Systems of Linear Equations

Concept of a Linear System

In 8th Grade Maths, you have already worked with linear equations. A linear equation is an equation whose graph is a straight line, such as \(y=2x+1\) or \(x+y=6\).

Sometimes, one equation is not enough to describe a situation. We may need two linear equations at the same time. When we study two linear equations together, we call them a system of linear equations, or simply a linear system.

The big idea is this: a solution to a linear system is an ordered pair \((x,y)\) that makes both equations true at the same time.

That means a point is not a solution just because it works in one equation. It must work in each equation in the system.

1. What is a system of linear equations?

A system of linear equations is a set of two or more linear equations with the same variables. In 8th Grade, you will usually work with two equations and two variables, like \(x\) and \(y\).

For example, this is a linear system:

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

Each equation describes a line. The system asks: Is there a point that lies on both lines?

If there is, that point is the solution to the system.

2. What does a solution mean?

A solution to a system is an ordered pair that satisfies both equations simultaneously. The word simultaneously means at the same time.

Suppose we want to check whether \((2,5)\) is a solution to this system:

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

Substitute \(x=2\) and \(y=5\) into the first equation:

$$ y=x+3 \Rightarrow 5=2+3 \Rightarrow 5=5 $$

This is true, so \((2,5)\) works for the first equation.

Now check the second equation:

$$ y=2x+1 \Rightarrow 5=2(2)+1 \Rightarrow 5=4+1 \Rightarrow 5=5 $$

This is also true, so \((2,5)\) works for the second equation too.

Because \((2,5)\) makes both equations true, it is a solution to the system.

3. Why does graphing help?

Each linear equation can be graphed as a straight line. The solution to the system is the point where the two lines intersect, or cross.

If two lines cross at one point, that point is the one ordered pair that satisfies both equations.

This means:

• Points on the first line satisfy the first equation.
• Points on the second line satisfy the second equation.
• The point where both lines meet satisfies both.

4. Ways a linear system can behave

When you graph two lines in a system, three main things can happen.

• One solution: The lines cross once. There is exactly one ordered pair that satisfies both equations.
• No solution: The lines are parallel and never meet. There is no ordered pair that works for both equations.
• Infinitely many solutions: The lines are actually the same line, so every point on that line works for both equations.

Most of the time in early examples, you will see systems with one solution.

5. How to tell if an ordered pair is a solution

To check whether an ordered pair is a solution:

1. Substitute the \(x\)-value and \(y\)-value into the first equation.
2. See if the equation is true.
3. Substitute the same values into the second equation.
4. If both are true, the ordered pair is a solution.

This is an important skill because even before solving a system, you can test possible answers.

Worked Example 1: Checking a possible solution

Is \((3,4)\) a solution to the system

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

Step 1: Check the first equation.

$$ x+y=7 \Rightarrow 3+4=7 \Rightarrow 7=7 $$

True.

Step 2: Check the second equation.

$$ y=x+1 \Rightarrow 4=3+1 \Rightarrow 4=4 $$

Also true.

Answer: Yes, \((3,4)\) is a solution because it satisfies both equations.

Worked Example 2: A point that is not a solution

Is \((2,3)\) a solution to the system

$$ \begin{cases} y=2x-1\\ x+y=5 \end{cases} $$

Step 1: Check the first equation.

$$ y=2x-1 \Rightarrow 3=2(2)-1 \Rightarrow 3=4-1 \Rightarrow 3=3 $$

True.

Step 2: Check the second equation.

$$ x+y=5 \Rightarrow 2+3=5 \Rightarrow 5=5 $$

True again.

So \((2,3)\) is a solution.

Now let us test \((2,4)\) in the same system.

First equation:

$$ y=2x-1 \Rightarrow 4=2(2)-1 \Rightarrow 4=3 $$

This is false.

Even if a point worked in the second equation, it would not be a solution, because it must satisfy both equations.

Worked Example 3: Understanding the graph of a system

Consider the system

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

These are two lines. The solution is where they meet.

We can find the intersection point by thinking about what both equations say about \(y\). Since both equal \(y\), set them equal:

$$ x+2=-x+4 $$

Add \(x\) to both sides:

$$ 2x+2=4 $$

Subtract 2 from both sides:

$$ 2x=2 $$

Divide by 2:

$$ x=1 $$

Substitute \(x=1\) into either equation:

$$ y=x+2=1+2=3 $$

So the solution is \((1,3)\).

This means the two lines intersect at \((1,3)\), and that point lies on both graphs.

Worked Example 4: Looking at different kinds of systems

Case A: One solution

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

The slopes are different, so the lines cross once. This system has one solution.

Case B: No solution

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

These lines have the same slope but different starting points, so they are parallel. They never intersect. This system has no solution.

Case C: Infinitely many solutions

$$ \begin{cases} y=x+4\\ 2y=2x+8 \end{cases} $$

If we simplify the second equation by dividing both sides by 2, we get

$$ y=x+4 $$

Now both equations are the same line. Every point on that line is a solution, so there are infinitely many solutions.

6. Real meaning of a linear system

A linear system can model a real situation where two conditions must both be true.

For example, imagine one equation shows the cost of one phone plan, and another equation shows the cost of a different phone plan. The solution to the system could tell you when the two plans cost the same amount.

So a system is useful because it helps us find where two relationships match.

7. Important ideas to remember

• A linear system is a set of two linear equations with the same variables.
• A solution is an ordered pair that makes both equations true.
• On a graph, the solution is the intersection point of the two lines.
• Some systems have one solution, no solution, or infinitely many solutions.
• To test a point, substitute it into both equations.

Brief Summary

A system of linear equations is two linear equations studied together. The solution to the system is the ordered pair that satisfies both equations at the same time. When graphed, the solution is usually the point where the two lines intersect. If the lines never meet, there is no solution, and if they are the same line, there are infinitely many solutions.

