there is one unique solution.
• If $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$, there are infinitely many solutions.
• If $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$$, there is no solution.

Remember: consistent means at least one solution, and inconsistent means no solution.

Solving Systems Graphically means finding the point where two lines cross on a graph. That crossing point is called the intersection, and it represents the solution to the system.

A system of linear equations in two variables is a pair of equations with the same variables, usually \(x\) and \(y\). A solution to the system is an ordered pair \((x,y)\) that makes both equations true at the same time.

When we solve a system graphically, we graph both lines on the same coordinate plane and look for where they meet. If they intersect at one point, that point is the solution.

For example, if the lines cross at \((2,3)\), then \((2,3)\) is the solution because it lies on both lines.

Why this works: Every point on a line satisfies that line’s equation. So the point that lies on both lines must satisfy both equations. That is exactly what a solution to a system means.

Steps for solving a system graphically

1. Write each equation in a form that is easy to graph, often slope-intercept form: \(y=mx+b\).
2. Graph the first line.
3. Graph the second line on the same axes.
4. Find the point where the lines intersect.
5. Check the point in both equations if needed.

Reminder about slope-intercept form:

In the equation

$$y=mx+b$$

the number \(m\) is the slope, and the number \(b\) is the y-intercept.

• The y-intercept tells where the line crosses the y-axis.
• The slope tells how steep the line is.

For example, in \(y=2x+1\):

• Slope = \(2\)
• y-intercept = \(1\)

That means the line crosses the y-axis at \((0,1)\). Then, using slope \(2\), you go up 2 and right 1 to get another point.

Possible results when graphing a system

• One solution: The lines intersect once.
• No solution: The lines are parallel and never meet.
• Infinitely many solutions: The lines are the same line, so every point on one is also on the other.

Now let’s work through examples.

Example 1: A system with one clear solution

Solve the system graphically:

$$ \begin{aligned} y&=x+1\\ y&=-x+5 \end{aligned} $$

Step 1: Graph the first line

For \(y=x+1\):

• y-intercept is \(1\), so plot \((0,1)\).
• Slope is \(1\), so go up 1 and right 1.
• Another point is \((1,2)\).

Step 2: Graph the second line

For \(y=-x+5\):

• y-intercept is \(5\), so plot \((0,5)\).
• Slope is \(-1\), so go down 1 and right 1.
• Another point is \((1,4)\).

Step 3: Find the intersection

When the lines are drawn, they intersect at \((2,3)\).

So the solution is

$$ (2,3) $$

Step 4: Check the solution

Substitute \((2,3)\) into both equations:

• First equation: \(y=x+1\) gives \(3=2+1\), true.
• Second equation: \(y=-x+5\) gives \(3=-2+5\), true.

Since the point works in both equations, it is the correct solution.

Example 2: One equation must be rewritten first

Solve the system graphically:

$$ \begin{aligned} 2x+y&=4\\ y&=x-2 \end{aligned} $$

Step 1: Rewrite the first equation in slope-intercept form

Start with

$$2x+y=4$$

Subtract \(2x\) from both sides:

$$y=-2x+4$$

Now the system is

$$ \begin{aligned} y&=-2x+4\\ y&=x-2 \end{aligned} $$

Step 2: Graph each line

For \(y=-2x+4\):

• y-intercept is \(4\), so plot \((0,4)\).
• Slope is \(-2\), so go down 2 and right 1.
• Another point is \((1,2)\).

For \(y=x-2\):

• y-intercept is \(-2\), so plot \((0,-2)\).
• Slope is \(1\), so go up 1 and right 1.
• Another point is \((1,-1)\).

Step 3: Locate the intersection

The lines meet at \((2,0)\).

So the solution is

$$ (2,0) $$

Check:

• In \(2x+y=4\): \(2(2)+0=4\), true.
• In \(y=x-2\): \(0=2-2\), true.

Example 3: No solution

Solve the system graphically:

$$ \begin{aligned} y&=2x+1\\ y&=2x-3 \end{aligned} $$

Both lines have the same slope, \(2\), but different y-intercepts, \(1\) and \(-3\).

That means the lines are parallel. Parallel lines never intersect, so there is no point that satisfies both equations.

Therefore, this system has

$$ \text{no solution} $$

What this means: There is no ordered pair \((x,y)\) that makes both equations true at the same time.

Example 4: Infinitely many solutions

Solve the system graphically:

$$ \begin{aligned} y&=3x-2\\ 2y&=6x-4 \end{aligned} $$

Step 1: Rewrite the second equation

Divide both sides by 2:

$$y=3x-2$$

Now both equations are the same:

$$ \begin{aligned} y&=3x-2\\ y&=3x-2 \end{aligned} $$

When graphed, both equations make the exact same line.

That means every point on the line is a solution. So the system has

$$ \text{infinitely many solutions} $$

Important idea: If two equations describe the same line, they do not intersect at just one point. They overlap completely.

How to read a graph carefully

• If the lines cross exactly on a grid point, the solution is usually exact.
• If the lines cross between grid points, the solution may be an estimate.
• Always write the solution as an ordered pair \((x,y)\).
• Be careful not to switch the x-coordinate and y-coordinate.

For example, if the lines seem to cross halfway between \(x=1\) and \(x=2\), and also halfway between \(y=3\) and \(y=4\), the approximate solution would be \((1.5, 3.5)\).

Common mistakes to avoid

• Graphing the slope incorrectly. Remember that slope is rise over run.
• Using the wrong y-intercept. In \(y=mx+b\), \(b\) is the y-intercept.
• Forgetting to rewrite the equation. If an equation is not already in \(y=mx+b\) form, solve for \(y\) first if needed.
• Reading the intersection wrong. Check the x-value and y-value carefully.
• Assuming all systems have one solution. Some have no solution or infinitely many solutions.

Real-world meaning of the intersection

Systems of equations are often used to compare two situations. The intersection point shows when both situations have the same value.

For example, imagine two taxi companies charge different rates. Each company’s cost can be shown by a line. The intersection tells the distance where both companies cost the same amount.

If one equation is

$$y=2x+5$$

and the other is

$$y=3x+2$$

then graphing them would show the distance \(x\) where the prices match, and the cost \(y\) at that point.

Quick strategy for solving systems graphically

1. Put both equations into slope-intercept form if possible.
2. Graph carefully using the y-intercept and slope.
3. Find the intersection point.
4. Decide whether the system has one solution, no solution, or infinitely many solutions.
5. Check the answer in both equations when you can.

Brief Summary

To solve a system graphically, graph both lines on the same coordinate plane and look for where they intersect. The intersection point is the solution because it satisfies both equations. If the lines are parallel, there is no solution. If the lines are the same, there are infinitely many solutions.

The Substitution Method is a way to solve a system of linear equations in two variables. A system of equations is a set of two or more equations that use the same variables. When we solve the system, we are looking for the ordered pair \\((x, y)\\) that makes both equations true at the same time.

