Linear Algebra in Two Variables

Linear equations describe straight lines on a coordinate plane. In Grade 11, it is important to recognize a line in different forms, understand what each form tells you, and know how to convert from one form to another.

This lesson focuses on the three most common forms of a linear equation in two variables:

• Slope-intercept form: \(y = mx + b\)
• Point-slope form: \(y - y_1 = m(x - x_1)\)
• Standard form: \(Ax + By = C\)

Each form is useful in a different situation. By the end of this lesson, you should be able to identify each form, write equations of lines from given information, and convert between forms correctly.

1. What makes an equation linear?

A linear equation in two variables is an equation that can be written so that the variables \(x\) and \(y\) both have degree 1. Its graph is always a straight line.

Examples of linear equations include:

• \(y = 2x + 5\)
• \(y - 3 = -4(x + 1)\)
• \(3x - 2y = 8\)

Non-examples include:

• \(y = x^2 + 1\) because of the squared term
• \(xy = 6\) because the variables are multiplied together

2. Slope and what it means

The slope of a line measures its steepness and direction. It tells you how much \(y\) changes when \(x\) increases by 1.

The slope formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Interpretation of slope:

• If \(m > 0\), the line rises from left to right.
• If \(m < 0\), the line falls from left to right.
• If \(m = 0\), the line is horizontal.
• If the denominator is 0, the slope is undefined, and the line is vertical.

3. Slope-intercept form

Slope-intercept form is:

$$y = mx + b$$

In this form:

• \(m\) is the slope
• \(b\) is the y-intercept, the value of \(y\) when \(x = 0\)

This form is especially useful when you want to graph a line quickly, because you can start at \((0,b)\) and use the slope to find more points.

For example, in \(y = 3x - 2\):

• Slope \(m = 3\)
• y-intercept \(b = -2\)

The line crosses the y-axis at \((0,-2)\) and rises 3 units for every 1 unit it moves right.

4. Point-slope form

Point-slope form is:

$$y - y_1 = m(x - x_1)$$

This form is useful when you know:

• the slope \(m\), and
• one point \((x_1, y_1)\) on the line

It comes directly from the idea of slope. If a line has slope \(m\) and passes through \((x_1, y_1)\), then any point \((x,y)\) on the line satisfies this equation.

For example, a line with slope 2 passing through \((3,4)\) can be written as:

$$y - 4 = 2(x - 3)$$

This is already a correct equation of the line, even if it is not simplified.

5. Standard form

Standard form is:

$$Ax + By = C$$

where \(A\), \(B\), and \(C\) are usually integers, and \(A\) is often taken to be positive.

This form is useful because:

• it shows both variables together in one equation,
• it is convenient for solving systems of equations,
• it makes x-intercepts and y-intercepts easier to find in some cases.

For example:

$$2x + 3y = 12$$

To find intercepts from standard form:

• Set \(x = 0\) to find the y-intercept.
• Set \(y = 0\) to find the x-intercept.

For \(2x + 3y = 12\):

• If \(x = 0\), then \(3y = 12\), so \(y = 4\). The y-intercept is \((0,4)\).
• If \(y = 0\), then \(2x = 12\), so \(x = 6\). The x-intercept is \((6,0)\).

6. How the forms are connected

All three forms can describe the same line. The difference is only how the information is written.

For example, these all represent the same line:

• Point-slope: \(y - 1 = 2(x - 2)\)
• Slope-intercept: \(y = 2x - 3\)
• Standard: \(2x - y = 3\)

Being able to move between forms is an important skill.

7. Converting between forms

From point-slope to slope-intercept

Start with:

$$y - y_1 = m(x - x_1)$$

Then expand and solve for \(y\).

Example:

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

Expand:

$$y - 5 = -3x + 6$$

Add 5 to both sides:

$$y = -3x + 11$$

From slope-intercept to standard form

Start with:

$$y = mx + b$$

Move the variable terms to one side so it looks like \(Ax + By = C\).

Example:

$$y = 4x - 7$$

Subtract \(4x\) from both sides:

$$-4x + y = -7$$

If you want \(A\) positive, multiply by \(-1\):

$$4x - y = 7$$

From standard form to slope-intercept

Solve for \(y\).

Example:

$$3x + 2y = 10$$

Subtract \(3x\):

$$2y = -3x + 10$$

Divide by 2:

$$y = -\frac{3}{2}x + 5$$

Now the slope is \(-\frac{3}{2}\) and the y-intercept is 5.