Solving Systems by Graphing

A system of linear equations is a set of two linear equations with the same variables. In 8th grade, you will often see systems with the variables \(x\) and \(y\).

When we solve a system, we are looking for the ordered pair \((x,y)\) that makes both equations true at the same time.

One way to solve a system is by graphing. We graph both lines on the same coordinate plane and look for the point where they cross. That point is called the intersection, and it is the solution to the system.

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

Why graphing works: Every point on a line is a solution to that equation. So if a point is on both lines, it must satisfy both equations. That is exactly what a solution to a system means.

Step-by-step: How to solve a system by graphing

1. Write each equation in a graph-friendly form, often slope-intercept form: \(y=mx+b\).
2. Graph the first line.
3. Graph the second line on the same coordinate plane.
4. Find the intersection point.
5. Check the point in both equations if needed.

Helpful reminder about slope-intercept form:

In the equation

$$y=mx+b$$

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

• The y-intercept tells where the line starts on the y-axis.
• The slope tells how the line moves: rise over run.

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

• slope \(=2=\frac{2}{1}\)
• y-intercept \(=1\)

That means start at \((0,1)\), then go up 2 and right 1 to get more points.

What the graph can show

When you graph a system, there are three possible results:

• One solution: The lines cross once.
• No solution: The lines are parallel and never meet.
• Infinitely many solutions: The lines are the same line, so every point on the line works.

Most 8th grade problems focus first on finding the single intersection point.

Worked Example 1: A simple system

Solve the system by graphing:

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

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

• y-intercept: \((0,1)\)
• slope: \(1=\frac{1}{1}\)

From \((0,1)\), go up 1 and right 1. Some points are \((1,2)\), \((2,3)\), and \((3,4)\).

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

• y-intercept: \((0,5)\)
• slope: \(-1=\frac{-1}{1}\)

From \((0,5)\), go down 1 and right 1. Some points are \((1,4)\), \((2,3)\), and \((3,2)\).

Step 3: Find the intersection.

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

So the solution is

$$ (2,3) $$

Check:

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

Because \((2,3)\) makes both equations true, it is the solution.

Worked Example 2: One equation needs to be rewritten

Solve the system by graphing:

$$ \begin{aligned} y&=2x-1 \\ x+y&=5 \end{aligned} $$

The first equation is already in slope-intercept form. The second is not, so we rewrite it.

Start with

$$x+y=5$$

Subtract \(x\) from both sides:

$$y=-x+5$$

Now graph both lines.

First line: \(y=2x-1\)

• y-intercept: \((0,-1)\)
• slope: \(2=\frac{2}{1}\)

Points on the line include \((1,1)\) and \((2,3)\).

Second line: \(y=-x+5\)

• y-intercept: \((0,5)\)
• slope: \(-1\)

Points on the line include \((1,4)\), \((2,3)\), and \((3,2)\).

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

So the solution is

$$ (2,3) $$

Check:

• In \(y=2x-1\): \(3=2(2)-1=3\)
• In \(x+y=5\): \(2+3=5\)

Worked Example 3: No solution

Solve the system by graphing:

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

Look at the slopes and y-intercepts.

• Both lines have slope \(\frac{1}{2}\).
• The y-intercepts are different: \(2\) and \(-3\).

Lines with the same slope but different y-intercepts are parallel. Parallel lines never intersect.

That means this system has no solution.

On a graph, you would see two lines going in the same direction, always staying the same distance apart.

Worked Example 4: Infinitely many solutions

Solve the system by graphing:

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

The second equation is not in slope-intercept form, so rewrite it by dividing every term by 2:

$$y=3x+2$$

Now both equations are the same:

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

If you graph them, both equations make the same line.

That means every point on the line is a solution, so the system has infinitely many solutions.

Tips for graphing accurately

• Use a ruler or graph paper if possible.
• Plot the y-intercept first.
• Use the slope carefully to find more points.
• Draw both lines on the same grid.
• Label your lines if the graph gets confusing.
• If the intersection is between grid points, graphing may only give an estimate.

How to tell what kind of answer a system has

• Different slopes \(\rightarrow\) one solution
• Same slope, different y-intercepts \(\rightarrow\) no solution
• Same slope, same y-intercept \(\rightarrow\) infinitely many solutions

Common mistakes to avoid

• Mixing up slope and y-intercept. In \(y=mx+b\), \(m\) is slope and \(b\) is the y-intercept.
• Forgetting to rewrite an equation. If an equation is not in an easy-to-graph form, rewrite it first.
• Graphing only one line. You need both lines on the same coordinate plane.
• Reading the intersection incorrectly. Double-check the \(x\)-value and \(y\)-value.
• Not checking the solution. Plug the point back into both equations when possible.

When graphing is useful

Graphing is a great way to see what is happening in a system. It helps you understand whether lines cross, are parallel, or are the same line.

It is especially useful when you want a visual answer. But if the lines intersect at a point that is not exactly on a grid point, graphing may not give a perfect answer. In those cases, other methods can sometimes be more exact.

Quick practice thinking

If you saw the system

$$ \begin{aligned} y&=x+4 \\ y&=-2x+1 \end{aligned} $$

you would:

• graph \(y=x+4\) using y-intercept \(4\) and slope \(1\)
• graph \(y=-2x+1\) using y-intercept \(1\) and slope \(-2\)
• look for the point where the two lines cross

That crossing point would be the solution to the system.

Summary

To solve a system by graphing, graph both linear equations on the same coordinate plane and find the point where they intersect. The intersection point is the ordered pair that makes both equations true.

If the lines cross once, there is one solution. If they never cross, there is no solution. If they are the same line, there are infinitely many solutions.

Always graph carefully, and check your answer when you can.

Solving Systems by Substitution

A system of linear equations is a set of two equations with the same variables. When we solve a system, we are looking for the ordered pair \\((x, y)\\) that makes both equations true at the same time.