The substitution method works by solving one equation for one variable, then replacing that variable in the other equation. This turns a two-variable problem into a one-variable problem, which is much easier to solve.

For example, if you know that \\(y = 2x + 1\\), then anywhere you see \\(y\\) in the other equation, you can substitute \\(2x + 1\\) in its place. That is why the method is called substitution.

Why this method makes sense: if two expressions are equal to the same variable, then they are equal to each other. Substitution lets us use that idea to combine two equations into one.

Goal: Solve for one variable first, then use that value to find the other variable.

Steps for the Substitution Method

1. Solve one equation for one variable.
   Choose the equation that is easiest to rearrange. Try to get either \\(x\\) or \\(y\\) by itself.
2. Substitute that expression into the other equation.
   Replace the variable with the expression you found.
3. Solve the new equation.
   You should now have an equation with only one variable.
4. Find the other variable.
   Substitute the value you found back into one of the original equations.
5. Check your answer.
   Make sure the ordered pair works in both original equations.

When is substitution a good choice?

• When one equation already has a variable isolated, like \\(y = 3x - 4\\).
• When it is easy to isolate one variable.
• When using substitution will avoid messy calculations.

Worked Example 1: A basic system

Solve the system:

$$ \begin{aligned} y &= x + 2 \\ x + y &= 8 \end{aligned} $$

Step 1: Solve one equation for one variable.

The first equation is already solved for \\(y\\):

$$y = x + 2$$

Step 2: Substitute into the other equation.

Replace \\(y\\) in \\(x + y = 8\\) with \\(x + 2\\):

$$x + (x + 2) = 8$$

Step 3: Solve for \\(x\\).

$$ \begin{aligned} 2x + 2 &= 8 \\ 2x &= 6 \\ x &= 3 \end{aligned} $$

Step 4: Find \\(y\\).

Substitute \\(x = 3\\) into \\(y = x + 2\\):

$$ \begin{aligned} y &= 3 + 2 \\ y &= 5 \end{aligned} $$

Solution: \\((3, 5)\\)

Step 5: Check.

First equation: \\(5 = 3 + 2\\) ✔

Second equation: \\(3 + 5 = 8\\) ✔

Worked Example 2: Solve for a variable first

Solve the system:

$$ \begin{aligned} 2x + y &= 11 \\ y &= 3x - 1 \end{aligned} $$

The second equation already gives \\(y\\), so we use it for substitution.

Step 1: Substitute \\(3x - 1\\) for \\(y\\).

$$2x + (3x - 1) = 11$$

Step 2: Solve for \\(x\\).

$$ \begin{aligned} 5x - 1 &= 11 \\ 5x &= 12 \\ x &= \frac{12}{5} \end{aligned} $$

Step 3: Find \\(y\\).

Use \\(y = 3x - 1\\):

$$ \begin{aligned} y &= 3\left(\frac{12}{5}\right) - 1 \\ y &= \frac{36}{5} - \frac{5}{5} \\ y &= \frac{31}{5} \end{aligned} $$

Solution: \\((\frac{12}{5}, \frac{31}{5})\\)

Check:

Second equation: \\(\frac{31}{5} = 3(\frac{12}{5}) - 1\\) ✔

First equation: \\(2(\frac{12}{5}) + \frac{31}{5} = \frac{24}{5} + \frac{31}{5} = \frac{55}{5} = 11\\) ✔

Important note: Answers do not always have to be whole numbers. Fractions and decimals can also be correct solutions.

Worked Example 3: Rearranging before substituting

Solve the system:

$$ \begin{aligned} 3x + y &= 14 \\ 2x - y &= 1 \end{aligned} $$

Neither equation has a variable isolated yet, so we choose one and solve for a variable first. Let's solve the first equation for \\(y\\).

Step 1: Isolate \\(y\\).

$$ \begin{aligned} 3x + y &= 14 \\ y &= 14 - 3x \end{aligned} $$

Step 2: Substitute into the second equation.

Replace \\(y\\) in \\(2x - y = 1\\) with \\(14 - 3x\\):

$$2x - (14 - 3x) = 1$$

Step 3: Solve carefully.

The minus sign in front of the parentheses changes the signs inside:

$$ \begin{aligned} 2x - 14 + 3x &= 1 \\ 5x - 14 &= 1 \\ 5x &= 15 \\ x &= 3 \end{aligned} $$

Step 4: Find \\(y\\).

Use \\(y = 14 - 3x\\):

$$ \begin{aligned} y &= 14 - 3(3) \\ y &= 14 - 9 \\ y &= 5 \end{aligned} $$

Solution: \\((3, 5)\\)

Worked Example 4: A real-world problem

A school sells tickets to a play. Student tickets cost \\$4\\) each, and adult tickets cost \\$7\\) each. In one hour, the school sold a total of \\(30\\) tickets and collected \\$156\\).

Let:

• \\(x\\) = number of student tickets
• \\(y\\) = number of adult tickets

Write a system:

$$ \begin{aligned} x + y &= 30 \\ 4x + 7y &= 156 \end{aligned} $$

Step 1: Solve one equation for one variable.

From \\(x + y = 30\\), solve for \\(x\\):

$$x = 30 - y$$

Step 2: Substitute into the second equation.

$$4(30 - y) + 7y = 156$$

Step 3: Solve for \\(y\\).

$$ \begin{aligned} 120 - 4y + 7y &= 156 \\ 120 + 3y &= 156 \\ 3y &= 36 \\ y &= 12 \end{aligned} $$

Step 4: Find \\(x\\).

$$ \begin{aligned} x &= 30 - 12 \\ x &= 18 \end{aligned} $$

Solution: \\(18\\) student tickets and \\(12\\) adult tickets.

Check:

Total tickets: \\(18 + 12 = 30\\) ✔

Total money: \\(4(18) + 7(12) = 72 + 84 = 156\\) ✔

Common Mistakes to Avoid

• Forgetting parentheses when substituting.
  If \\(y = 14 - 3x\\), then substitute as \\(2x - (14 - 3x)\\), not \\(2x - 14 - 3x\\).
• Not solving all the way.
  After finding one variable, always substitute back to find the other one.
• Arithmetic errors.
  Be careful with negative signs, fractions, and combining like terms.
• Not checking the answer.
  An ordered pair is only correct if it works in both equations.

How substitution connects to graphs

Each linear equation can be drawn as a line on a graph. The solution to the system is the point where the two lines intersect. When you solve a system using substitution, you are finding the same point algebraically instead of graphically.

If the lines cross once, the system has one solution.

If the lines are the same line, the system has infinitely many solutions.

If the lines are parallel and never meet, the system has no solution.

