Systems of Linear Equations and Inequalities

Geometric Meaning of a System

When you solve a system, you are looking for values that make all equations or inequalities true at the same time.

In 9th Grade maths, this usually means a system of linear equations or inequalities. The geometric meaning of a system is what the solution looks like on a graph.

For linear equations, each equation makes a line. The solution to the system is the point where the lines meet, because that point lies on both lines.

For inequalities, each inequality makes a region of the graph. The solution to the system is the area where the shaded regions overlap.

This idea is very important: a solution to a system must satisfy every condition in the system.

1. What is a system?

A system is a set of two or more equations or inequalities considered together.

For example, this is a system of equations:

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

And this is a system of inequalities:

$$\begin{cases} y \ge x - 1\\ y < 3 \end{cases}$$

In both cases, we want the values of \(x\) and \(y\) that make both statements true.

2. Geometric meaning for a system of equations

Each linear equation in two variables can be drawn as a straight line on the coordinate plane.

If two lines cross, their intersection point is the solution to the system. That point gives one ordered pair \((x,y)\) that works in both equations.

There are three possible geometric cases for two linear equations:

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

So, solving a system is really about asking: Where do these graphs have points in common?

3. Geometric meaning for a system of inequalities

For an inequality, the graph is not just a line. Instead, it is a half-plane, which means one side of the boundary line is shaded.

The boundary line shows where the inequality changes.

• If the inequality uses \(\le\) or \(\ge\), draw a solid boundary line, because points on the line are included.
• If the inequality uses \(<\) or \(>\), draw a dashed boundary line, because points on the line are not included.

When you have a system of inequalities, the solution is the set of points that are in all shaded regions at once.

This overlapping region is often called the feasible region or solution region.

4. Why the intersection matters

Suppose one equation says a point must be on Line A, and another says it must be on Line B. A point that is only on one line is not enough. It must be on both.

The same is true for inequalities. A point in only one shaded region is not a solution. It has to be in the overlap.

You can think of a system as a set of rules. A solution follows every rule at the same time.

Worked Example 1: One solution from intersecting lines

Solve the system and explain its geometric meaning:

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

Step 1: Set the equations equal to each other.

Since both equal \(y\), we can write

$$x + 1 = -x + 5$$

Step 2: Solve for \(x\).

$$2x = 4$$

$$x = 2$$

Step 3: Find \(y\).

Substitute into \(y = x + 1\):

$$y = 2 + 1 = 3$$

So the solution is

$$ (2,3) $$

Geometric meaning: The two lines intersect at the point \((2,3)\). That point lies on both lines, so it satisfies both equations.

Worked Example 2: No solution from parallel lines

Consider the system:

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

Both lines have slope \(3\), so they rise at the same rate.

But they have different \(y\)-intercepts: one crosses at \(-2\) and the other at \(4\).

That means the lines are parallel.

Parallel lines never meet, so there is no intersection point.

Geometric meaning: There is no solution because there is no point that lies on both lines.

Worked Example 3: Infinitely many solutions

Consider the system:

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

Step 1: Rewrite the second equation.

Divide both sides by \(2\):

$$y = 2x + 1$$

Now both equations are exactly the same.

So the two graphs are not different lines. They are the same line.

Geometric meaning: Every point on the line works for both equations, so there are infinitely many solutions.

Worked Example 4: System of inequalities

Graph and describe the solution to:

$$\begin{cases} y \ge x - 2\\ y < 2 \end{cases}$$

Step 1: Graph the boundary lines.

• For \(y \ge x - 2\), graph the line \(y = x - 2\) as a solid line.
• For \(y < 2\), graph the line \(y = 2\) as a dashed line.

Step 2: Shade the correct sides.

• \(y \ge x - 2\) means shade above the line.
• \(y < 2\) means shade below the horizontal line \(y = 2\).

Step 3: Find the overlap.

The solution is all points that are above or on \(y = x - 2\) and also below \(y = 2\).

Geometric meaning: The solution is not one point. It is the overlapping region between the two shaded areas.

For example, the point \((0,0)\) is a solution because:

$$0 \ge 0 - 2 \quad \text{true}$$

$$0 < 2 \quad \text{true}$$

But the point \((5,3)\) is not a solution because:

$$3 \ge 5 - 2 \quad \text{true}$$

$$3 < 2 \quad \text{false}$$

Since it does not satisfy both inequalities, it is not in the solution region.

5. How to recognize the geometric meaning quickly

1. Graph each equation or inequality.
2. Look for where the graphs overlap.
3. For equations, overlap usually means an intersection point or the same line.
4. For inequalities, overlap means a shared shaded region.
5. Check that any point you choose satisfies every equation or inequality.

6. Important ideas to remember

• A solution to a system must satisfy all parts of the system.
• For two linear equations, the solution is the intersection point of the lines, if one exists.
• If lines are parallel, the system has no solution.
• If the equations represent the same line, the system has infinitely many solutions.
• For inequalities, the solution is the overlapping shaded region.
• Solid lines include the boundary; dashed lines do not.

7. Common mistakes

• Using a point that works in only one equation: a system requires the point to work in all equations or inequalities.
• Forgetting solid vs. dashed lines: this changes whether boundary points count as solutions.
• Thinking every system has one answer: some have no solution, and some have infinitely many.
• Shading the wrong side of an inequality line: test a simple point like \((0,0)\) if needed.

8. Final summary

The geometric meaning of a system is the set of points that satisfy every condition in the system at the same time.

For systems of linear equations, this usually means the point where the lines intersect. For systems of inequalities, it means the region where the shaded parts overlap.

If you remember that a solution must be common to all graphs, you will understand the geometric meaning of a system.

Solving via Graphing means finding the solution to a system of linear equations by drawing the lines and looking for where they meet.

A system of linear equations is a set of two or more linear equations. In 9th Grade, you usually work with two equations and two variables, such as:

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

The solution to the system is the point that makes both equations true at the same time. On a graph, that solution is the intersection point of the two lines.

This method is called solving via graphing because you use the picture of the lines to estimate or confirm the answer.

Why graphing helps:

• You can see where the two equations meet.
• You can check whether your answer makes sense.
• You can tell if there is one solution, no solution, or infinitely many solutions.

Important idea: Every point on a line is a solution to that line’s equation. The point where two lines cross is the one point that works for both equations.