8. Writing equations from geometric information

You are often asked to write the equation of a line from facts about the line, such as:

• its slope and one point,
• two points,
• parallel or perpendicular relationships.

If you know the slope and one point, point-slope form is usually the fastest method.

If you know two points, first find the slope using the slope formula, then use point-slope form.

If a line is parallel to another line, the slopes are equal.

If a line is perpendicular to another line, the slope is the negative reciprocal.

For example:

• If one line has slope \(\frac{3}{4}\), a parallel line also has slope \(\frac{3}{4}\).
• A perpendicular line has slope \(-\frac{4}{3}\).

Worked Example 1: Write a line in point-slope and slope-intercept form

Find the equation of the line with slope \(5\) passing through \((2,-1)\).

Step 1: Use point-slope form.

$$y - y_1 = m(x - x_1)$$

Substitute \(m = 5\), \((x_1,y_1) = (2,-1)\):

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

This is the point-slope form.

Step 2: Convert to slope-intercept form.

$$y + 1 = 5x - 10$$ $$y = 5x - 11$$

So the equation is:

• Point-slope: \(y + 1 = 5(x - 2)\)
• Slope-intercept: \(y = 5x - 11\)

Worked Example 2: Convert standard form to slope-intercept form

Convert \(4x - 3y = 12\) to slope-intercept form.

Step 1: Isolate the \(y\)-term.

$$-3y = -4x + 12$$

Step 2: Divide by \(-3\).

$$y = \frac{4}{3}x - 4$$

So the slope-intercept form is:

$$y = \frac{4}{3}x - 4$$

This tells us the slope is \(\frac{4}{3}\) and the y-intercept is \(-4\).

Worked Example 3: Find an equation from two points

Find the equation of the line through \((1,3)\) and \((5,11)\).

Step 1: Find the slope.

$$m = \frac{11 - 3}{5 - 1} = \frac{8}{4} = 2$$

Step 2: Use point-slope form with one of the points.

Using \((1,3)\):

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

Step 3: Convert if needed.

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

So the line can be written as:

• Point-slope: \(y - 3 = 2(x - 1)\)
• Slope-intercept: \(y = 2x + 1\)
• Standard: \(2x - y = -1\)

Worked Example 4: Use parallel and perpendicular slopes

Write the equation of the line that passes through \((4,2)\) and is perpendicular to the line

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

Step 1: Identify the given slope.

The given line has slope \(\frac{1}{2}\).

Step 2: Find the perpendicular slope.

The negative reciprocal of \(\frac{1}{2}\) is \(-2\).

Step 3: Use point-slope form.

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

Step 4: Convert to slope-intercept form.

$$y - 2 = -2x + 8$$ $$y = -2x + 10$$

So the equation is:

• Point-slope: \(y - 2 = -2(x - 4)\)
• Slope-intercept: \(y = -2x + 10\)

9. Special cases: horizontal and vertical lines

Not every line can be written in slope-intercept form in the usual way.

Horizontal lines have slope 0. Their equations look like:

$$y = c$$

Example: \(y = 4\)

This means every point on the line has y-coordinate 4.

Vertical lines have undefined slope. Their equations look like:

$$x = c$$

Example: \(x = -2\)

This means every point on the line has x-coordinate \(-2\).

A vertical line cannot be written as \(y = mx + b\), because its slope is undefined.

10. Common mistakes to avoid

• Mixing up slope and intercept: In \(y = mx + b\), \(m\) is the slope and \(b\) is the y-intercept.
• Sign errors in point-slope form: If the point is \((2,-3)\), then \(y - (-3) = y + 3\).
• Forgetting to simplify standard form correctly: Move terms carefully and watch negatives.
• Using the wrong perpendicular slope: Perpendicular does not mean just changing the sign. You must take the negative reciprocal.
• Assuming all lines have a defined slope: Vertical lines do not.

11. Quick strategy guide

When deciding which form to use, ask yourself what information you have.

• If you know slope and y-intercept, use slope-intercept form.
• If you know slope and one point, use point-slope form.
• If you need an equation useful for systems or integer coefficients, use standard form.
• If you know two points, first find slope, then use point-slope form.

12. Brief summary

Linear equations represent straight lines and can be written in slope-intercept form, point-slope form, or standard form. Slope-intercept form shows the slope and y-intercept directly, point-slope form is useful when you know a point and a slope, and standard form is useful for algebraic work such as systems.