Quick Practice Thinking

• If one equation is \\(y = 2x + 6\\), substitute \\(2x + 6\\) wherever you see \\(y\\) in the other equation.
• If no variable is isolated, first rearrange one equation.
• Always write the final answer as an ordered pair \\((x, y)\\), unless the problem asks for the meaning in a real-world context.

Summary

The substitution method solves a system of linear equations by replacing one variable with an equal expression from the other equation. This changes a two-variable system into a one-variable equation. After solving for one variable, substitute back to find the second variable, then check your answer in both equations.

Lesson: The Elimination Method

When you have a system of linear equations, you are working with two or more equations that share the same variables. In 10th Grade, you often solve systems with two variables, such as \(x\) and \(y\).

The elimination method is a way to solve a system by combining the equations so that one variable disappears. Once one variable is eliminated, you can solve for the other variable more easily.

This method is especially useful when the equations are already lined up in standard form:

$$ax + by = c$$ $$dx + ey = f$$

The goal is to make the coefficients of one variable the same or opposites, so that when you add or subtract the equations, that variable is eliminated.

What does “eliminate” mean?

To eliminate a variable means to remove it by adding or subtracting the equations. For example, if you have \(+3x\) in one equation and \(-3x\) in the other, adding the equations will make the \(x\)-terms cancel out.

Steps for the Elimination Method

1. Write both equations clearly. It helps if the variables are lined up, like \(x\) above \(x\) and \(y\) above \(y\).
2. Choose a variable to eliminate. Look for a variable whose coefficients are already opposites, or can easily be made opposites.
3. Multiply one or both equations if needed. This creates matching or opposite coefficients.
4. Add or subtract the equations. One variable should cancel out.
5. Solve the new equation. You will get the value of one variable.
6. Substitute back into one original equation. Then solve for the other variable.
7. Check your answer. Make sure your ordered pair works in both equations.

When do you add, and when do you subtract?

• If the coefficients are opposites, add the equations.
• If the coefficients are the same, subtract one equation from the other.

For example:

• \(2y\) and \(-2y\) cancel when you add.
• \(5x\) and \(5x\) cancel when you subtract.

Worked Example 1: Elimination without multiplying

Solve the system:

$$x + y = 7$$ $$x - y = 1$$

The \(y\)-coefficients are opposites: \(+y\) and \(-y\). So we add the equations.

$$ \begin{aligned} x + y &= 7 \\ x - y &= 1 \\ \hline 2x &= 8 \end{aligned} $$

Now solve for \(x\):

$$x = 4$$

Substitute \(x=4\) into one of the original equations, such as \(x+y=7\):

$$4 + y = 7$$ $$y = 3$$

So the solution is:

$$ (4, 3) $$

Check in the second equation:

$$4 - 3 = 1$$

The answer is correct.

Worked Example 2: Multiply one equation first

Solve the system:

$$2x + y = 9$$ $$x - y = 0$$

The \(y\)-coefficients are already opposites: \(+1\) and \(-1\). So we can add the equations right away.

$$ \begin{aligned} 2x + y &= 9 \\ x - y &= 0 \\ \hline 3x &= 9 \end{aligned} $$

Solve for \(x\):

$$x = 3$$

Now substitute into \(x-y=0\):

$$3 - y = 0$$ $$y = 3$$

So the solution is:

$$ (3, 3) $$

Worked Example 3: Multiply to create opposite coefficients

Solve the system:

$$3x + 2y = 16$$ $$x + y = 6$$

The coefficients do not match yet, so we choose a variable to eliminate. Let’s eliminate \(y\).

The first equation has \(2y\), and the second has \(y\). If we multiply the second equation by \(-2\), the \(y\)-terms will become opposites.

$$-2(x + y = 6)$$

This gives:

$$-2x - 2y = -12$$

Now add this new equation to the first equation:

$$ \begin{aligned} 3x + 2y &= 16 \\ -2x - 2y &= -12 \\ \hline x &= 4 \end{aligned} $$

Substitute \(x=4\) into \(x+y=6\):

$$4 + y = 6$$ $$y = 2$$

So the solution is:

$$ (4, 2) $$

Worked Example 4: Multiply both equations

Solve the system:

$$2x + 3y = 13$$ $$4x - 5y = -3$$

Neither variable has matching coefficients yet. We can eliminate either \(x\) or \(y\). Let’s eliminate \(x\).

The coefficients of \(x\) are \(2\) and \(4\). The least common multiple is \(4\), so we multiply the first equation by \(-2\). This will make the \(x\)-terms opposites.

$$-2(2x + 3y = 13)$$

This gives:

$$-4x - 6y = -26$$

Now add it to the second equation:

$$ \begin{aligned} -4x - 6y &= -26 \\ 4x - 5y &= -3 \\ \hline -11y &= -29 \end{aligned} $$

Solve for \(y\):

$$y = \frac{29}{11}$$

Now substitute into one original equation, such as \(2x + 3y = 13\):

$$2x + 3\left(\frac{29}{11}\right) = 13$$ $$2x + \frac{87}{11} = 13$$

Write \(13\) as \(\frac{143}{11}\):

$$2x = \frac{143}{11} - \frac{87}{11}$$ $$2x = \frac{56}{11}$$ $$x = \frac{28}{11}$$

So the solution is:

$$\left(\frac{28}{11}, \frac{29}{11}\right)$$

This example shows that solutions are not always whole numbers. They can also be fractions.

How to choose which variable to eliminate

• Choose the variable with coefficients that are already opposites or the same.
• If not, choose the one that needs the least multiplying.
• Try to avoid large numbers if possible, because they make the work longer.

Real-world meaning of the solution

In a real-world problem, the solution to a system tells you the values that satisfy both conditions at the same time. That is why the ordered pair must work in both equations.

For example, if \(x\) is the number of notebooks and \(y\) is the number of pens, then the solution tells you how many of each make both equations true.

Common mistakes to avoid

• Forgetting to multiply every term in an equation by the chosen number.
• Adding when you should subtract, or subtracting when you should add.
• Sign mistakes, especially with negative numbers.
• Not substituting back correctly after solving for one variable.
• Not checking the final answer in both original equations.

Quick Practice Thinking

Before solving, ask yourself:

• Which variable is easier to eliminate?
• Do I need to multiply one equation, or both?
• Will I add or subtract after multiplying?

Brief Summary

The elimination method solves a system by removing one variable so the other can be found. Sometimes the equations can be added or subtracted right away, and sometimes you first multiply one or both equations by a constant. After finding one variable, substitute back to find the other, and always check that your solution works in both equations.

Lesson: The Cross-Multiplication Method

When we solve a system of linear equations in two variables, we are looking for the values of the variables that make both equations true at the same time.

One direct algebraic method for solving such systems is the cross-multiplication method. This method is especially useful when both equations are written in the general form:

$$ax + by + c = 0$$

$$a'x + b'y + c' = 0$$

Here, the letters represent constants, and the apostrophe marks in \(a'\), \(b'\), and \(c'\) just mean they belong to the second equation.