How to solve a system by graphing:

1. Write each equation in a form that is easy to graph, usually slope-intercept form: \(y=mx+b\).
2. Graph the first line.
3. Graph the second line on the same coordinate plane.
4. Find the point where the lines intersect.
5. Write the solution as an ordered pair \((x,y)\).
6. Check by substituting the point into both equations, if needed.

Review: slope-intercept form

In the equation \(y=mx+b\):

• \(m\) is the slope, which tells how steep the line is.
• \(b\) is the y-intercept, which tells where the line crosses the y-axis.

To graph a line in slope-intercept form:

• Plot the y-intercept \((0,b)\).
• Use the slope to find another point.
• Draw the line through the points.

For example, if the slope is \(\frac{2}{3}\), move up 2 and right 3. If the slope is \(-1\), move down 1 and right 1.

What the graph can show:

• One solution: The lines cross once.
• No solution: The lines are parallel and never meet.
• Infinitely many solutions: The lines are the same line.

Now let’s look at some worked examples.

Example 1: One clear intersection

Solve the system by graphing:

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

Step 1: Graph the first line

For \(y=x+1\), the slope is \(1\) and the y-intercept is \(1\).

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

Step 2: Graph the second line

For \(y=-x+5\), the slope is \(-1\) and the y-intercept is \(5\).

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

Step 3: Find the intersection

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

So the solution is:

$$ (2,3) $$

Step 4: Check the solution

Substitute \(x=2\) into both equations:

First equation: \(y=2+1=3\)

Second equation: \(y=-2+5=3\)

The point \((2,3)\) works in both equations, so it is correct.

Example 2: Equations not already in slope-intercept form

Solve the system by graphing:

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

Step 1: Rewrite the first equation

To graph easily, solve \(x+y=4\) for \(y\):

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

Now the system is:

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

Step 2: Graph each line

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

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

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

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

Step 3: Find the intersection

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

So the solution is:

$$ (2,2) $$

Check:

In \(x+y=4\): \(2+2=4\)

In \(y=2x-2\): \(2=2(2)-2=2\)

The point is correct.

Example 3: No solution

Solve the system by graphing:

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

Step 1: Look at the slopes

Both lines have slope \(\frac{1}{2}\), so they rise at the same rate.

Step 2: Look at the y-intercepts

One line crosses the y-axis at \(1\), and the other crosses at \(-3\).

Since the lines have the same slope but different y-intercepts, they are parallel.

Parallel lines never intersect, so there is no solution.

We write:

$$ \text{No solution} $$

Example 4: Infinitely many solutions

Solve the system by graphing:

$$ \begin{aligned} y&=3x-1\\ 2y&=6x-2 \end{aligned} $$

Step 1: Rewrite the second equation

Divide both sides by 2:

$$ \begin{aligned} 2y&=6x-2\\ y&=3x-1 \end{aligned} $$

Now both equations are exactly the same:

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

Step 2: Graph the lines

Both equations make the same line, so when you graph them, one line lies directly on top of the other.

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

So the system has:

$$ \text{Infinitely many solutions} $$

Using graphing to estimate

Sometimes the lines intersect at a point that is not easy to read exactly from the graph. In that case, graphing helps you estimate the solution.

For example, if two lines appear to cross near \((1.5,2.7)\), then that is your approximate solution. A graph is very useful for seeing where the solution is, even if the exact coordinates are not whole numbers.

Tips for graphing accurately

• Use a ruler if possible to draw straight lines.
• Label your axes clearly.
• Plot at least two points for each line.
• If the graph is unclear, plot a third point to make sure the line is correct.
• Be careful with negative slopes and negative intercepts.
• Read the intersection point carefully from the grid.

Common mistakes to avoid

• Mixing up slope and y-intercept: In \(y=mx+b\), \(m\) is slope and \(b\) is the y-intercept.
• Graphing the slope incorrectly: A slope of \(-2\) means down 2 and right 1, not up 2 and right 1.
• Forgetting to rewrite the equation: If the equation is not in graph-friendly form, solve for \(y\) first.
• Writing only the x-value: The solution is an ordered pair, like \((2,3)\), not just \(2\).
• Thinking lines always intersect: Some systems have no solution or infinitely many solutions.

How graphing connects to inequalities

In systems of linear inequalities, graphing is also important. Instead of looking only for one intersection point, you look for the region where the shaded parts overlap.

For now, with systems of equations, the main idea is simpler: graph the lines and find the point where they cross.

Quick checklist for solving via graphing

• Are both equations in a form I can graph?
• Did I graph each line correctly?
• Did I identify the intersection point?
• Did I write the answer as \((x,y)\)?
• Did I notice if the lines are parallel or the same line?

Summary

To solve a system of linear equations by graphing, draw both lines on the same coordinate plane and find where they intersect. The intersection point is the solution because it satisfies both equations. If the lines do not meet, there is no solution. If the lines are the same, there are infinitely many solutions.

Solving via Substitution is a method for solving a system of linear equations. A system of linear equations is a set of two or more equations with the same variables, such as \(x\) and \(y\).

When we solve a system, we are looking for the solution that makes both equations true at the same time. On a graph, this solution is the point where the two lines intersect.

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

This method is especially helpful when one equation already has a variable by itself, or when it is easy to isolate one variable.

Example of a system:

In this system, the first equation already tells us what \(y\) equals. That makes substitution a great choice.

Why substitution works: if \(y = 2x + 1\), then anywhere we see \(y\), we can replace it with \(2x + 1\). Since both expressions are equal, the value of the equation does not change.

Steps for solving by substitution

1. Choose one equation and solve it for one variable if needed.
2. Substitute that expression into the other equation.
3. Solve the new equation for the remaining variable.
4. Substitute back to find the value of the other variable.
5. Check your solution in both original equations.

Let us work through some examples from easier to harder.

Worked Example 1: A variable is already isolated

The first equation already has \(y\) by itself, so we can substitute \(3x - 2\) for \(y\) in the second equation.

Now solve for \(x\).

Next, substitute \(x = 3\) into the first equation to find \(y\).

So the solution is:

Check:

• First equation: \(y = 3x - 2\) becomes \(7 = 3(3) - 2 = 7\) ✓
• Second equation: \(x + y = 10\) becomes \(3 + 7 = 10\) ✓

The point \((3, 7)\) makes both equations true.