One useful way to solve a system is called substitution. This method works by solving one equation for one variable, then replacing that variable in the other equation.

Substitution is especially helpful when one equation already has a variable by itself, or when it is easy to get a variable alone.

What substitution means

If one equation tells you that \\(y = x + 3\\), then wherever you see \\(y\\), you can replace it with \\(x + 3\\). That replacement is called substituting.

For example, if another equation is \\(2x + y = 9\\), then substitute \\(x + 3\\) for \\(y\\):

$$2x + (x + 3) = 9$$

Now the equation has only one variable, so it is easier to solve.

Why substitution works

If two expressions are equal to the same variable, then they are equal to each other. So if \\(y = x + 3\\), then in any equation containing \\(y\\), we can use \\(x + 3\\) instead. This lets us turn two equations into one equation with one unknown.

Steps for solving by substitution

1. Choose one equation and solve it for one variable.
2. Substitute that expression into the other equation.
3. Solve the new equation.
4. Plug your answer back in to find the other variable.
5. Check your ordered pair in both original equations.

Example 1: A variable is already isolated

Solve the system:

$$ \begin{cases} y = x + 4 \\ 2x + y = 10 \end{cases} $$

The first equation already has \\(y\\) by itself:

$$y = x + 4$$

Substitute \\(x + 4\\) for \\(y\\) in the second equation:

$$2x + y = 10$$ $$2x + (x + 4) = 10$$

Now solve:

$$3x + 4 = 10$$ $$3x = 6$$ $$x = 2$$

Now find \\(y\\) by using \\(y = x + 4\\):

$$y = 2 + 4 = 6$$

So the solution is:

$$ (2, 6) $$

Check in both equations:

• First equation: \\(6 = 2 + 4\\) ✔
• Second equation: \\(2(2) + 6 = 10\\), so \\(4 + 6 = 10\\) ✔

Example 2: Solve one equation first

Solve the system:

$$ \begin{cases} x + y = 9 \\ y = 2x \end{cases} $$

The second equation already tells us \\(y = 2x\\). Substitute \\(2x\\) for \\(y\\) in the first equation:

$$x + y = 9$$ $$x + 2x = 9$$

Solve for \\(x\\):

$$3x = 9$$ $$x = 3$$

Now use \\(y = 2x\\):

$$y = 2(3) = 6$$

The solution is:

$$ (3, 6) $$

Check:

• First equation: \\(3 + 6 = 9\\) ✔
• Second equation: \\(6 = 2(3)\\) ✔

Example 3: You must isolate a variable first

Solve the system:

$$ \begin{cases} 3x + y = 11 \\ x - y = 1 \end{cases} $$

Neither equation has a variable alone yet. Let’s solve the first equation for \\(y\\):

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

Now substitute \\(11 - 3x\\) for \\(y\\) in the second equation:

$$x - y = 1$$ $$x - (11 - 3x) = 1$$

Be careful with the minus sign. Distribute it correctly:

$$x - 11 + 3x = 1$$ $$4x - 11 = 1$$ $$4x = 12$$ $$x = 3$$

Now substitute \\(x = 3\\) into \\(y = 11 - 3x\\):

$$y = 11 - 3(3)$$ $$y = 11 - 9 = 2$$

The solution is:

$$ (3, 2) $$

Check:

• First equation: \\(3(3) + 2 = 11\\), so \\(9 + 2 = 11\\) ✔
• Second equation: \\(3 - 2 = 1\\) ✔

Example 4: A word problem using substitution

Jada bought 2 notebooks and 1 pen for \\$7\\). She also bought 1 notebook and 1 pen for \\$5\\). How much does each item cost?

Let:

• \\(n\\) = cost of one notebook
• \\(p\\) = cost of one pen

Write a system:

$$ \begin{cases} 2n + p = 7 \\ n + p = 5 \end{cases} $$

Solve the second equation for \\(p\\):

$$n + p = 5$$ $$p = 5 - n$$

Substitute \\(5 - n\\) for \\(p\\) in the first equation:

$$2n + p = 7$$ $$2n + (5 - n) = 7$$

Solve:

$$n + 5 = 7$$ $$n = 2$$

Now find \\(p\\):

$$p = 5 - 2 = 3$$

So:

• A notebook costs \\$2\\).
• A pen costs \\$3\\).

Important tips

• Pick the easier equation to solve first. If a variable already has a coefficient of 1, that is often the best choice.
• Use parentheses when substituting, especially if the expression has more than one term.
• Watch negative signs carefully. A mistake with signs can change the whole answer.
• Always check your solution in both original equations.

Common mistakes to avoid

• Forgetting parentheses
  Example: if \\(y = 4 - x\\), then substitute as \\(2x + (4 - x)\\), not \\(2x + 4 - x\\) without thinking about the grouping.
• Not solving all the way
  Finding only \\(x\\) or only \\(y\\) is not enough. A system’s solution is an ordered pair.
• Checking in only one equation
  Your answer must make both equations true.
• Mixing up variables
  Be sure you replace the correct variable with the correct expression.

When substitution is a good choice

Substitution is often the best method when:

• one variable is already isolated, like \\(y = 3x - 2\\)
• it is easy to solve for a variable
• you want an algebraic method instead of graphing

Quick practice thinking

For the system

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

You would substitute \\(x - 1\\) for \\(y\\):

$$x + (x - 1) = 7$$

Then solve to find \\(x\\), and use that value to find \\(y\\).

Summary

To solve a system by substitution, first get one variable alone. Then replace that variable in the other equation with its equal expression. Solve the new equation, substitute back to find the second variable, and check your answer in both equations.

Substitution helps you find the point where two lines meet without drawing a graph. With careful work and checking, it is a powerful way to solve systems of linear equations.

Solving Systems by Elimination

Sometimes in math, you have two equations with the same two variables. This is called a system of linear equations.