To write equations of lines, identify what information is given, find the slope if needed, and choose the form that fits best. With practice, you can convert between forms and describe the same line in multiple correct ways.

Parallel and Perpendicular Dynamics is about understanding how lines behave on the coordinate plane using their slopes. In 11th Grade Maths, this idea helps you analyze shapes, prove geometric facts, and solve problems involving straight lines in two variables.

When you look at two lines, one of the first questions to ask is: are they parallel, perpendicular, or neither? The answer often comes from comparing their slopes.

This lesson will show you how to use slope to identify line relationships, write equations of lines, and prove properties of figures on the coordinate plane.

1. Review: What is slope?

The slope of a line measures its steepness and direction. It tells how much the line rises or falls as you move from left to right.

If a line passes through two points \((x_1, y_1)\) and \((x_2, y_2)\), its slope is

$$m = \frac{y_2-y_1}{x_2-x_1}$$

This is often described as rise over run.

• If \(m > 0\), the line rises from left to right.
• If \(m < 0\), the line falls from left to right.
• If \(m = 0\), the line is horizontal.
• If the denominator is 0, the line is vertical and its slope is undefined.

2. Parallel lines

Two distinct lines are parallel if they never intersect. On the coordinate plane, parallel lines have the same slope.

If two lines have equations

$$y = m_1x + b_1 \quad \text{and} \quad y = m_2x + b_2,$$

then they are parallel when

$$m_1 = m_2$$

To be different lines, they must have different y-intercepts. If both slope and intercept are the same, then the equations describe the same line, not two different parallel lines.

For example, the lines

$$y = 2x + 1 \quad \text{and} \quad y = 2x - 4$$

are parallel because both have slope 2.

Special case: all vertical lines are parallel to each other. Their equations look like \(x = a\), where \(a\) is a constant.

3. Perpendicular lines

Two lines are perpendicular if they meet at a right angle, or \(90^\circ\).

For non-vertical and non-horizontal lines, perpendicular slopes are negative reciprocals of each other.

This means that if one slope is \(m\), then the perpendicular slope is

$$-\frac{1}{m}$$

So if \(m_1\) and \(m_2\) are slopes of perpendicular lines, then

$$m_1m_2 = -1$$

Example slope pairs for perpendicular lines:

• \(2\) and \(-\frac{1}{2}\)
• \(\frac{3}{4}\) and \(-\frac{4}{3}\)
• \(-5\) and \(\frac{1}{5}\)

Special case: a horizontal line and a vertical line are perpendicular. For example, \(y=3\) and \(x=-1\) intersect at a right angle.

4. Why negative reciprocal?

You do not need advanced geometry to use the rule, but it helps to notice the pattern. If one line has slope \(\frac{2}{3}\), it rises 2 units for every 3 units to the right. A perpendicular line must turn to make a right angle, so its movement switches those numbers and changes the direction. That gives slope \(-\frac{3}{2}\).

The reciprocal flips the fraction, and the negative changes the direction.

5. Finding slope from equations

The easiest way to compare lines is to write each equation in slope-intercept form:

$$y = mx + b$$

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

For example:

• In \(y = -3x + 7\), the slope is \(-3\).
• In \(2y = 8x - 6\), first solve for \(y\): \(y = 4x - 3\), so the slope is \(4\).
• In \(3x + 2y = 10\), solve for \(y\): \(2y = -3x + 10\), so \(y = -\frac{3}{2}x + 5\), and the slope is \(-\frac{3}{2}\).

6. Using points to determine whether lines are parallel or perpendicular

If line equations are not given, you can find the slope from points. Then compare the slopes.

Suppose one segment has endpoints \((x_1,y_1)\) and \((x_2,y_2)\), and another has endpoints \((x_3,y_3)\) and \((x_4,y_4)\). Compute both slopes and compare:

• If the slopes are equal, the segments are parallel.
• If the slopes are negative reciprocals, the segments are perpendicular.

7. Using slope to prove properties of shapes

Coordinate geometry lets you prove whether a figure is a rectangle, parallelogram, or other shape by checking slopes.

Some useful ideas are:

• Parallelogram: both pairs of opposite sides are parallel.
• Rectangle: opposite sides are parallel and one angle is a right angle.
• Square: opposite sides are parallel, one angle is a right angle, and all sides have equal length.

In this lesson, we focus on the slope part of the proof:

• Equal slopes show parallel sides.
• Negative reciprocal slopes show right angles.