Worked Example 2: Solve one equation first

Neither equation has a variable isolated yet. Let us solve the second equation for \(x\).

Add \(y\) to both sides:

Now substitute \(y + 1\) for \(x\) in the first equation.

Distribute and combine like terms.

Now substitute \(y = 3\) into \(x = y + 1\).

The solution is:

Check:

• First equation: \(2(4) + 3 = 8 + 3 = 11\) ✓
• Second equation: \(4 - 3 = 1\) ✓

Worked Example 3: Watch out for distribution

The first equation already isolates \(y\), so substitute \(-2x + 5\) for \(y\) in the second equation.

Now distribute the \(2\) carefully.

Substitute \(x = 6\) into the first equation.

The solution is:

Check:

• First equation: \(-7 = -2(6) + 5 = -7\) ✓
• Second equation: \(3(6) + 2(-7) = 18 - 14 = 4\) ✓

This example shows why parentheses and distribution are very important in substitution.

Worked Example 4: Fractions can appear

Since the second equation gives \(y\) directly, substitute it into the first equation.

Combine like terms.

Multiply both sides by \(\frac{2}{3}\).

Now find \(y\).

The solution is:

What the solution means

If a system has solution \((a, b)\), then:

• \(x = a\)
• \(y = b\)
• both original equations are true when those values are used

On a graph, \((a, b)\) is where the two lines cross.

Common mistakes to avoid

• Not isolating a variable correctly. If you make an error in this step, the rest of the work will be wrong.
• Forgetting parentheses. When substituting an expression like \(-2x + 5\), write it as \((-2x + 5)\).
• Forgetting to distribute. For example, \(2(-2x + 5) = -4x + 10\), not \(-2x + 5\).
• Stopping too early. After finding one variable, you still need to find the other.
• Not checking the answer. Always test your solution in both original equations.

When substitution is a good method

• One equation already has \(x\) or \(y\) isolated.
• It is easy to solve one equation for a variable.
• You want an algebraic method instead of graphing.

Quick practice thinking

For the system

You would replace \(y\) in the second equation with \(x + 4\):

Then solve for \(x\), and substitute back to find \(y\).

Summary

Substitution is a way to solve a system by turning two equations into one. You do this by solving one equation for a variable and replacing that variable in the other equation.

After solving for one variable, substitute back to find the other. Then check your answer in both original equations to make sure your solution is correct.

Solving via Elimination is a method for solving a system of linear equations. A system of linear equations is a set of two equations with the same variables, such as \(x\) 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. On a graph, this solution is the point where the two lines intersect.

The elimination method works by adding or subtracting the equations so that one variable disappears. Then we can solve for the remaining variable, and finally substitute back to find the other one.

This method is especially useful when the equations are already lined up neatly, or when it is easy to make the coefficients of one variable opposites.

Important idea: We can multiply an equation by a number without changing its solutions, as long as we multiply every term on both sides of the equation.

For example, if we start with

$$x + y = 5$$

and multiply every term by \(2\), we get

$$2x + 2y = 10$$

These two equations are equivalent because they represent the same line.

Why elimination works: If one equation has \(+3y\) and the other has \(-3y\), then adding the equations makes the \(y\)-terms cancel:

$$3y + (-3y) = 0$$

That leaves an equation with just one variable, which is much easier to solve.

Steps for solving by elimination:

1. Write both equations in standard form if possible: \(Ax + By = C\).
2. Line up like terms so the \(x\)-terms are above each other and the \(y\)-terms are above each other.
3. Choose a variable to eliminate.
4. Multiply one or both equations if needed so the coefficients of that variable are opposites.
5. Add or subtract the equations to eliminate the chosen variable.
6. Solve the resulting one-variable equation.
7. Substitute the value back into one of the original equations to find the other variable.
8. Check your solution in both original equations.

Tip: If the coefficients are already opposites, you can add the equations right away. If the coefficients are the same, you can subtract one equation from the other.

Example 1: Elimination without multiplying first

Solve the system:

$$x + y = 7$$

$$x - y = 1$$

The \(y\)-coefficients are \(+1\) and \(-1\), so the \(y\)-terms will cancel if we add the equations.

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

Add the equations:

$$2x = 8$$

Solve for \(x\):

$$x = 4$$

Now substitute \(x=4\) into one of the original equations. Use \(x+y=7\):

$$4 + y = 7$$

$$y = 3$$

So the solution is

$$\boxed{(4,3)}$$

Check:

In \(x+y=7\): \(4+3=7\) ✓

In \(x-y=1\): \(4-3=1\) ✓

Example 2: Multiply one equation first

Solve the system:

$$2x + y = 9$$

$$x - y = 3$$

Notice the \(y\)-coefficients are \(+1\) and \(-1\). They are already opposites, so we can add the equations.

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

Add the equations:

$$3x = 12$$

$$x = 4$$

Substitute into \(x-y=3\):

$$4 - y = 3$$

$$-y = -1$$

$$y = 1$$

So the solution is

$$\boxed{(4,1)}$$

Check:

In \(2x+y=9\): \(2(4)+1=9\) ✓

In \(x-y=3\): \(4-1=3\) ✓

Example 3: Multiply to create opposite coefficients

Solve the system:

$$2x + 3y = 13$$

$$x + y = 5$$

The coefficients of \(x\) are \(2\) and \(1\), and the coefficients of \(y\) are \(3\) and \(1\). Nothing will cancel right away, so we need to multiply one equation.

Let’s eliminate \(x\). If we multiply the second equation by \(-2\), its \(x\)-coefficient becomes \(-2\), which is the opposite of \(+2\).

Start with:

$$x + y = 5$$

Multiply every term by \(-2\):

$$-2x - 2y = -10$$

Now combine with the first equation:

$$\begin{aligned} 2x + 3y &= 13\\ -2x - 2y &= -10 \end{aligned}$$

Add the equations:

$$y = 3$$

Substitute \(y=3\) into \(x+y=5\):

$$x + 3 = 5$$

$$x = 2$$

So the solution is

$$\boxed{(2,3)}$$

Check:

In \(2x+3y=13\): \(2(2)+3(3)=4+9=13\) ✓

In \(x+y=5\): \(2+3=5\) ✓

Example 4: Multiply both equations

Solve the system:

$$3x + 2y = 16$$

$$2x - 5y = -1$$

This time, neither variable can be eliminated right away. We will create opposite coefficients.