The cross-multiplication method gives a formula for finding \(x\) and \(y\) directly, without first using substitution or elimination.

When can we use this method?

• When there are two linear equations.
• When both equations are written in the form \(ax + by + c = 0\).
• When the system has a unique solution.

The formula

If the equations are:

$$ax + by + c = 0$$

$$a'x + b'y + c' = 0$$

then:

$$\frac{x}{bc' - b'c} = \frac{y}{ca' - c'a} = \frac{1}{ab' - a'b}$$

This is the cross-multiplication formula.

From it, we can write:

$$x = \frac{bc' - b'c}{ab' - a'b}$$

$$y = \frac{ca' - c'a}{ab' - a'b}$$

Important idea: Before using the formula, make sure each equation is arranged so that everything is on one side and the other side is \(0\).

How to use the cross-multiplication method

1. Write both equations in the form \(ax + by + c = 0\).
2. Identify the coefficients:
   • First equation: \(a, b, c\)
   • Second equation: \(a', b', c'\)
3. Substitute these values into:
   • \(x = \dfrac{bc' - b'c}{ab' - a'b}\)
   • \(y = \dfrac{ca' - c'a}{ab' - a'b}\)
4. Simplify carefully.
5. Check your answer in both original equations.

Why does this method work?

The method comes from eliminating variables in a structured way. Instead of doing elimination separately every time, the formula combines the needed calculations into one standard pattern.

You do not need to memorize the full derivation right away, but you should understand that it is based on the same algebra used in elimination.

Worked Example 1: Simple whole-number solution

Solve:

$$2x + 3y - 13 = 0$$

$$3x - 2y - 4 = 0$$

First, compare with:

$$ax + by + c = 0$$

$$a'x + b'y + c' = 0$$

So we identify:

• \(a = 2\), \(b = 3\), \(c = -13\)
• \(a' = 3\), \(b' = -2\), \(c' = -4\)

Now find \(x\):

$$x = \frac{bc' - b'c}{ab' - a'b}$$

$$x = \frac{3(-4) - (-2)(-13)}{2(-2) - 3(3)}$$

$$x = \frac{-12 - 26}{-4 - 9} = \frac{-38}{-13}$$

$$x = \frac{38}{13}$$

Now find \(y\):

$$y = \frac{ca' - c'a}{ab' - a'b}$$

$$y = \frac{(-13)(3) - (-4)(2)}{2(-2) - 3(3)}$$

$$y = \frac{-39 + 8}{-13} = \frac{-31}{-13}$$

$$y = \frac{31}{13}$$

So the solution is:

$$x = \frac{38}{13}, \quad y = \frac{31}{13}$$

Let us check one equation:

$$2\left(\frac{38}{13}\right) + 3\left(\frac{31}{13}\right) - 13$$

$$= \frac{76}{13} + \frac{93}{13} - 13 = \frac{169}{13} - 13 = 13 - 13 = 0$$

The solution is correct.

Worked Example 2: Rearranging first

Solve:

$$x + 2y = 7$$

$$3x - y = 5$$

These are not yet in the required form, so rewrite them:

$$x + 2y - 7 = 0$$

$$3x - y - 5 = 0$$

Now identify the coefficients:

• \(a = 1\), \(b = 2\), \(c = -7\)
• \(a' = 3\), \(b' = -1\), \(c' = -5\)

Find \(x\):

$$x = \frac{bc' - b'c}{ab' - a'b}$$

$$x = \frac{2(-5) - (-1)(-7)}{1(-1) - 3(2)}$$

$$x = \frac{-10 - 7}{-1 - 6} = \frac{-17}{-7}$$

$$x = \frac{17}{7}$$

Find \(y\):

$$y = \frac{ca' - c'a}{ab' - a'b}$$

$$y = \frac{(-7)(3) - (-5)(1)}{-7}$$

$$y = \frac{-21 + 5}{-7} = \frac{-16}{-7}$$

$$y = \frac{16}{7}$$

So the solution is:

$$x = \frac{17}{7}, \quad y = \frac{16}{7}$$

Worked Example 3: A system with integer answer

Solve:

$$4x + 5y - 6 = 0$$

$$2x - 3y + 8 = 0$$

Identify the coefficients:

• \(a = 4\), \(b = 5\), \(c = -6\)
• \(a' = 2\), \(b' = -3\), \(c' = 8\)

Find the denominator first:

$$ab' - a'b = 4(-3) - 2(5) = -12 - 10 = -22$$

Now find \(x\):

$$x = \frac{bc' - b'c}{ab' - a'b}$$

$$x = \frac{5(8) - (-3)(-6)}{-22}$$

$$x = \frac{40 - 18}{-22} = \frac{22}{-22} = -1$$

Now find \(y\):

$$y = \frac{ca' - c'a}{ab' - a'b}$$

$$y = \frac{(-6)(2) - 8(4)}{-22}$$

$$y = \frac{-12 - 32}{-22} = \frac{-44}{-22} = 2$$

So the solution is:

$$x = -1, \quad y = 2$$

Check in the first equation:

$$4(-1) + 5(2) - 6 = -4 + 10 - 6 = 0$$

Check in the second equation:

$$2(-1) - 3(2) + 8 = -2 - 6 + 8 = 0$$

Both checks work.

Worked Example 4: A word problem

At a school event, 2 notebooks and 3 pens cost Rs. 36. Also, 4 notebooks and 1 pen cost Rs. 34. Find the cost of one notebook and one pen.

Let:

• \(x\) = cost of one notebook
• \(y\) = cost of one pen

Then the equations are:

$$2x + 3y = 36$$

$$4x + y = 34$$

Write them in standard form:

$$2x + 3y - 36 = 0$$

$$4x + y - 34 = 0$$

Identify the coefficients:

• \(a = 2\), \(b = 3\), \(c = -36\)
• \(a' = 4\), \(b' = 1\), \(c' = -34\)

Find \(x\):

$$x = \frac{bc' - b'c}{ab' - a'b}$$

$$x = \frac{3(-34) - 1(-36)}{2(1) - 4(3)}$$

$$x = \frac{-102 + 36}{2 - 12} = \frac{-66}{-10}$$

$$x = \frac{33}{5} = 6.6$$

Find \(y\):

$$y = \frac{ca' - c'a}{ab' - a'b}$$

$$y = \frac{(-36)(4) - (-34)(2)}{-10}$$

$$y = \frac{-144 + 68}{-10} = \frac{-76}{-10}$$

$$y = \frac{38}{5} = 7.6$$

So:

• One notebook costs Rs. \(6.60\)
• One pen costs Rs. \(7.60\)

Common mistakes to avoid

• Not writing the equations in standard form first. If one equation is written as \(x + y = 5\), you must change it to \(x + y - 5 = 0\).
• Mixing up the coefficients. Be careful when identifying \(a, b, c, a', b', c'\).
• Sign errors. Negative signs are very important in this method.
• Using the wrong denominator. The denominator is \(ab' - a'b\), not something else.
• Forgetting to check the answer. A quick check can catch mistakes.