Worked Example 1: Determine whether two lines are parallel, perpendicular, or neither

Consider the lines

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

Step 1: Identify the slopes.

The first line has slope \(3\). The second line also has slope \(3\).

Step 2: Compare the slopes.

Since the slopes are equal, the lines are parallel.

Answer: The lines are parallel.

Worked Example 2: Check perpendicularity from equations

Determine the relationship between

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

Step 1: Find the slopes.

The slopes are \(\frac{1}{2}\) and \(-2\).

Step 2: Test whether they are negative reciprocals.

The negative reciprocal of \(\frac{1}{2}\) is \(-2\).

Also,

$$\frac{1}{2} \cdot (-2) = -1$$

Answer: The lines are perpendicular.

Worked Example 3: Find the equation of a parallel line through a given point

Write the equation of the line that is parallel to

$$y = -4x + 7$$

and passes through \((2, -1)\).

Step 1: Find the slope of the given line.

The slope is \(-4\).

Step 2: A parallel line has the same slope.

So the new line also has slope \(-4\).

Step 3: Use point-slope form.

$$y - y_1 = m(x - x_1)$$

Substitute \((2,-1)\) and \(m=-4\):

$$y - (-1) = -4(x - 2)$$ $$y + 1 = -4x + 8$$ $$y = -4x + 7$$

This result matches the original line, which means the point \((2,-1)\) lies on that line.

Answer: The equation is

$$y = -4x + 7$$

This example is a good reminder: sometimes a “new” parallel line is actually the same line if the point already lies on it.

Worked Example 4: Prove a quadrilateral is a rectangle using slopes

Let the vertices of quadrilateral \(ABCD\) be

$$A(1,1), \quad B(5,3), \quad C(3,7), \quad D(-1,5)$$

Show that opposite sides are parallel and one angle is a right angle.

Step 1: Find the slope of \(AB\).

$$m_{AB} = \frac{3-1}{5-1} = \frac{2}{4} = \frac{1}{2}$$

Step 2: Find the slope of \(CD\).

$$m_{CD} = \frac{5-7}{-1-3} = \frac{-2}{-4} = \frac{1}{2}$$

So \(AB\) is parallel to \(CD\).

Step 3: Find the slope of \(BC\).

$$m_{BC} = \frac{7-3}{3-5} = \frac{4}{-2} = -2$$

Step 4: Find the slope of \(AD\).

$$m_{AD} = \frac{5-1}{-1-1} = \frac{4}{-2} = -2$$

So \(BC\) is parallel to \(AD\).

Step 5: Check for a right angle.

The slopes of \(AB\) and \(BC\) are \(\frac{1}{2}\) and \(-2\). These are negative reciprocals because

$$\frac{1}{2} \cdot (-2) = -1$$

So \(AB\) is perpendicular to \(BC\).

Conclusion:

• Opposite sides are parallel.
• One angle is a right angle.

Therefore, \(ABCD\) is a rectangle.

8. Common mistakes to avoid

• Mixing up parallel and perpendicular: parallel means same slope; perpendicular means negative reciprocal slopes.
• Forgetting the negative sign: the reciprocal alone is not enough. For example, the reciprocal of \(2\) is \(\frac{1}{2}\), but the perpendicular slope is \(-\frac{1}{2}\).
• Not simplifying slopes: \(\frac{2}{4}\) and \(\frac{1}{2}\) represent the same slope.
• Ignoring vertical and horizontal lines: vertical lines have undefined slope, horizontal lines have slope 0, and they are perpendicular to each other.
• Using y-intercepts instead of slopes: y-intercepts do not determine whether lines are parallel or perpendicular.

9. Quick strategy for solving problems

1. Find the slope of each line.
2. If needed, rewrite equations into \(y=mx+b\) form first.
3. Compare the slopes:

• Same slope \(\rightarrow\) parallel
• Negative reciprocals \(\rightarrow\) perpendicular
• Otherwise \(\rightarrow\) neither

If the problem involves a shape, compute the slopes of its sides and use those results to justify your conclusion.

10. Brief summary

Slope is the key to understanding line relationships on the coordinate plane. Parallel lines have equal slopes, while perpendicular lines have slopes that are negative reciprocals.

By calculating and comparing slopes, you can classify line pairs, write equations of new lines, and prove geometric facts about figures such as parallelograms and rectangles. Always watch for special cases like horizontal and vertical lines, and check your signs carefully.