Let’s eliminate \(x\). The least common multiple of \(3\) and \(2\) is \(6\).

Multiply the first equation by \(2\):

$$2(3x + 2y = 16)$$

$$6x + 4y = 32$$

Multiply the second equation by \(-3\):

$$-3(2x - 5y = -1)$$

$$-6x + 15y = 3$$

Now add the equations:

$$\begin{aligned} 6x + 4y &= 32\\ -6x + 15y &= 3 \end{aligned}$$

$$19y = 35$$

$$y = \frac{35}{19}$$

Now substitute into one of the original equations. Use \(3x+2y=16\):

$$3x + 2\left(\frac{35}{19}\right) = 16$$

$$3x + \frac{70}{19} = 16$$

Write \(16\) as a fraction with denominator \(19\):

$$16 = \frac{304}{19}$$

So:

$$3x = \frac{304}{19} - \frac{70}{19}$$

$$3x = \frac{234}{19}$$

$$x = \frac{234}{57} = \frac{78}{19}$$

So the solution is

$$\boxed{\left(\frac{78}{19},\frac{35}{19}\right)}$$

This example shows that solutions are not always whole numbers. Elimination still works the same way.

Common mistakes to avoid:

• Forgetting to multiply every term in an equation. If you multiply one term, you must multiply all terms on both sides.
• Adding incorrectly with negative numbers. Be careful with signs like \(-y\), \(-2x\), or \(-5y\).
• Stopping too early. After finding one variable, you still need to substitute back to find the other one.
• Not checking the answer. A quick check can catch small mistakes.

How to decide which variable to eliminate:

• Look for coefficients that are already opposites.
• If not, choose the variable that will need the least multiplying.
• Try to make the arithmetic as simple as possible.

Connection to graphs: Each equation in a system represents a line. Solving by elimination finds the point where the two lines cross. Even though elimination is an algebra method, it gives the same answer you would get from graphing.

Practice thinking: Before doing any calculations, ask yourself:

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

Summary: Elimination solves a system by making one variable disappear. Then you solve the simpler one-variable equation, substitute back, and check your answer. The main skill is choosing how to multiply equations so that one pair of coefficients becomes opposites.

Consistent vs. Inconsistent Systems

When you solve a system of linear equations, you are looking for a point that makes both equations true at the same time. On a graph, that means you are looking for a point where the two lines meet.

Some systems have one solution. Some have no solution. Some have infinitely many solutions. Learning to tell the difference helps you classify a system as consistent or inconsistent.

In this lesson, you will learn what these words mean, how they look on a graph, and how to recognize them from the equations.

1. What is a system?

A system of linear equations is a set of two or more linear equations that are considered together. In 9th Grade, you will usually work with two equations and two variables, such as

The solution is the ordered pair \((x,y)\) that satisfies both equations.

2. What does consistent mean?

A system is consistent if it has at least one solution.

That means the lines either:

• intersect at exactly one point, or
• lie on top of each other, giving infinitely many solutions.

So there are two kinds of consistent systems:

• One-solution system: the lines cross once.
• Infinitely-many-solutions system: the lines are the same line.

3. What does inconsistent mean?

A system is inconsistent if it has no solution.

This happens when the lines are parallel. Parallel lines never meet, so there is no point that makes both equations true.

4. The three possible results

For two linear equations, there are only three possible situations:

1. One solution → the lines intersect once → consistent
2. No solution → the lines are parallel → inconsistent
3. Infinitely many solutions → the lines are identical → consistent

5. How to recognize each type from slope and intercept

Many linear equations can be written in slope-intercept form:

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

You can compare the slopes and intercepts of the two lines.

• If the slopes are different, the lines intersect once. The system is consistent.
• If the slopes are the same but the intercepts are different, the lines are parallel. The system is inconsistent.
• If the slopes are the same and the intercepts are the same, the equations represent the same line. The system is consistent with infinitely many solutions.

6. How to recognize each type by solving

You can also classify a system by using substitution or elimination.

When you solve:

• If you get one ordered pair, the system is consistent.
• If you get a false statement like \(3 = 7\), the system is inconsistent.
• If you get a true statement like \(0 = 0\), the equations are the same line, so the system is consistent with infinitely many solutions.

7. Worked Example 1: One solution

Classify the system:

Step 1: Compare the slopes.

The first line has slope \(1\). The second line has slope \(-1\).

Since the slopes are different, the lines intersect once.

Conclusion: The system is consistent and has one solution.

We can also find the solution by setting the right sides equal:

Substitute back:

So the solution is \((2,4)\).

8. Worked Example 2: No solution

Classify the system:

Step 1: Compare the slopes and intercepts.

Both lines have slope \(3\), so they rise at the same rate.

The intercepts are different: one is \(-1\) and the other is \(5\).

That means the lines are parallel and never meet.

Conclusion: The system is inconsistent and has no solution.

Let us check by solving:

This is false, so there is no solution.

9. Worked Example 3: Infinitely many solutions

Classify the system:

Step 1: Check whether one equation is a multiple of the other.

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

This is exactly the second equation.

So both equations describe the same line.

Conclusion: The system is consistent and has infinitely many solutions.

We can also see this with elimination. Subtract the first equation multiplied by \(2\) from the second:

Because \(0 = 0\) is always true, the equations are identical.

10. Worked Example 4: Classify from standard form

Classify the system:

Step 1: Rewrite each equation in slope-intercept form.

First equation:

Second equation:

Step 2: Compare slopes and intercepts.

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

So the lines are parallel.

Conclusion: The system is inconsistent and has no solution.

11. Quick visual guide

• Intersecting lines → one crossing point → consistent
• Parallel lines → no crossing point → inconsistent
• Same line → every point on the line works → consistent

12. Common mistakes to avoid

• Mistake: Thinking “consistent” means exactly one solution.
  Remember: Consistent means at least one solution, so it can be one solution or infinitely many.
• Mistake: Looking only at the slopes.
  Remember: If slopes are equal, you must also compare the intercepts.
• Mistake: Seeing \(0 = 0\) and thinking there is no solution.
  Remember: \(0 = 0\) means the equations are the same line, so there are infinitely many solutions.
• Mistake: Seeing a false statement like \(2 = -3\) and continuing to solve.
  Remember: A false statement means no solution.