Your goal is to find the one pair of values that makes both equations true at the same time.

One useful way to solve a system is called elimination. In elimination, you add or subtract the equations to make one variable disappear. Then you solve for the remaining variable.

This method works especially well when the equations are lined up neatly and the variable terms can cancel out.

What does eliminate mean?

To eliminate a variable means to get rid of it for a moment. For example, if you add \(+3y\) and \(-3y\), they make \(0y\), so the \(y\)-terms disappear.

That leaves you with an equation in only one variable, which is much easier to solve.

When does elimination work?

Elimination works best when:

• the coefficients of one variable are already opposites, like \(+2x\) and \(-2x\)
• or you can multiply one or both equations so the coefficients become opposites

Step-by-step process for elimination

1. Write the equations in standard form if possible, such as \(Ax + By = C\).
2. Line up like terms so the \(x\)-terms are above each other and the \(y\)-terms are above each other.
3. Check whether one variable can be eliminated by adding or subtracting the equations.
4. If not, multiply one or both equations to create opposite coefficients for one variable.
5. Add or subtract the equations to eliminate a variable.
6. Solve the new equation.
7. Substitute your answer back into one of the original equations to find the other variable.
8. Check your solution in both equations.

Important idea: adding vs. subtracting

If the coefficients are already opposites, you can add the equations.

If the coefficients are the same, you can subtract one equation from the other.

For example:

• \(3x + 2y = 10\) and \(-3x + y = 4\) → add the equations to eliminate \(x\)
• \(4x + 5y = 9\) and \(4x - 2y = 12\) → subtract the equations to eliminate \(x\)

Worked Example 1: One variable is already easy to eliminate

Solve the system:

$$ \begin{aligned} 2x + y &= 7 \\ 3x - y &= 8 \end{aligned} $$

The \(y\)-terms are \(+y\) and \(-y\). They are opposites, so we can add the equations.

$$ \begin{aligned} 2x + y &= 7 \\ 3x - y &= 8 \\ \hline 5x &= 15 \end{aligned} $$

Now solve for \(x\):

$$ 5x = 15 $$ $$ x = 3 $$

Substitute \(x=3\) into one of the original equations. Use \(2x + y = 7\):

$$ 2(3) + y = 7 $$ $$ 6 + y = 7 $$ $$ y = 1 $$

So the solution is:

$$ (x,y) = (3,1) $$

Check in the second equation:

$$ 3(3) - 1 = 8 $$ $$ 9 - 1 = 8 $$

It works.

Worked Example 2: Subtracting to eliminate

Solve the system:

$$ \begin{aligned} 4x + 2y &= 18 \\ 4x - 3y &= 3 \end{aligned} $$

The \(x\)-terms are both \(4x\). Since they are the same, we can subtract one equation from the other.

Subtract the second equation from the first:

$$ \begin{aligned} 4x + 2y &= 18 \\ -(4x - 3y &= 3) \end{aligned} $$

This gives:

$$ 2y - (-3y) = 18 - 3 $$ $$ 5y = 15 $$ $$ y = 3 $$

Now substitute \(y=3\) into one of the original equations. Use \(4x - 3y = 3\):

$$ 4x - 3(3) = 3 $$ $$ 4x - 9 = 3 $$ $$ 4x = 12 $$ $$ x = 3 $$

So the solution is:

$$ (x,y) = (3,3) $$

Worked Example 3: Multiply first, then eliminate

Solve the system:

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

The \(y\)-terms are \(+2y\) and \(-2y\), so they are already opposites. That means we can add the equations right away.

$$ \begin{aligned} x + 2y &= 11 \\ 3x - 2y &= 1 \\ \hline 4x &= 12 \end{aligned} $$ $$ x = 3 $$

Substitute into \(x + 2y = 11\):

$$ 3 + 2y = 11 $$ $$ 2y = 8 $$ $$ y = 4 $$

So the solution is:

$$ (x,y) = (3,4) $$

Now let’s look at a system where we must multiply first.

Solve:

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

Actually, in this system, the \(y\)-terms are already opposites, so we can add:

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

Then:

$$ 3 + y = 9 $$ $$ y = 6 $$

So the solution is \((3,6)\).

Let’s now do an example where multiplication is truly needed.

Worked Example 4: Multiplication is necessary

Solve the system:

$$ \begin{aligned} 2x + y &= 10 \\ 3x + 2y &= 18 \end{aligned} $$

The coefficients of \(y\) are \(1\) and \(2\). They are not opposites yet. We can multiply the first equation by \(-2\) so the \(y\)-terms become opposites.

Multiply the first equation by \(-2\):

$$ -2(2x + y = 10) $$ $$ -4x - 2y = -20 $$

Now use the new system:

$$ \begin{aligned} -4x - 2y &= -20 \\ 3x + 2y &= 18 \end{aligned} $$

Add the equations:

$$ \begin{aligned} -4x - 2y &= -20 \\ 3x + 2y &= 18 \\ \hline -x &= -2 \end{aligned} $$ $$ x = 2 $$

Substitute \(x=2\) into \(2x + y = 10\):

$$ 2(2) + y = 10 $$ $$ 4 + y = 10 $$ $$ y = 6 $$

So the solution is:

$$ (x,y) = (2,6) $$

Why multiplication helps

Sometimes the variable terms do not match in a way that lets them cancel. Multiplying changes the coefficients but keeps the equation balanced.

For example, if you have \(y\) in one equation and \(2y\) in the other, you can multiply the first equation by \(2\) or \(-2\). This creates matching or opposite coefficients so elimination can happen.

Common mistakes to avoid

• Do not forget to line up like terms. Make sure \(x\)-terms, \(y\)-terms, and constants are in the correct columns.
• If you multiply an equation, multiply every term. For example, multiplying \(2x + y = 10\) by \(-2\) gives \(-4x - 2y = -20\), not just \(-4x + y = 10\).
• Be careful with negative signs. When subtracting equations, signs often change.
• Always substitute back. After finding one variable, use an original equation to find the other.
• Check your answer in both equations. This helps catch errors.