Algebraic Systems of Equations are sets of two or more equations that use the same variables. A solution to a system is a value, or set of values, that makes every equation in the system true at the same time.

In 11th Grade Maths, systems of equations are important because they connect algebra, graphing, and real-world situations. They help us find where lines intersect, compare relationships, and solve problems with multiple conditions.

In this lesson, you will learn how to solve systems of equations in two and three variables using substitution and elimination. You will also learn how to recognize when a system has one solution, no solution, or infinitely many solutions.

1. What is a system of equations?

A system of equations is a group of equations involving the same variables. For example, this is a system in two variables:

$$ \begin{cases} x + y = 7 \\ 2x - y = 5 \end{cases} $$

We want to find the pair \((x,y)\) that satisfies both equations.

If we graph both equations, the solution is the point where the two lines intersect. Algebraically, we solve to find the exact coordinates.

2. Possible types of solutions

When solving a system, there are three possible outcomes:

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

For systems in three variables, the same ideas apply. The equations may meet at one point, may not all meet together, or may describe the same relationship in more than one way.

3. Solving by substitution

The substitution method works best when one equation is already solved for one variable, or can easily be rearranged. The idea is simple: solve one equation for a variable, then substitute that expression into the other equation.

Steps for substitution:

1. Solve one equation for one variable.
2. Substitute that expression into the other equation.
3. Solve the resulting equation.
4. Substitute back to find the remaining variable.
5. Check your answer in both equations.

Worked Example 1: A basic system in two variables

Solve the system:

$$ \begin{cases} y = 2x + 1 \\ x + y = 10 \end{cases} $$

Step 1: The first equation already gives us \(y\):

$$y = 2x + 1$$

Step 2: Substitute \(2x+1\) for \(y\) in the second equation:

$$x + y = 10$$ $$x + (2x + 1) = 10$$

Step 3: Solve for \(x\):

$$3x + 1 = 10$$ $$3x = 9$$ $$x = 3$$

Step 4: Substitute \(x=3\) into \(y=2x+1\):

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

So the solution is:

$$\boxed{(3,7)}$$

Step 5: Check

$$x+y = 3+7 = 10$$

The answer works in both equations.

4. Solving by elimination

The elimination method works by adding or subtracting equations so that one variable disappears. This is often the fastest method when the coefficients line up nicely.

Steps for elimination:

1. Write the equations in standard form if possible.
2. Choose a variable to eliminate.
3. Multiply one or both equations if needed so the coefficients of that variable are opposites.
4. Add or subtract the equations.
5. Solve for the remaining variable.
6. Substitute back to find the other variable.

Worked Example 2: Elimination in two variables

Solve the system:

$$ \begin{cases} 2x + 3y = 12 \\ 4x - 3y = 6 \end{cases} $$

Notice that the \(y\)-coefficients are already opposites: \(3y\) and \(-3y\). So we can add the equations.

$$ (2x + 3y) + (4x - 3y) = 12 + 6 $$ $$6x = 18$$ $$x = 3$$

Now substitute \(x=3\) into one of the original equations. Use the first equation:

$$2x + 3y = 12$$ $$2(3) + 3y = 12$$ $$6 + 3y = 12$$ $$3y = 6$$ $$y = 2$$

So the solution is:

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

5. Recognizing no solution and infinitely many solutions

Sometimes, solving a system leads to a statement that is either always false or always true.

If you get a false statement such as:

$$0 = 5$$

then the system is inconsistent, which means it has no solution.

If you get a true statement such as:

$$0 = 0$$

then the system is dependent, which means it has infinitely many solutions.

Worked Example 3: Special cases

(a) No solution

Solve the system:

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

Multiply the first equation by \(2\):

$$4x + 2y = 8$$

Now compare with the second equation:

$$4x + 2y = 10$$

The left sides are the same, but the right sides are different. Subtracting gives:

$$0 = 2$$

This is false, so the system has no solution.

Graphically, these are parallel lines.

(b) Infinitely many solutions

Solve the system:

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

Multiply the first equation by \(2\):

$$2x - 4y = 6$$

This is exactly the second equation. So both equations describe the same line.

Therefore, the system has infinitely many solutions.

6. Systems of three variables

A system of three variables usually looks like this:

$$ \begin{cases} ax + by + cz = d \\ ex + fy + gz = h \\ ix + jy + kz = l \end{cases} $$

The solution is an ordered triple \((x,y,z)\) that makes all three equations true.