13. Steps for classifying a system

1. Write both equations in a form you can compare, often \(y = mx + b\).
2. Compare the slopes.
3. If the slopes are different, the system is consistent with one solution.
4. If the slopes are the same, compare the intercepts.
5. If the intercepts are different, the system is inconsistent with no solution.
6. If the intercepts are the same, the system is consistent with infinitely many solutions.

14. Summary

A system of linear equations is consistent if there is at least one solution, and inconsistent if there is no solution.

For two lines, there are three possibilities: one intersection, no intersection, or the same line. Different slopes give one solution. Equal slopes with different intercepts give no solution. Equal slopes with the same intercept give infinitely many solutions.

If you remember how lines behave on a graph, it becomes much easier to classify any system correctly.

Modeling with Linear Systems means turning a real-world situation into two linear equations with the same two variables, then solving the system to find values that make both equations true at the same time.

This is useful when two different facts describe the same situation. For example, you might know the total number of items and the total cost, or the total distance and the total time. Each fact gives one equation. Together, they form a system of linear equations.

In this lesson, you will learn how to:

• choose variables for an unknown situation,
• write two linear equations from a word problem,
• solve the system, and
• check whether the answer makes sense in the real-world context.

1. What is a linear system?

A linear equation in two variables has a form like:

$$ax + by = c$$

where the variables are to the first power only.

A system of linear equations is a set of two or more linear equations that use the same variables. In 9th Grade, we usually work with two equations and two variables.

For example:

$$\begin{cases} x + y = 10 \\ 2x + y = 16 \end{cases}$$

The solution to a system is the pair of numbers that makes both equations true.

On a graph, each equation is a line. The solution is the point where the two lines intersect.

2. How to model a situation with a linear system

When you read a word problem, use these steps:

1. Choose variables. Decide what each variable represents.
2. Find two relationships. Look for two facts in the problem that connect the variables.
3. Write two equations. Turn each relationship into a linear equation.
4. Solve the system. Use substitution, elimination, or graphing.
5. Interpret the answer. State what the numbers mean in the context of the problem.
6. Check. Substitute your values back into both equations and see if they fit the situation.

3. Clues that a situation can be modeled by a system

• There are two unknowns.
• You are given two different pieces of information about those unknowns.
• The relationships are linear, meaning they involve adding, subtracting, or multiplying by constants, but not powers like \(x^2\).

Common types of problems include:

• Mixture problems involving amounts and totals,
• Distance-rate-time problems, and
• Cost or value problems involving quantities and prices.

4. Important modeling ideas

When writing equations, pay close attention to the meaning of each number in the problem.

• If a problem gives a total number of items, that often becomes an equation like \(x + y = \text{total}\).
• If a problem gives a total cost, multiply each quantity by its price and add them.
• If a problem gives a distance, use the idea:

$$\text{distance} = \text{rate} \times \text{time}$$

Also remember that your answer should make sense. For example, the number of tickets sold cannot be negative, and a time value usually should not be negative either.

Worked Example 1: Tickets at a school event

A school play sold 30 tickets. Student tickets cost \(\$4\) each, and adult tickets cost \(\$7\) each. The total money collected was \(\$171\). How many student tickets and adult tickets were sold?

Step 1: Choose 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 money is 171 dollars:

$$4s + 7a = 171$$

Step 3: Solve the system

Use substitution from the first equation:

$$s = 30 - a$$

Substitute into the second equation:

$$4(30 - a) + 7a = 171$$

$$120 - 4a + 7a = 171$$

$$120 + 3a = 171$$

$$3a = 51$$

$$a = 17$$

Now find \(s\):

$$s = 30 - 17 = 13$$

Step 4: Interpret the answer

There were 13 student tickets and 17 adult tickets sold.

Step 5: Check

Number of tickets: \(13 + 17 = 30\)

Money: \(4(13) + 7(17) = 52 + 119 = 171\)

The answer is correct.

Worked Example 2: Mixture problem

A science class needs 20 liters of a salt solution. They will mix a 10% salt solution with a 25% salt solution to get a mixture that is 16% salt. How many liters of each solution are needed?

Step 1: Choose variables

Let \(x\) = liters of 10% solution.

Let \(y\) = liters of 25% solution.

Step 2: Write the equations

The total amount of solution is 20 liters:

$$x + y = 20$$

Now write an equation for the amount of pure salt.

The 10% solution contributes \(0.10x\) liters of salt.

The 25% solution contributes \(0.25y\) liters of salt.

The final mixture is 16% of 20 liters, so the amount of salt is:

$$0.16(20) = 3.2$$

So the second equation is:

$$0.10x + 0.25y = 3.2$$

Step 3: Solve the system

From the first equation:

$$x = 20 - y$$

Substitute into the second equation:

$$0.10(20 - y) + 0.25y = 3.2$$

$$2 - 0.10y + 0.25y = 3.2$$

$$2 + 0.15y = 3.2$$

$$0.15y = 1.2$$

$$y = 8$$

Then:

$$x = 20 - 8 = 12$$

Step 4: Interpret the answer

The class needs 12 liters of the 10% solution and 8 liters of the 25% solution.

Step 5: Check

Total liters: \(12 + 8 = 20\)

Salt amount: \(0.10(12) + 0.25(8) = 1.2 + 2 = 3.2\)

And \(3.2\) is 16% of 20, so the answer makes sense.

Worked Example 3: Distance-rate-time

Two cyclists start at the same place and travel in opposite directions. After 3 hours, they are 66 miles apart. One cyclist rides at 8 miles per hour faster than the other. What are their speeds?

Step 1: Choose variables

Let \(x\) = speed of the slower cyclist in miles per hour.

Let \(y\) = speed of the faster cyclist in miles per hour.

Step 2: Write the equations

The faster cyclist is 8 mph faster:

$$y = x + 8$$

Distance is rate times time. Since they ride in opposite directions, their distances add.