Quick strategy tips

• Look first for coefficients that are already opposites.
• If the coefficients are the same, subtraction may be easiest.
• If nothing matches, decide which variable will be easiest to eliminate by multiplying.
• Choose the multiplication that keeps numbers as small as possible.

How elimination connects to the meaning of the system

When you solve a system, you are finding the point where the two lines meet. That point is the ordered pair that makes both equations true.

Elimination is an algebra method for finding that same point without drawing the graph.

Brief Summary

To solve a system by elimination, you combine equations so one variable disappears. Then you solve for the remaining variable and substitute back to find the other one.

If needed, multiply one or both equations first to create opposite coefficients. Always check your final answer in both original equations.

Special Cases in Systems of Linear Equations

When you solve a system of linear equations, you are looking for a point that makes both equations true at the same time.

Usually, two lines cross once, so a system has one solution. But sometimes a system has a special case. The lines might never cross, or they might lie on top of each other.

In this lesson, you will learn how to recognize the two special cases in systems:

• No solution: the lines are parallel and never meet
• Infinitely many solutions: the two equations represent the same line

Understanding these cases helps you know what a system means before you even finish solving it.

1. Review: What is a solution to a system?

A system of linear equations has two equations with the same variables, such as \(x\) and \(y\).

Example:

$$ \begin{aligned} y &= 2x + 1\\ y &= -x + 4 \end{aligned} $$

A solution is an ordered pair \((x,y)\) that works in both equations.

Graphically, the solution is where the two lines intersect.

There are three possible outcomes for a system of linear equations:

• One solution: the lines intersect once
• No solution: the lines never intersect
• Infinitely many solutions: the lines are the same line

2. No solution: Parallel lines

Two lines are parallel if they have the same slope but different y-intercepts.

That means they rise or fall at the same rate, so they stay the same distance apart and never meet.

For example:

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

Both lines have slope \(3\), but their y-intercepts are different: \(2\) and \(-5\).

Since the lines are parallel, there is no point where they intersect.

So this system has no solution.

How to spot no solution algebraically

When solving by substitution or elimination, a no-solution system often ends with a false statement.

For example, you might get:

$$2 = 7$$

This is never true, so the system has no solution.

3. Infinitely many solutions: Same line

Sometimes two equations look different at first, but they actually represent the same line.

If that happens, every point on the line is a solution to both equations.

That means the system has infinitely many solutions.

Example:

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

If we divide the second equation by \(2\), we get:

$$y = 2x + 3$$

Now both equations are exactly the same.

So the two lines lie on top of each other, and the system has infinitely many solutions.

How to spot infinitely many solutions algebraically

When solving, you may end with a true statement such as:

$$0 = 0$$

or

$$5 = 5$$

This means the two equations are equivalent, so the system has infinitely many solutions.

4. Comparing the three types of systems

• One solution: different slopes, so the lines cross once
• No solution: same slope, different y-intercepts, so the lines are parallel
• Infinitely many solutions: same slope and same y-intercept, so the equations are the same line

If the equations are in slope-intercept form, \(y = mx + b\):

• \(m\) is the slope
• \(b\) is the y-intercept

You can often decide the type of solution by comparing \(m\) and \(b\).

5. Worked Examples

Example 1: Identify a system with no solution from slope-intercept form

Determine whether the system has one solution, no solution, or infinitely many solutions:

$$ \begin{aligned} y &= -2x + 4\\ y &= -2x - 1 \end{aligned} $$

Step 1: Compare the slopes.

Both lines have slope \(-2\).

Step 2: Compare the y-intercepts.

The y-intercepts are \(4\) and \(-1\), which are different.

Conclusion: The lines are parallel, so the system has no solution.

Example 2: Identify a system with infinitely many solutions

Determine whether the system has one solution, no solution, or infinitely many solutions:

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

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

$$3x + y = 9$$

Subtract \(3x\) from both sides:

$$y = -3x + 9$$

Step 2: Compare the two equations.

Both equations are:

$$y = -3x + 9$$

Conclusion: They are the same line, so the system has infinitely many solutions.

Example 3: Use elimination to recognize no solution

Solve the system:

$$ \begin{aligned} 2x + y &= 5\\ 2x + y &= 11 \end{aligned} $$

Step 1: Subtract the first equation from the second.

$$ (2x + y) - (2x + y) = 11 - 5 $$

$$0 = 6$$

Step 2: Interpret the result.

The statement \(0 = 6\) is false.

Conclusion: The system has no solution.

This makes sense because both equations have the same left side but different right sides, so they cannot both be true for the same point.

Example 4: Use elimination to recognize infinitely many solutions

Solve the system:

$$ \begin{aligned} 4x - 2y &= 8\\ 2x - y &= 4 \end{aligned} $$

Step 1: Compare the equations.

If we multiply the second equation by \(2\), we get:

$$2(2x - y = 4)$$

$$4x - 2y = 8$$

This is exactly the first equation.

Step 2: Use elimination if needed.

Subtracting one equation from the other leads to a true statement like:

$$0 = 0$$

Conclusion: The equations are the same line, so the system has infinitely many solutions.

6. Tips for recognizing special cases quickly

1. Put both equations in the same form, if possible. Slope-intercept form is often easiest to compare.
2. Check the slopes.
   • Different slopes \(rightarrow\) one solution
   • Same slopes \(rightarrow\) could be no solution or infinitely many solutions
3. Check the y-intercepts.
   • Same slope + different y-intercepts \(rightarrow\) no solution
   • Same slope + same y-intercept \(rightarrow\) infinitely many solutions
4. Watch for false or true statements when solving.
   • False statement, like \(3 = 8\) \(rightarrow\) no solution
   • True statement, like \(0 = 0\) \(rightarrow\) infinitely many solutions

7. Common mistakes to avoid

• Mistake: Thinking same slope always means same line.
  Fix: Check the y-intercepts too.
• Mistake: Stopping after getting \(0 = 0\) and saying the solution is \(0,0\).
  Fix: \(0 = 0\) means infinitely many solutions, not the point \((0,0)\).
• Mistake: Stopping after getting \(0 = 5\) and trying to solve more.
  Fix: A false statement means there is no solution.