What if the denominator is zero?

If:

$$ab' - a'b = 0$$

then the cross-multiplication formula cannot give a unique solution.

This usually means one of two things:

• The lines are parallel, so there is no solution.
• The equations represent the same line, so there are infinitely many solutions.

So this method works best when the two equations intersect at exactly one point.

Quick comparison with other methods

• Substitution: Good when one variable is already easy to isolate.
• Elimination: Good when coefficients can be matched easily.
• Cross-multiplication: Good when both equations are in general form and you want a direct formula.

Steps to remember

1. Write each equation as \(ax + by + c = 0\).
2. Mark the coefficients carefully.
3. Use:
   • $$x = \frac{bc' - b'c}{ab' - a'b}$$
   • $$y = \frac{ca' - c'a}{ab' - a'b}$$
4. Simplify.
5. Check in the original equations.

Brief Summary

The cross-multiplication method is a formula-based way to solve two linear equations in two variables. First, write both equations in the form \(ax + by + c = 0\). Then use the formulas for \(x\) and \(y\) carefully, paying close attention to signs. This method is fast and useful, especially when the equations are already in general form.

Equations Reducible to Linear Systems

Sometimes a system of equations does not look linear at first. It may include fractions with variables in the denominator, or expressions such as \(\frac{1}{x}\) and \(\frac{1}{y}\). Even though the equations look non-linear, we can often rewrite them so they become a linear system.

This idea is called solving equations reducible to linear systems. The main strategy is to replace complicated variable expressions with new variables. Then we solve the easier linear system, and finally change back to the original variables.

This is useful because you already know how to solve linear systems by substitution or elimination. So instead of trying to solve a difficult-looking system directly, you turn it into a form you already understand.

Main Idea

If the same repeated expression appears in more than one equation, we can let that expression equal a new variable.

For example:

• If \(\frac{1}{x}\) appears often, let \(u = \frac{1}{x}\).
• If \(\frac{1}{y}\) appears often, let \(v = \frac{1}{y}\).
• If \(x+y\) appears often, let \(u = x+y\).
• If \(xy\) appears often, let \(v = xy\).

After making the substitution, the equations may become linear in the new variables such as \(u\) and \(v\).

Important Reminder

When solving equations with fractions like \(\frac{1}{x}\), the variable cannot be zero. For example, if \(u = \frac{1}{x}\), then \(x \neq 0\). Always check that your final answers are allowed in the original equations.

General Steps

1. Look for repeated expressions in the system.
2. Replace those expressions with new variables.
3. Solve the resulting linear system.
4. Substitute back to find the original variables.
5. Check the answers in the original equations.

Why This Works

A linear system is easier to solve because each variable appears only to the first power and is not multiplied by another variable. By using substitution cleverly, we temporarily hide the complicated parts and reveal a linear pattern.

Let us work through examples.

Worked Example 1: Fractions with variables in the denominator

Solve the system

$$ \frac{1}{x} + \frac{1}{y} = 5 $$ $$ \frac{2}{x} - \frac{1}{y} = 4 $$

Step 1: Choose new variables

The repeated expressions are \(\frac{1}{x}\) and \(\frac{1}{y}\). Let

$$ u = \frac{1}{x}, \qquad v = \frac{1}{y} $$

Then the system becomes

$$ u + v = 5 $$ $$ 2u - v = 4 $$

This is now a linear system in \(u\) and \(v\).

Step 2: Solve the linear system

Add the two equations:

$$ (u+v) + (2u-v) = 5 + 4 $$ $$ 3u = 9 $$ $$ u = 3 $$

Substitute into \(u + v = 5\):

$$ 3 + v = 5 $$ $$ v = 2 $$

Step 3: Change back to the original variables

Since \(u = \frac{1}{x} = 3\), we get

$$ x = \frac{1}{3} $$

Since \(v = \frac{1}{y} = 2\), we get

$$ y = \frac{1}{2} $$

Answer: \(\left(\frac{1}{3}, \frac{1}{2}\right)\)

Check

$$ \frac{1}{1/3} + \frac{1}{1/2} = 3 + 2 = 5 $$ $$ \frac{2}{1/3} - \frac{1}{1/2} = 6 - 2 = 4 $$

The solution works.

Worked Example 2: A slightly harder fraction system

Solve the system

$$ \frac{3}{x} + \frac{2}{y} = 7 $$ $$ \frac{1}{x} - \frac{1}{y} = 1 $$

Step 1: Substitute

Let

$$ u = \frac{1}{x}, \qquad v = \frac{1}{y} $$

Then the system becomes

$$ 3u + 2v = 7 $$ $$ u - v = 1 $$

Step 2: Solve the linear system

From \(u - v = 1\), we get

$$ u = v + 1 $$

Substitute into the first equation:

$$ 3(v+1) + 2v = 7 $$ $$ 3v + 3 + 2v = 7 $$ $$ 5v + 3 = 7 $$ $$ 5v = 4 $$ $$ v = \frac{4}{5} $$

Now find \(u\):

$$ u = v + 1 = \frac{4}{5} + 1 = \frac{9}{5} $$

Step 3: Return to \(x\) and \(y\)

$$ \frac{1}{x} = \frac{9}{5} \Rightarrow x = \frac{5}{9} $$ $$ \frac{1}{y} = \frac{4}{5} \Rightarrow y = \frac{5}{4} $$

Answer: \(\left(\frac{5}{9}, \frac{5}{4}\right)\)

Worked Example 3: Using expressions like \(x+y\) and \(x-y\)

Not all reducible systems involve fractions. Sometimes a system can be simplified by replacing repeated groups of terms.