After 3 hours, the total distance is 66 miles:

$$3x + 3y = 66$$

You can simplify by dividing by 3:

$$x + y = 22$$

Step 3: Solve the system

Substitute \(y = x + 8\) into \(x + y = 22\):

$$x + (x + 8) = 22$$

$$2x + 8 = 22$$

$$2x = 14$$

$$x = 7$$

Then:

$$y = 7 + 8 = 15$$

Step 4: Interpret the answer

The slower cyclist rides at 7 mph, and the faster cyclist rides at 15 mph.

Step 5: Check

In 3 hours, the slower cyclist travels \(3 \cdot 7 = 21\) miles.

The faster cyclist travels \(3 \cdot 15 = 45\) miles.

Together they are \(21 + 45 = 66\) miles apart, so the answer is correct.

Worked Example 4: Using elimination in a cost problem

A bookstore sold notebooks for \(\$3\) each and folders for \(\$2\) each. In one day, 40 items were sold for a total of \(\$94\). How many notebooks and folders were sold?

Step 1: Choose variables

Let \(n\) = number of notebooks.

Let \(f\) = number of folders.

Step 2: Write the equations

Total items:

$$n + f = 40$$

Total cost:

$$3n + 2f = 94$$

Step 3: Solve using elimination

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

$$-2n - 2f = -80$$

Add this to the second equation:

$$3n + 2f = 94$$

$$-2n - 2f = -80$$

$$n = 14$$

Now substitute into \(n + f = 40\):

$$14 + f = 40$$

$$f = 26$$

Step 4: Interpret the answer

The bookstore sold 14 notebooks and 26 folders.

Step 5: Check

Items: \(14 + 26 = 40\)

Cost: \(3(14) + 2(26) = 42 + 52 = 94\)

The solution works.

5. How graphing connects to modeling

Every system can also be viewed on a graph. Each equation forms a line, and the point where they cross is the solution.

For example, in Example 1, the equations were:

$$s + a = 30$$

$$4s + 7a = 171$$

If you graph these lines, they intersect at \((13, 17)\). This means \(s = 13\) and \(a = 17\).

Graphing is useful because it shows the meaning of the solution visually. However, in many word problems, substitution or elimination gives a more exact answer more quickly.

6. Common mistakes to avoid

• Mixing up the variables. Write clearly what each variable means before starting.
• Using the wrong total. Make sure you know whether a number represents total items, total cost, total distance, or total amount of salt.
• Forgetting to multiply. In cost problems, use price times quantity. In distance problems, use rate times time.
• Not checking the answer. A correct calculation can still be the wrong model if the equations were written incorrectly.
• Ignoring reasonableness. Negative answers or impossible amounts usually mean something went wrong.

7. Quick strategy guide

• If the problem talks about two kinds of objects and gives a total number and total cost, use one equation for the count and one for the cost.
• If the problem is about mixing two solutions, use one equation for total amount and one for total pure ingredient.
• If the problem is about travel, use \(\text{distance} = \text{rate} \times \text{time}\).
• If one quantity is described in terms of another, such as “8 more than” or “5 less than,” write an equation like \(y = x + 8\) or \(y = x - 5\).

8. Practice thinking: setting up equations

Before solving, practice identifying the equations.

Situation A: A farmer has chickens and goats. There are 18 animals total and 50 legs total.

• Let \(c\) = chickens, \(g\) = goats
• Total animals: \(c + g = 18\)
• Total legs: \(2c + 4g = 50\)

Situation B: A student buys pens for \(\$1.50\) each and markers for \(\$2.50\) each. She buys 12 items for \(\$24\).

• Let \(p\) = pens, \(m\) = markers
• Total items: \(p + m = 12\)
• Total cost: \(1.5p + 2.5m = 24\)

Situation C: A car travels for 2 hours at one speed and 3 hours at another speed, covering 290 miles total. The second speed is 10 mph faster than the first.

• Let \(x\) = first speed, \(y\) = second speed
• Speed relationship: \(y = x + 10\)
• Total distance: \(2x + 3y = 290\)

9. Summary

Modeling with linear systems is about translating a real situation into two equations with two variables. The key is to identify the unknowns and the two facts that connect them.

After writing the equations, solve the system using substitution, elimination, or graphing. Then always check your answer in the original context to make sure it is reasonable and correct.

When you get better at modeling, word problems become easier because you learn to organize the information clearly. If you can define variables carefully and write accurate equations, you are already doing the hardest part well.

Systems of Linear Inequalities help us describe situations where more than one rule must be true at the same time.

A linear inequality compares two expressions using symbols like \(<\), \(>\), \(\le\), or \(\ge\). For example, \(y > 2x - 1\) is a linear inequality.

A system of linear inequalities is a set of two or more linear inequalities. To solve the system, we find all points that make every inequality true. On a graph, this means finding the overlapping shaded region.

This overlapping region is sometimes called the solution region or feasible region.

Why this matters: In real life, we often have several limits at once. For example, a budget must stay below a certain amount, while time spent must stay above a minimum. Systems of inequalities help model these kinds of situations.

1. Review: What does one linear inequality look like on a graph?

If you graph a linear equation such as \(y = x + 2\), you get a line. But if you graph an inequality such as \(y > x + 2\), you shade one side of that line.

• If the inequality is \(y >\) or \(y <\), use a dashed line because points on the line are not included.
• If the inequality is \(y \ge\) or \(y \le\), use a solid line because points on the line are included.

To decide which side to shade:

1. Graph the boundary line.
2. Pick a test point, usually \((0,0)\) if it is not on the line.
3. Substitute the test point into the inequality.
4. If the inequality is true, shade the side containing the test point. If it is false, shade the other side.

2. How to solve a system of linear inequalities

Here is the basic process:

1. Graph each inequality one at a time.
2. Use dashed or solid boundary lines correctly.
3. Shade the correct half-plane for each inequality.
4. Find the area where all the shaded parts overlap.

The overlap is the set of all solutions.

3. Important idea: “And” means overlap

In a system, the inequalities are connected by the idea of and. A point must satisfy the first inequality and the second inequality and any others in the system.

So even if a point works for one inequality, it is not a solution unless it works for all of them.

4. Writing inequalities in graphing form

The easiest form to graph is slope-intercept form:

For inequalities, this becomes something like:

If the inequality is not already solved for \(y\), rewrite it first.

For example:

Solve for \(y\):

Now it is easier to graph.