8. Summary

A system of linear equations can have one solution, no solution, or infinitely many solutions.

If the lines are parallel, they never intersect, so there is no solution.

If the two equations describe the same line, then every point on that line works, so there are infinitely many solutions.

You can recognize these special cases by comparing slopes and y-intercepts or by noticing whether solving gives a false statement like \(0 = 6\) or a true statement like \(0 = 0\).

Systems with Fractional and Decimal Coefficients

Sometimes a system of linear equations includes fractions or decimals in front of the variables. That can make the equations look harder than they really are.

The good news is that you can often make the system easier first. Before solving, you can clear the fractions or clear the decimals. Then you solve the system using the same methods you already know: substitution or elimination.

In this lesson, you will learn how to simplify systems with fractional and decimal coefficients and then solve them step by step.

What is a system of linear equations?

A system of linear equations is a set of two equations with the same variables. A solution is an ordered pair, like \((x,y)\), that makes both equations true.

For example, this is a system:

$$ \begin{cases} \frac{1}{2}x + y = 5 \\ 2x - y = 1 \end{cases} $$

The fractions do not change what the solution means. We still want to find the values of \(x\) and \(y\) that work in both equations.

Why clear fractions or decimals first?

Fractions and decimals can make arithmetic slower and more confusing. If you rewrite the equations so the coefficients are whole numbers, the solving process is usually much easier.

You are not changing the solution if you multiply every term in one equation by the same nonzero number. You are just rewriting the equation in an easier form.

Clearing fractions

To clear fractions in an equation, find the least common denominator (LCD) of all the fractions in that equation. Then multiply every term in the equation by that LCD.

Example of clearing fractions:

$$ \frac{1}{2}x + \frac{3}{4}y = 6 $$

The denominators are 2 and 4, so the LCD is 4. Multiply every term by 4:

$$ 4\left(\frac{1}{2}x\right) + 4\left(\frac{3}{4}y\right) = 4(6) $$ $$ 2x + 3y = 24 $$

Now the equation has whole-number coefficients.

Clearing decimals

To clear decimals, multiply every term by a power of 10:

• Multiply by 10 if the greatest number of decimal places is 1.
• Multiply by 100 if the greatest number of decimal places is 2.
• Multiply by 1000 if the greatest number of decimal places is 3.

Example of clearing decimals:

$$ 0.3x + 1.2y = 4.5 $$

There is 1 decimal place in each number, so multiply every term by 10:

$$ 10(0.3x) + 10(1.2y) = 10(4.5) $$ $$ 3x + 12y = 45 $$

Now the equation is easier to use in substitution or elimination.

Important rule

When you clear fractions or decimals, you must multiply every term on both sides of the equation. If you miss a term, the equation will no longer be equivalent.

For example, in

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

if you multiply by 3, you must multiply all terms:

$$ 3\left(\frac{1}{3}x\right) + 3(y) = 3(7) $$ $$ x + 3y = 21 $$

After clearing, choose a solving method

Once the equations are rewritten, solve the system the same way you solve any other system.

• Use substitution when one variable is easy to isolate.
• Use elimination when the coefficients can be made opposites easily.

Worked Example 1: One equation has fractions

Solve the system:

$$ \begin{cases} \frac{1}{2}x + y = 4 \\ x - y = 2 \end{cases} $$

Step 1: Clear the fraction in the first equation.

The LCD is 2, so multiply every term in the first equation by 2:

$$ 2\left(\frac{1}{2}x + y = 4\right) $$ $$ x + 2y = 8 $$

Now the system is:

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

Step 2: Use elimination.

Subtract the second equation from the first:

$$ (x + 2y) - (x - y) = 8 - 2 $$ $$ 3y = 6 $$ $$ y = 2 $$

Step 3: Substitute to find \(x\).

Use \(x - y = 2\):

$$ x - 2 = 2 $$ $$ x = 4 $$

Solution: \((4,2)\)

Check:

$$ \frac{1}{2}(4) + 2 = 2 + 2 = 4 $$ $$ 4 - 2 = 2 $$

Both equations are true.

Worked Example 2: Decimals in both equations

Solve the system:

$$ \begin{cases} 0.5x + 0.2y = 3.4 \\ 0.3x - 0.2y = 0.6 \end{cases} $$

Step 1: Clear decimals.

Each number has 1 decimal place, so multiply every term in both equations by 10.

$$ \begin{cases} 10(0.5x + 0.2y = 3.4) \\ 10(0.3x - 0.2y = 0.6) \end{cases} $$ $$ \begin{cases} 5x + 2y = 34 \\ 3x - 2y = 6 \end{cases} $$

Step 2: Use elimination.

Add the equations because \(2y\) and \(-2y\) are opposites:

$$ (5x + 2y) + (3x - 2y) = 34 + 6 $$ $$ 8x = 40 $$ $$ x = 5 $$

Step 3: Find \(y\).

Substitute into \(5x + 2y = 34\):

$$ 5(5) + 2y = 34 $$ $$ 25 + 2y = 34 $$ $$ 2y = 9 $$ $$ y = 4.5 $$

Solution: \((5,4.5)\)

Check:

$$ 0.5(5) + 0.2(4.5) = 2.5 + 0.9 = 3.4 $$ $$ 0.3(5) - 0.2(4.5) = 1.5 - 0.9 = 0.6 $$

Both equations are true.