Solve the system

$$ 2(x+y) + 3(x-y) = 13 $$ $$ (x+y) - (x-y) = 2 $$

Step 1: Substitute repeated expressions

Let

$$ u = x+y, \qquad v = x-y $$

Then the system becomes

$$ 2u + 3v = 13 $$ $$ u - v = 2 $$

Step 2: Solve the linear system

From \(u - v = 2\), we get

$$ u = v + 2 $$

Substitute into \(2u + 3v = 13\):

$$ 2(v+2) + 3v = 13 $$ $$ 2v + 4 + 3v = 13 $$ $$ 5v = 9 $$ $$ v = \frac{9}{5} $$

Then

$$ u = v + 2 = \frac{9}{5} + \frac{10}{5} = \frac{19}{5} $$

Step 3: Change back to the original variables

We now have

$$ x+y = \frac{19}{5} $$ $$ x-y = \frac{9}{5} $$

Add the equations:

$$ 2x = \frac{28}{5} $$ $$ x = \frac{14}{5} $$

Substitute into \(x+y = \frac{19}{5}\):

$$ \frac{14}{5} + y = \frac{19}{5} $$ $$ y = 1 $$

Answer: \(\left(\frac{14}{5}, 1\right)\)

Worked Example 4: Product and sum as new variables

Solve the system

$$ xy + x + y = 11 $$ $$ xy - x - y = 1 $$

Step 1: Look for repeated expressions

Both equations contain \(xy\) and \(x+y\). Let

$$ u = xy, \qquad v = x+y $$

Then the system becomes

$$ u + v = 11 $$ $$ u - v = 1 $$

Step 2: Solve the linear system

Add the equations:

$$ 2u = 12 $$ $$ u = 6 $$

Then

$$ 6 + v = 11 \Rightarrow v = 5 $$

So now we know

$$ xy = 6, \qquad x+y = 5 $$

Step 3: Find \(x\) and \(y\)

We need two numbers whose sum is 5 and product is 6. Those numbers are 2 and 3.

So the solutions are

$$ (x,y) = (2,3) \text{ or } (3,2) $$

Check

For \((2,3)\):

$$ xy + x + y = 6 + 2 + 3 = 11 $$ $$ xy - x - y = 6 - 2 - 3 = 1 $$

It works. The same is true for \((3,2)\).

What to Watch Out For

• Forgetting restrictions: If the original equation has \(\frac{1}{x}\), then \(x\neq 0\).
• Not substituting back: Solving for \(u\) and \(v\) is not the final answer. You must return to \(x\) and \(y\).
• Not checking: A value might work in the transformed system but fail in the original one if it breaks a restriction.
• Mixing old and new variables: Stay organized. Write clearly what \(u\) and \(v\) stand for.

How This Connects to Linear Systems

In a usual linear system, you solve directly for \(x\) and \(y\). In a reducible system, you first create a new linear system in variables like \(u\) and \(v\). Then you solve that system using the same methods you already know.

So this topic is really about recognizing structure. When you see repeated expressions, ask yourself: “Can I rename these parts and turn the system into something linear?”

Quick Practice Tips

• Circle repeated expressions before you start.
• Choose simple replacement letters such as \(u\) and \(v\).
• Solve the linear system carefully.
• Always write a final answer in terms of \(x\) and \(y\), not just \(u\) and \(v\).
• Check your answer in the original equations.

Summary

Some systems look non-linear, but they can be turned into linear systems by substituting repeated expressions with new variables. Common substitutions include \(u=\frac{1}{x}\), \(v=\frac{1}{y}\), \(u=x+y\), or \(v=xy\). After solving the linear system, substitute back to find the original variables and check that the answers fit the original equations.

Modeling Word Problems: Age, Money, and Fractions

Many real-world problems involve two unknown quantities. In 10th Grade Maths, we often model these situations using a system of linear equations in two variables.

Word problems about ages, money, and fractions are common examples. The key skill is turning words into equations, then solving those equations and checking whether the answer makes sense in the context.

In this lesson, you will learn how to:

• identify the unknowns,
• define variables clearly,
• translate word statements into linear equations,
• solve the system algebraically,
• and interpret the solution in the real-world situation.

1. The General Strategy

Most word problems of this type can be solved using the same steps.

1. Choose variables. Decide what each variable represents.
2. Write two equations. Use the information in the problem.
3. Solve the system. Use substitution, elimination, or graphing.
4. Check the answer. Make sure it fits the original story.
5. Write a sentence answer. Always include units like years, dollars, or values.

A very important habit is to define variables carefully. For example, write: “Let Ann’s age be \(a\) and Ben’s age be \(b\).” This helps prevent confusion later.

2. Recognizing Common Clue Words

Word problems often use phrases that translate directly into algebra.

• “sum” means add
• “difference” means subtract
• “is” means equals
• “twice” means multiply by 2
• “three times” means multiply by 3
• “more than” means add after the quantity
• “less than” means subtract from the quantity
• “in 5 years” means add 5 to an age
• “5 years ago” means subtract 5 from an age

For money problems, phrases like “total value,” “cost,” and “amount spent” often lead to equations involving multiplication and addition.

For fraction problems, you may need to represent a number using its digits or compare a fraction to another expression.

3. Age Problems

Age problems usually involve two facts:

• a relationship between two people’s present ages, and
• a second relationship involving the past or future.

Remember: if a person is 4 years older than another person now, then that age difference stays 4 years in the future and in the past.

Worked Example 1: Present and Future Ages

A mother is 24 years older than her son. In 6 years, the mother’s age will be 3 times the son’s age. Find their present ages.

Step 1: Define variables

Let the mother’s present age be \(m\), and the son’s present age be \(s\).

Step 2: Write the equations

“A mother is 24 years older than her son” gives:

$$m = s + 24$$

“In 6 years, the mother’s age will be 3 times the son’s age” gives:

$$m + 6 = 3(s + 6)$$

Step 3: Solve

Substitute \(m = s + 24\) into the second equation:

$$s + 24 + 6 = 3(s + 6)$$

$$s + 30 = 3s + 18$$

$$12 = 2s$$

$$s = 6$$

Now find \(m\):

$$m = s + 24 = 6 + 24 = 30$$

Step 4: Check

In 6 years, the son will be \(12\) and the mother will be \(36\). Since \(36 = 3 \times 12\), the answer is correct.

Answer: The son is \(6\) years old, and the mother is \(30\) years old.

Age Tip: When the problem says “in 6 years,” add 6 to both ages. When it says “5 years ago,” subtract 5 from both ages.

4. Money Problems

Money problems often involve two types of items or two groups of bills/coins. The system usually comes from:

• the total number of items, and
• the total value or cost.

Worked Example 2: Coins

A jar contains only \(10\)-rupee coins and \(5\)-rupee coins. There are 18 coins in total, worth 135 rupees altogether. How many of each coin are there?

Step 1: Define variables

Let \(x\) be the number of \(10\)-rupee coins and \(y\) be the number of \(5\)-rupee coins.

Step 2: Write the equations

Total number of coins:

$$x + y = 18$$

Total value:

$$10x + 5y = 135$$

Step 3: Solve

We can simplify the second equation by dividing by 5:

$$2x + y = 27$$

Now the system is:

$$x + y = 18$$

$$2x + y = 27$$

Subtract the first equation from the second:

$$x = 9$$

Substitute into \(x + y = 18\):

$$9 + y = 18$$

$$y = 9$$

Step 4: Check

There are 9 coins of each type. Their total value is:

$$10(9) + 5(9) = 90 + 45 = 135$$

Answer: There are \(9\) ten-rupee coins and \(9\) five-rupee coins.

Money Tip: If the values are large, look for a common factor to simplify the equations before solving.