Worked Example 1: A basic system

Graph and solve the system:

Step 1: Graph \(y > x - 1\)

The boundary is the line:

Because the inequality is \(>\), use a dashed line.

Shade above the line, because \(y\) is greater than \(x-1\).

Step 2: Graph \(y \le 3\)

The boundary is the horizontal line:

Because the inequality is \(\le\), use a solid line.

Shade on or below the line.

Step 3: Find the overlap

The solution is all points that are:

• above the dashed line \(y = x - 1\), and
• on or below the solid line \(y = 3\)

So the answer is the overlapping region between those two boundaries.

Check a point: Is \((1,2)\) a solution?

Test in both inequalities:

Since both are true, \((1,2)\) is in the solution region.

Worked Example 2: Rewriting before graphing

Graph and solve the system:

Step 1: Rewrite the first inequality

Now the system is:

Step 2: Graph \(y \ge -2x + 4\)

The boundary line is solid because of \(\ge\).

Shade above or on the line.

Step 3: Graph \(y < -x + 5\)

The boundary line is dashed because of \(<\).

Shade below the line.

Step 4: Find the overlap

The solution is the region that is above the first line and below the second line.

Check a point: Is \((1,3)\) a solution?

Test the first inequality:

Test the second inequality:

Both are true, so \((1,3)\) is a solution.

Worked Example 3: A system with three inequalities

Graph and solve the system:

This system includes the x-axis, the y-axis, and a diagonal line.

Step 1: Graph \(y \ge 0\)

This means all points on or above the x-axis.

Step 2: Graph \(x \ge 0\)

This means all points on or to the right of the y-axis.

Step 3: Graph \(y \le -x + 4\)

The boundary line is solid because of \(\le\).

Shade on or below the line.

Step 4: Find the overlap

The solution region is the triangular area in the first quadrant that lies under the line \(y = -x + 4\).

The corner points are:

• \((0,0)\)
• \((4,0)\)
• \((0,4)\)

These are the vertices of the feasible region.

Why corner points matter: When a system forms a closed region, the corner points show exactly where the boundary lines meet.

Worked Example 4: No overlapping region

Graph and solve the system:

Step 1: Graph the boundary lines

Both inequalities use lines with slope \(1\):

• \(y = x + 2\)
• \(y = x - 1\)

Both are dashed because the inequalities are strict.

Step 2: Shade the correct sides

• \(y > x + 2\): shade above the higher line
• \(y < x - 1\): shade below the lower line

Step 3: Look for overlap

There is no region that is both above \(y = x + 2\) and below \(y = x - 1\).

So this system has no solution.

5. How to test whether a point is a solution

If you are given a point and asked whether it solves a system, substitute it into every inequality.

For example, test \((2,1)\) in:

First inequality:

Second inequality:

Since both are true, \((2,1)\) is a solution.

6. Common mistakes to avoid

• Using the wrong line style: Dashed for \(<\) or \(>\), solid for \(\le\) or \(\ge\).
• Shading the wrong side: Use a test point if you are unsure.
• Forgetting to rewrite the inequality: If it is not solved for \(y\), graphing is harder.
• Thinking one shaded region is enough: The solution must satisfy all inequalities.
• Not checking boundary lines: A solid line means boundary points count; a dashed line means they do not.

7. Quick tips for graphing systems of inequalities

• Graph neatly and label each line if possible.
• Shade lightly at first so overlap is easy to see.
• If the overlap is small, outline it clearly.
• Use test points when the direction of shading is not obvious.
• Remember: the answer is a region, not usually a single point.

8. What if the system forms a closed shape?

Sometimes the overlapping region is enclosed, like a triangle or quadrilateral. This is called a bounded region.

Other times the region continues forever in one or more directions. This is an unbounded region.

Both kinds can still be correct solution regions.

Summary

A system of linear inequalities is solved by graphing each inequality and finding the overlapping shaded region.

Use dashed lines for strict inequalities \((<, >)\) and solid lines for inclusive inequalities \((\le, \ge)\).

The solution is all points that make every inequality true. You can check a point by substituting it into each inequality.

When you graph carefully and shade correctly, the solution region becomes easy to see.

Linear Programming Fundamentals is about finding the largest or smallest value of a quantity, while following a set of rules called constraints.

These constraints are usually written as linear inequalities, and when we graph them together, they form a region of possible answers called the feasible region.

In 9th Grade maths, the main goal is to learn how to:

• Graph linear inequalities
• Identify the feasible region
• Find the corner points, also called vertices
• Test those vertices in an objective function
• Decide which vertex gives the maximum or minimum value

This is important because many real-life problems involve making the best choice with limited resources. For example, a business may want to maximize profit or minimize cost.

1. Key Vocabulary

• Constraint: a rule written as a linear inequality, such as \(x + y \le 10\)
• Feasible region: the area where all constraints overlap
• Objective function: the expression you want to maximize or minimize, such as \(P = 3x + 2y\)
• Vertex: a corner point of the feasible region
• Maximum: the greatest value
• Minimum: the smallest value

2. Why Do We Check the Vertices?

A major idea in linear programming is this:

The maximum or minimum value of the objective function will occur at a vertex of the feasible region.

This means we do not need to test every point in the shaded region. We only need to find the corner points and evaluate the objective function there.

3. General Steps for Solving a Linear Programming Problem

1. Write the constraints as inequalities.
2. Graph each inequality on the coordinate plane.
3. Shade the correct side of each boundary line.
4. Find the overlapping shaded area. This is the feasible region.
5. Locate all vertices of the feasible region.
6. Substitute each vertex into the objective function.
7. Choose the greatest value for a maximum problem or the smallest value for a minimum problem.

4. Understanding the Feasible Region

The feasible region is the set of all points that satisfy every constraint at the same time.

For example, if the constraints are:

\(x \ge 0\), \(y \ge 0\), and \(x + y \le 6\)

then the feasible region is the triangular area in the first quadrant that lies on or below the line \(x + y = 6\).

The lines form the edges, and the corner points are where the boundary lines meet.

5. Worked Example 1: Finding a Maximum from a Simple Feasible Region

Problem: Maximize \(P = x + 2y\) subject to:

• \(x \ge 0\)
• \(y \ge 0\)
• \(x + y \le 4\)

Step 1: Graph the constraints.

Because \(x \ge 0\) and \(y \ge 0\), we stay in the first quadrant.