Worked Example 3: Fractions in both equations

Solve the system:

$$ \begin{cases} \frac{1}{3}x + \frac{1}{2}y = 4 \\ \frac{2}{3}x - \frac{1}{2}y = 1 \end{cases} $$

Step 1: Clear fractions in each equation.

In both equations, the denominators are 3 and 2, so the LCD is 6. Multiply each equation by 6.

First equation:

$$ 6\left(\frac{1}{3}x + \frac{1}{2}y = 4\right) $$ $$ 2x + 3y = 24 $$

Second equation:

$$ 6\left(\frac{2}{3}x - \frac{1}{2}y = 1\right) $$ $$ 4x - 3y = 6 $$

Now the system is:

$$ \begin{cases} 2x + 3y = 24 \\ 4x - 3y = 6 \end{cases} $$

Step 2: Use elimination.

Add the equations:

$$ (2x + 3y) + (4x - 3y) = 24 + 6 $$ $$ 6x = 30 $$ $$ x = 5 $$

Step 3: Find \(y\).

Substitute into \(2x + 3y = 24\):

$$ 2(5) + 3y = 24 $$ $$ 10 + 3y = 24 $$ $$ 3y = 14 $$ $$ y = \frac{14}{3} $$

Solution: \((5,\frac{14}{3})\)

It is okay if a solution includes a fraction. Clearing fractions from the equations does not mean the answer must be a whole number.

Worked Example 4: Use substitution after clearing decimals

Solve the system:

$$ \begin{cases} 1.2x + 0.5y = 7.1 \\ y = 2x + 1 \end{cases} $$

Step 1: Clear decimals in the first equation.

The greatest number of decimal places is 1, so multiply the first equation by 10:

$$ 10(1.2x + 0.5y = 7.1) $$ $$ 12x + 5y = 71 $$

Now the system is:

$$ \begin{cases} 12x + 5y = 71 \\ y = 2x + 1 \end{cases} $$

Step 2: Use substitution.

Substitute \(2x + 1\) for \(y\):

$$ 12x + 5(2x + 1) = 71 $$ $$ 12x + 10x + 5 = 71 $$ $$ 22x + 5 = 71 $$ $$ 22x = 66 $$ $$ x = 3 $$

Step 3: Find \(y\).

$$ y = 2(3) + 1 = 7 $$

Solution: \((3,7)\)

Common mistakes to avoid

• Forgetting to multiply every term. If you clear fractions or decimals, every term must be multiplied.
• Using the wrong LCD. Check the denominators carefully before multiplying.
• Only clearing one equation when both need it. Simplify each equation as needed.
• Sign mistakes. Be careful with negative signs when adding or subtracting equations.
• Not checking the solution. Substitute your answer back into the original equations.

Helpful strategy

1. Look at each equation.
2. If there are fractions, multiply by the LCD.
3. If there are decimals, multiply by a power of 10.
4. Rewrite the system with simpler coefficients.
5. Choose substitution or elimination.
6. Solve for both variables.
7. Check your answer in the original system.

Brief Summary

Systems with fractional and decimal coefficients can be solved just like other systems, but it helps to simplify first. Clear fractions by multiplying each equation by the least common denominator. Clear decimals by multiplying by a power of 10. Then use substitution or elimination, and always check your solution in the original equations.

Contextual Applications of Systems means using systems of linear equations to solve real-world problems. A system has two equations with the same two variables. The solution is the pair of values that makes both equations true at the same time.

In 8th grade, you will often use systems to model situations with costs, mixtures, rates, and break-even points. These problems may sound like stories, but they can be solved by carefully defining variables, writing equations, and solving the system.

This lesson will show you how to turn words into equations, solve the system, and explain what the answer means in the context of the problem.

Why systems are useful

Sometimes one equation is not enough. If a problem has two unknown values, you usually need two equations. Systems help us find those two unknowns together.

For example, if you buy some adult tickets and some student tickets, there are two unknowns: how many of each type were bought. If you know the total number of tickets and the total cost, you can write a system.

How to model a real-world problem with a system

1. Read carefully. Decide what the unknown quantities are.
2. Define variables. For example, let \(a\) be the number of adult tickets and \(s\) be the number of student tickets.
3. Write two equations. Look for two facts in the problem, such as a total number and a total cost.
4. Solve the system. You can use graphing, substitution, or elimination.
5. Check the answer. Substitute your values back into both equations.
6. Interpret the answer. Write a sentence that explains what the numbers mean in the real situation.

Clue words to look for

• Total number often leads to adding variables, like \(x + y = 20\).
• Total cost often leads to a money equation, like \(5x + 3y = 76\).
• Rate problems often use distance, time, or speed relationships.
• Mixture problems often involve amount and value.
• Break-even means two costs or earnings are equal.

Common forms of contextual systems

1. Cost problems

These problems involve prices of two different items. You may know the total number of items and the total amount spent.

A common structure is:

$$x + y = \text{total number of items}$$$$ax + by = \text{total cost}$$

Here, \(a\) and \(b\) are the prices of the two items.

2. Mixture problems

These problems combine two groups or ingredients. In 8th grade, this often means mixing items with different values, such as coins or different prices.

You may know:

• the total number of items, and
• the total value

That creates a system similar to a cost problem.

3. Rate problems

Rate problems compare how fast things happen or how far things travel. You may have one situation for each equation.

For example, if two people travel different distances at different rates but for the same amount of time, you can write equations using the information given.

4. Break-even problems

A break-even point is when two amounts are equal. For example, one company may charge a starting fee plus a smaller cost per hour, while another company charges no starting fee but a larger cost per hour.

You can model each company with a linear equation. The point where the graphs meet is the break-even point.

Ways to solve systems

Graphing

Graph both equations on the same coordinate plane. The point where the lines intersect is the solution.

This method is helpful for seeing what the solution means, especially in break-even problems. But if the intersection is not at an exact grid point, graphing may only give an estimate.