5. Fraction and Digit Problems

Some fraction word problems involve a two-digit number. In these problems, it helps to remember how place value works.

If a two-digit number has tens digit \(x\) and ones digit \(y\), then the number is:

$$10x + y$$

If the digits are reversed, the new number is:

$$10y + x$$

The sum of the digits is:

$$x + y$$

Worked Example 3: Digit Sum and Reversed Number

The sum of the digits of a two-digit number is 11. When the digits are reversed, the new number is 27 less than the original number. Find the number.

Step 1: Define variables

Let \(x\) be the tens digit and \(y\) be the ones digit.

Step 2: Write the equations

“The sum of the digits is 11” gives:

$$x + y = 11$$

The original number is \(10x + y\).

The reversed number is \(10y + x\).

“The new number is 27 less than the original number” gives:

$$10y + x = 10x + y - 27$$

Step 3: Simplify the second equation

$$10y + x = 10x + y - 27$$

$$9y - 9x = -27$$

$$y - x = -3$$

So the system is:

$$x + y = 11$$

$$y - x = -3$$

Step 4: Solve

Add the equations:

$$x + y + y - x = 11 + (-3)$$

$$2y = 8$$

$$y = 4$$

Then:

$$x + 4 = 11$$

$$x = 7$$

Step 5: Interpret

The original number is:

$$10x + y = 10(7) + 4 = 74$$

Check: Reversing the digits gives 47, and \(47\) is 27 less than \(74\).

Answer: The number is \(74\).

6. Fraction Relationship Problems

Some problems involve two numbers and their fractional relationship. In these problems, turn the statements into equations carefully.

Worked Example 4: Two Numbers and Fractions

Two numbers have a sum of 35. One-half of the first number is equal to one-third of the second number. Find the numbers.

Step 1: Define variables

Let the first number be \(x\) and the second number be \(y\).

Step 2: Write the equations

The sum is 35:

$$x + y = 35$$

“One-half of the first number is equal to one-third of the second number” gives:

$$\frac{x}{2} = \frac{y}{3}$$

Step 3: Clear fractions

Multiply both sides by 6:

$$3x = 2y$$

So the system is:

$$x + y = 35$$

$$3x = 2y$$

Step 4: Solve

From \(3x = 2y\), we get:

$$y = \frac{3x}{2}$$

Substitute into the first equation:

$$x + \frac{3x}{2} = 35$$

$$\frac{2x}{2} + \frac{3x}{2} = 35$$

$$\frac{5x}{2} = 35$$

$$5x = 70$$

$$x = 14$$

Now find \(y\):

$$y = 35 - 14 = 21$$

Check:

$$\frac{14}{2} = 7, \quad \frac{21}{3} = 7$$

The condition is true.

Answer: The numbers are \(14\) and \(21\).

Fraction Tip: Fractions in equations can look difficult, but they become easier when you multiply by the least common denominator to remove the fractions.

7. Solving Graphically vs Algebraically

These systems can be solved in two main ways:

• Algebraically using substitution or elimination
• Graphically by plotting both equations and finding their point of intersection

For word problems, algebra is usually faster and more exact. Graphing is helpful for seeing what the solution means: the intersection point represents the values that satisfy both conditions at the same time.

For example, in the coin problem:

$$x + y = 18$$

$$10x + 5y = 135$$

If these equations are graphed, the point where the two lines meet is \((9, 9)\). This means 9 ten-rupee coins and 9 five-rupee coins.

8. Common Mistakes to Avoid

• Not defining variables clearly. Always say what \(x\) and \(y\) represent.
• Mixing up “more than” and “less than.” Read carefully.
• Forgetting place value in digit problems. A two-digit number is \(10x + y\), not \(x + y\).
• Changing only one age in past/future problems. Time affects both people.
• Ignoring the context. A digit must be a whole number from 0 to 9, and ages or numbers of coins should not be negative.

9. Quick Checklist for Word Problems

Before solving, ask yourself:

• What are the two unknowns?
• What two facts give me equations?
• Did I translate the words correctly?
• Did I solve the system correctly?
• Does my answer make sense in the situation?

10. Summary

To model word problems involving age, money, and fractions, start by choosing variables and translating the information into a system of linear equations. Then solve the system using substitution, elimination, or graphing.

In age problems, pay attention to past and future relationships. In money problems, use total number and total value. In fraction and digit problems, use place value and clear fraction relationships carefully.

The most important idea is that the solution must satisfy both equations and must also make sense in the real-world context.

Modeling Word Problems: Distance, Speed, and Time

Many real-world problems involve how distance, speed, and time are connected. In 10th Grade maths, these problems often become more interesting when there is more than one unknown. That is when we can model the situation using a system of linear equations in two variables.

This lesson focuses on word problems involving upstream and downstream river travel, relative motion, and different speeds. You will learn how to translate the words into equations, solve the system, and explain what the solution means in the context of the problem.

The key relationship to remember is:

$$ \text{Distance} = \text{Speed} \times \text{Time} $$

This can also be rearranged as:

$$ \text{Speed} = \frac{\text{Distance}}{\text{Time}}, \qquad \text{Time} = \frac{\text{Distance}}{\text{Speed}} $$

When a problem gives two different situations involving the same unknowns, you can often write two linear equations and solve them as a system.

Step-by-step strategy for modeling

1. Read carefully and decide what quantities are unknown.
2. Define variables clearly. For example, let \(b\) be the speed of the boat in still water and \(c\) be the speed of the current.
3. Use the relationship \(d = rt\) to write equations.
4. Look for how the situations compare. Words like upstream, downstream, toward each other, and same distance are clues.
5. Solve the system algebraically or graphically.
6. Check the answer by substituting back into the original situation.
7. Interpret the solution using units such as miles per hour, kilometers per hour, or hours.

Important modeling ideas

• Still-water speed means the speed of a boat in water that is not moving.
• Current speed means the speed of the river.
• Downstream speed is faster because the current helps the boat:

$$ \text{Downstream speed} = \text{boat speed} + \text{current speed} $$

• Upstream speed is slower because the current pushes against the boat:

$$ \text{Upstream speed} = \text{boat speed} - \text{current speed} $$

• When two objects move toward each other, their speeds add.
• When one object is chasing another in the same direction, the difference in their speeds matters.

How systems appear in these problems

Suppose a boat travels downstream for one trip and upstream for another. The unknowns might be the boat's speed in still water and the current's speed. Since the problem gives information about two trips, you can usually write two equations.

In other problems, two people, cars, or trains travel at different speeds. If the problem gives two relationships, such as total distance and total time, or two separate trips, you can form a system with two variables.

Worked Example 1: Finding two speeds from total time

A car travels for 5 hours. Part of the trip is on city roads at 40 miles per hour, and the rest is on the highway at 60 miles per hour. The total distance traveled is 260 miles. How many hours did the car travel at each speed?