The line \(x + y = 4\) passes through \((4,0)\) and \((0,4)\). Since the inequality is \(x + y \le 4\), we shade below the line.

Step 2: Identify the feasible region.

The shaded region is a triangle with vertices:

• \((0,0)\)
• \((4,0)\)
• \((0,4)\)

Step 3: Test the vertices in the objective function.

Objective function:

$$P = x + 2y$$

• At \((0,0)\): \(P = 0 + 2(0) = 0\)
• At \((4,0)\): \(P = 4 + 2(0) = 4\)
• At \((0,4)\): \(P = 0 + 2(4) = 8\)

Answer: The maximum value is \(8\), and it occurs at \((0,4)\).

6. Worked Example 2: Finding a Minimum Value

Problem: Minimize \(C = 2x + y\) subject to:

• \(x \ge 1\)
• \(y \ge 2\)
• \(x + y \ge 5\)

Step 1: Graph the boundary lines.

• \(x = 1\)
• \(y = 2\)
• \(x + y = 5\)

Since \(x + y \ge 5\), we shade above that line.

Step 2: Find the feasible region.

The feasible region is the area that satisfies all three inequalities.

Step 3: Find the vertices.

We find where the boundary lines intersect:

• \(x = 1\) and \(y = 2\) gives \((1,2)\), but check: \(1 + 2 = 3\), which does not satisfy \(x + y \ge 5\), so this point is not in the feasible region.
• \(x = 1\) and \(x + y = 5\): then \(1 + y = 5\), so \(y = 4\). Vertex: \((1,4)\)
• \(y = 2\) and \(x + y = 5\): then \(x + 2 = 5\), so \(x = 3\). Vertex: \((3,2)\)

Step 4: Evaluate the objective function.

$$C = 2x + y$$

• At \((1,4)\): \(C = 2(1) + 4 = 6\)
• At \((3,2)\): \(C = 2(3) + 2 = 8\)

Answer: The minimum value is \(6\), and it occurs at \((1,4)\).

7. How to Find Vertices Algebraically

Sometimes it is not enough to just look at the graph. You may need to calculate the exact intersection points of the boundary lines.

To do this, solve the two equations together as a system of linear equations.

For example, to find the intersection of:

\(x + y = 6\)

and

\(2x + y = 8\)

subtract the first equation from the second:

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

$$ x = 2 $$

Now substitute into \(x + y = 6\):

$$2 + y = 6$$

$$y = 4$$

So the lines intersect at \((2,4)\).

This point could be a vertex of the feasible region if it satisfies all the constraints.

8. Worked Example 3: Using Intersections to Find the Best Value

Problem: Maximize \(Z = 3x + 2y\) subject to:

• \(x \ge 0\)
• \(y \ge 0\)
• \(x + y \le 6\)
• \(2x + y \le 8\)

Step 1: Find the vertices of the feasible region.

Since \(x \ge 0\) and \(y \ge 0\), the region is in the first quadrant.

Now find important intersection points:

• With the axes, \(2x + y = 8\) has intercepts \((4,0)\) and \((0,8)\)
• \(x + y = 6\) has intercepts \((6,0)\) and \((0,6)\)
• The two lines intersect where:

$$x + y = 6$$

$$2x + y = 8$$

Subtract the first from the second:

$$x = 2$$

Then:

$$2 + y = 6$$

$$y = 4$$

So the lines meet at \((2,4)\).

Step 2: List the vertices of the feasible region.

The feasible region has vertices:

• \((0,0)\)
• \((4,0)\)
• \((2,4)\)
• \((0,6)\)

Step 3: Evaluate the objective function.

$$Z = 3x + 2y$$

• At \((0,0)\): \(Z = 0\)
• At \((4,0)\): \(Z = 12\)
• At \((2,4)\): \(Z = 3(2) + 2(4) = 6 + 8 = 14\)
• At \((0,6)\): \(Z = 12\)

Answer: The maximum value is \(14\), and it occurs at \((2,4)\).

9. Worked Example 4: A Word Problem

Problem: A school club is making bracelets and keychains to raise money.

Let:

• \(x\) = number of bracelets
• \(y\) = number of keychains

They must follow these rules:

• They can make at most 10 total items: \(x + y \le 10\)
• They can make at most 6 bracelets: \(x \le 6\)
• They cannot make negative items: \(x \ge 0\), \(y \ge 0\)

Each bracelet earns \(\$4\), and each keychain earns \(\$3\).

What is the maximum money they can earn?

Step 1: Write the objective function.

$$M = 4x + 3y$$

Step 2: Find the vertices of the feasible region.

The constraints are:

• \(x + y \le 10\)
• \(x \le 6\)
• \(x \ge 0\)
• \(y \ge 0\)

The corner points are:

• \((0,0)\)
• \((6,0)\)
• \((6,4)\) because if \(x = 6\), then \(6 + y = 10\), so \(y = 4\)
• \((0,10)\)

Step 3: Evaluate the objective function.

• At \((0,0)\): \(M = 0\)
• At \((6,0)\): \(M = 4(6) + 3(0) = 24\)
• At \((6,4)\): \(M = 4(6) + 3(4) = 24 + 12 = 36\)
• At \((0,10)\): \(M = 4(0) + 3(10) = 30\)

Answer: The maximum money is \(\$36\), when the club makes 6 bracelets and 4 keychains.

10. Common Mistakes to Avoid

• Forgetting to check all constraints: A point may be an intersection, but it must satisfy every inequality.
• Using the wrong shaded side: Always test a point like \((0,0)\) if it is not on the line.
• Missing a vertex: Check all corners of the feasible region carefully.
• Testing points inside the region instead of vertices: In linear programming, the best value occurs at a vertex.
• Mixing up maximum and minimum: Read the question carefully.

11. Quick Strategy Checklist

• Write the inequalities clearly.
• Graph each boundary line.
• Shade the correct side.
• Find the feasible region.
• List every vertex.
• Substitute each vertex into the objective function.
• Choose the largest or smallest value asked for.

12. Brief Summary

Linear programming helps us find the best possible value of an objective function, such as maximum profit or minimum cost, while following given constraints.

After graphing the inequalities, we look at the feasible region and find its vertices. Then we test the objective function at each vertex. The greatest or smallest result at those vertices gives the answer.