Substitution

Solve one equation for one variable, then substitute that expression into the other equation.

This method works well when one equation is already written as something like \(y = x + 4\) or \(x = 12 - y\).

Elimination

Add or subtract equations to remove one variable. This method is often useful in cost and mixture problems.

Worked Example 1: Ticket Cost Problem

A school play sold 30 tickets. Student tickets cost \(\$4\) each and adult tickets cost \(\$6\) each. The total money collected was \(\$156\). How many student tickets and adult tickets were sold?

Step 1: Define variables.

Let \(s\) = number of student tickets.

Let \(a\) = number of adult tickets.

Step 2: Write the equations.

The total number of tickets is 30:

$$s + a = 30$$

The total cost is 156 dollars:

$$4s + 6a = 156$$

Step 3: Solve the system.

Use substitution. From the first equation:

$$s = 30 - a$$

Substitute into the cost equation:

$$4(30 - a) + 6a = 156$$$$120 - 4a + 6a = 156$$$$120 + 2a = 156$$$$2a = 36$$$$a = 18$$

Now find \(s\):

$$s = 30 - 18 = 12$$

Step 4: Write the answer.

There were 12 student tickets and 18 adult tickets sold.

Check:

$$12 + 18 = 30$$$$4(12) + 6(18) = 48 + 108 = 156$$

Both equations are true.

Worked Example 2: Coin Mixture Problem

A jar contains only nickels and dimes. There are 40 coins in all, and their total value is \(\$3.10\). How many nickels and how many dimes are in the jar?

Step 1: Define variables.

Let \(n\) = number of nickels.

Let \(d\) = number of dimes.

Step 2: Write the equations.

Total number of coins:

$$n + d = 40$$

Total value in cents: \(\$3.10 = 310\) cents

$$5n + 10d = 310$$

Step 3: Solve the system.

It helps to simplify the second equation by dividing by 5:

$$n + 2d = 62$$

Now subtract the first equation:

$$n + 2d = 62$$$$n + d = 40$$

Subtract:

$$d = 22$$

Then:

$$n = 40 - 22 = 18$$

Step 4: Write the answer.

The jar has 18 nickels and 22 dimes.

Check:

$$18 + 22 = 40$$$$5(18) + 10(22) = 90 + 220 = 310$$

Worked Example 3: Rate Problem

A cyclist travels at 12 miles per hour, and a runner travels at 6 miles per hour. Together they travel 54 miles in one afternoon. The cyclist rides for \(x\) hours and the runner runs for \(y\) hours. If together they spend 6 hours exercising, how long does each person exercise?

Step 1: Define variables.

Let \(x\) = hours the cyclist rides.

Let \(y\) = hours the runner runs.

Step 2: Write the equations.

Total time:

$$x + y = 6$$

Total distance:

$$12x + 6y = 54$$

Step 3: Solve the system.

Divide the second equation by 6:

$$2x + y = 9$$

Now use the first equation:

$$x + y = 6$$

Subtract the first equation from the simplified second equation:

$$2x + y = 9$$$$x + y = 6$$

Subtract:

$$x = 3$$

Then:

$$y = 6 - 3 = 3$$

Step 4: Write the answer.

The cyclist rides for 3 hours, and the runner runs for 3 hours.

Check:

$$12(3) + 6(3) = 36 + 18 = 54$$

Worked Example 4: Break-Even Problem

Two music streaming plans have different pricing.

• Plan A charges a \(\$10\) monthly fee plus \(\$2\) per download.
• Plan B charges no monthly fee and \(\$4\) per download.

At how many downloads do the two plans cost the same? What is that cost?

Step 1: Define variables.

Let \(x\) = number of downloads.

Let \(y\) = total cost.

Step 2: Write equations for each plan.

Plan A:

$$y = 2x + 10$$

Plan B:

$$y = 4x$$

Step 3: Solve the system.

At the break-even point, the costs are equal, so set the equations equal:

$$2x + 10 = 4x$$$$10 = 2x$$$$x = 5$$

Now find the cost:

$$y = 4(5) = 20$$

Step 4: Write the answer.

The plans cost the same at 5 downloads, and the cost is \(\$20\).

What the graph means: If you graph both lines, they intersect at \((5, 20)\). That point is the break-even point.

How to choose the right method

• Use graphing when you want to see the situation clearly, especially for break-even points.
• Use substitution when one equation is easy to solve for one variable.
• Use elimination when the equations line up nicely and one variable can be removed by adding or subtracting.

Tips for writing equations from word problems

• Always define your variables first.
• Use one equation for one fact from the problem.
• Be careful with units. For money, decide whether to use dollars or cents.
• Read the question at the end. Make sure your answer matches what is being asked.
• Check whether your answer makes sense. For example, a negative number of tickets would not make sense.

Common mistakes to avoid

• Mixing up variables: If \(x\) is adult tickets, do not use it later for student tickets.
• Forgetting total equations: Many problems include a total number and a total value.
• Using the wrong rates or costs: Match each variable with the correct number.
• Not checking the solution: A quick check can catch mistakes.
• Ignoring context: The solution to the equations must fit the real-world situation.

Quick practice thinking

If a problem says, “There are 25 animals made up of chickens and rabbits, and there are 70 legs total,” you can define variables and write:

$$c + r = 25$$$$2c + 4r = 70$$

That is a system because there are two unknowns and two facts.

If a problem says, “A gym charges \(\$15\) to sign up and \(\$8\) per class, while another gym charges \(\$11\) per class with no sign-up fee,” you can write:

$$y = 8x + 15$$$$y = 11x$$

The solution tells when the total cost is the same.

Summary

Real-world systems problems usually involve two unknowns and two relationships. To solve them, define variables, write two equations, solve the system, and explain what the answer means.

Systems can model many everyday situations, including costs, coin values, rates, and break-even points. The key is turning words into equations carefully and checking that your solution makes sense in the real situation.