Step 1: Define variables.

Let:

• \(x\) = number of hours on city roads
• \(y\) = number of hours on the highway

Step 2: Write the equations.

The total time is 5 hours, so:

$$ x + y = 5 $$

The total distance is 260 miles. Using \(d = rt\):

$$ 40x + 60y = 260 $$

So the system is:

$$ \begin{cases} x + y = 5 \\ 40x + 60y = 260 \end{cases} $$

Step 3: Solve the system.

From \(x + y = 5\), solve for \(x\):

$$ x = 5 - y $$

Substitute into the distance equation:

$$ 40(5 - y) + 60y = 260 $$ $$ 200 - 40y + 60y = 260 $$ $$ 20y = 60 $$ $$ y = 3 $$

Then:

$$ x = 5 - 3 = 2 $$

Answer: The car traveled 2 hours at 40 mph and 3 hours at 60 mph.

Check:

$$ 40(2) + 60(3) = 80 + 180 = 260 $$

The answer makes sense.

Worked Example 2: Upstream and downstream travel

A boat travels 24 miles downstream in 2 hours. The same boat travels 18 miles upstream in 3 hours. Find the speed of the boat in still water and the speed of the current.

Step 1: Define variables.

Let:

• \(b\) = speed of the boat in still water
• \(c\) = speed of the current

Step 2: Find the actual speeds for each trip.

Downstream speed:

$$ \frac{24}{2} = 12 $$

So:

$$ b + c = 12 $$

Upstream speed:

$$ \frac{18}{3} = 6 $$

So:

$$ b - c = 6 $$

The system is:

$$ \begin{cases} b + c = 12 \\ b - c = 6 \end{cases} $$

Step 3: Solve the system.

Add the equations:

$$ (b + c) + (b - c) = 12 + 6 $$ $$ 2b = 18 $$ $$ b = 9 $$

Substitute into \(b + c = 12\):

$$ 9 + c = 12 $$ $$ c = 3 $$

Answer: The boat's speed in still water is 9 mph, and the current's speed is 3 mph.

Interpretation: Downstream, the boat moves at \(9 + 3 = 12\) mph. Upstream, it moves at \(9 - 3 = 6\) mph.

Worked Example 3: Relative motion in opposite directions

Two cyclists start 90 miles apart and ride toward each other. One cyclist rides at 12 miles per hour, and the other rides at 18 miles per hour. How long will it take before they meet?

This problem can be solved directly, but we can also model it as a system to understand the distances traveled by each cyclist.

Step 1: Define variables.

Let:

• \(x\) = distance traveled by the first cyclist
• \(y\) = distance traveled by the second cyclist

Because they start 90 miles apart and meet, the distances add to 90:

$$ x + y = 90 $$

Since time is the same for both cyclists:

$$ \frac{x}{12} = \frac{y}{18} $$

Multiply both sides by 36:

$$ 3x = 2y $$

So the system is:

$$ \begin{cases} x + y = 90 \\ 3x = 2y \end{cases} $$

Step 2: Solve the system.

From \(3x = 2y\), solve for \(y\):

$$ y = \frac{3}{2}x $$

Substitute into \(x + y = 90\):

$$ x + \frac{3}{2}x = 90 $$ $$ \frac{5}{2}x = 90 $$ $$ x = 36 $$

Then:

$$ y = 90 - 36 = 54 $$

Step 3: Find the time.

Use either cyclist:

$$ \text{time} = \frac{36}{12} = 3 $$

or

$$ \text{time} = \frac{54}{18} = 3 $$

Answer: They meet after 3 hours.

Important idea: When two objects move toward each other, you can also add their speeds:

$$ 12 + 18 = 30 $$ $$ \text{time} = \frac{90}{30} = 3 $$

Both methods give the same result.

Worked Example 4: A system with river travel and times

A motorboat travels 30 miles downstream in the same amount of time that it travels 18 miles upstream. The current is 2 miles per hour. Find the speed of the boat in still water.

This problem has one unknown if the current is already known, but we can still model the two situations using equations.

Step 1: Define the variable.

Let \(b\) be the speed of the boat in still water.

Then:

• Downstream speed = \(b + 2\)
• Upstream speed = \(b - 2\)

Step 2: Use equal times.

Time equals distance divided by speed, so:

$$ \frac{30}{b + 2} = \frac{18}{b - 2} $$

Step 3: Solve.

Cross multiply:

$$ 30(b - 2) = 18(b + 2) $$ $$ 30b - 60 = 18b + 36 $$ $$ 12b = 96 $$ $$ b = 8 $$

Answer: The boat's speed in still water is 8 mph.

Check:

• Downstream speed: \(8 + 2 = 10\) mph
• Upstream speed: \(8 - 2 = 6\) mph

Times:

$$ \frac{30}{10} = 3 \text{ hours}, \qquad \frac{18}{6} = 3 \text{ hours} $$

The times are equal, so the answer is correct.

Solving graphically

Systems can also be solved by graphing. For example, in Worked Example 1 the system was:

$$ \begin{cases} x + y = 5 \\ 40x + 60y = 260 \end{cases} $$

Rewrite each equation in slope-intercept or easy-to-graph form.

First equation:

$$ y = 5 - x $$

Second equation:

$$ 40x + 60y = 260 $$ $$ 2x + 3y = 13 $$ $$ 3y = 13 - 2x $$ $$ y = \frac{13 - 2x}{3} $$

If you graph both lines, their intersection is \((2, 3)\). This means:

• \(x = 2\) hours at 40 mph
• \(y = 3\) hours at 60 mph

Graphing is useful because it shows the solution visually. The point where the lines meet represents the values that make both equations true at the same time.

Common mistakes to avoid

• Mixing up distance, speed, and time. Always use \(d = rt\) carefully.
• Forgetting units. If speed is in miles per hour, time should be in hours.
• Adding instead of subtracting for upstream travel. Upstream means the current slows the boat down.
• Using the wrong relationship in relative motion. Toward each other means speeds add; same direction often means speeds subtract.
• Not defining variables. A clear variable definition helps avoid confusion.
• Stopping too early. Sometimes the system gives distances first, and you still need to find the time.

Quick checklist for word problems

• What are the unknowns?
• What formula connects the quantities?
• What two equations can I write?
• Do my equations match the meaning of the words?
• Did I solve and label the answer with units?
• Did I check that the answer makes sense in the story?

Brief Summary

Distance, speed, and time problems can often be modeled with systems of linear equations when there are two unknown quantities. The most important formula is \(d = rt\), along with the ideas that downstream speed is \(b + c\) and upstream speed is \(b - c\).

To solve these problems, define variables, write equations from the information given, solve the system, and interpret the solution in context. Whether you solve algebraically or graphically, the goal is the same: find values that make both parts of the situation true at the same time.