To solve a system in three variables, we usually use elimination to reduce the system to two equations in two variables. Then we solve that smaller system, and finally substitute back to find the third variable.

Basic strategy for three variables:

1. Choose one variable to eliminate from two pairs of equations.
2. Create two new equations with only two variables.
3. Solve that two-variable system.
4. Substitute the values back into one original equation to find the third variable.
5. Check the solution in all three equations.

Worked Example 4: Solving a system of three variables

Solve:

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

We will eliminate \(z\) first.

Step 1: Eliminate \(z\) using the first and second equations:

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

This gives us:

$$-x + 2y = 3 \qquad (A)$$

Step 2: Eliminate \(z\) using the first and third equations:

$$ (x + y + z) + (x + 2y - z) = 6 + 3 $$ $$2x + 3y = 9$$

This gives us:

$$2x + 3y = 9 \qquad (B)$$

Now solve the two-variable system:

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

Step 3: Solve equations \((A)\) and \((B)\).

Multiply \((A)\) by \(2\):

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

Add this to \((B)\):

$$(-2x + 4y) + (2x + 3y) = 6 + 9$$ $$7y = 15$$ $$y = \frac{15}{7}$$

Now substitute into \((A)\):

$$-x + 2\left(\frac{15}{7}\right) = 3$$ $$-x + \frac{30}{7} = 3$$ $$-x = 3 - \frac{30}{7} = \frac{21}{7} - \frac{30}{7} = -\frac{9}{7}$$ $$x = \frac{9}{7}$$

Step 4: Substitute \(x\) and \(y\) into the first original equation:

$$x + y + z = 6$$ $$\frac{9}{7} + \frac{15}{7} + z = 6$$ $$\frac{24}{7} + z = 6 = \frac{42}{7}$$ $$z = \frac{18}{7}$$

So the solution is:

$$\boxed{\left(\frac{9}{7},\frac{15}{7},\frac{18}{7}\right)}$$

7. How to check your solution

Always substitute your answer back into the original equations. This helps catch arithmetic mistakes.

For a two-variable system, check the ordered pair in both equations. For a three-variable system, check the ordered triple in all three equations.

8. Choosing the best method

There is not just one correct method. Good algebra students choose the method that makes the work simplest.

• Use substitution when a variable is already isolated, such as \(y=3x-2\).
• Use elimination when coefficients are already opposites or can easily be made opposites.
• For three variables, elimination is usually the most efficient starting method.

9. Common mistakes to avoid

• Forgetting to substitute into every needed equation. After finding one variable, continue until all variables are found.
• Sign errors. Be careful when adding, subtracting, or multiplying negative numbers.
• Stopping too early. In a three-variable system, reducing to two variables is only part of the solution.
• Not recognizing special cases. If variables disappear and you get \(0=0\), there are infinitely many solutions. If you get something false like \(0=4\), there is no solution.
• Not checking. A quick check can save you from losing marks on simple arithmetic mistakes.

10. Real meaning of solutions

In many applications, each equation represents a condition or constraint. The solution is the value that satisfies all the conditions at once.

For example, two lines may represent two different costs, distances, or rates. Their intersection tells you the point where both situations match. In three variables, a solution can represent a combination of quantities that fits three separate conditions.

Summary

A system of equations is solved by finding values that make all equations true at the same time. In two variables, you can solve systems using substitution or elimination. In three variables, elimination helps reduce the system step by step.

You should also be able to identify whether a system has one solution, no solution, or infinitely many solutions. With practice, you will get better at choosing the most efficient method and avoiding common mistakes.

Linear Inequalities and Half-Planes

In algebra, you already know how to graph a linear equation such as \(y = 2x + 1\). Its graph is a straight line. But many real problems do not ask for just one line. Instead, they ask for all points that satisfy a condition like “greater than,” “less than,” “at least,” or “at most.” These conditions are called linear inequalities.

A linear inequality in two variables describes a whole region of the coordinate plane, not just a single line. That region is called a half-plane. In this lesson, you will learn how to graph linear inequalities, decide whether the boundary line should be solid or dashed, and shade the correct side of the line.

1. What is a linear inequality in two variables?

A linear inequality in two variables is an inequality that can be written in a form such as:

• \(y > mx + b\)
• \(y < mx + b\)
• \(y \ge mx + b\)
• \(y \le mx + b\)

It can also appear in standard form, such as \(2x + 3y \le 6\). The graph of the related equation gives the boundary, and the inequality tells you which side of the boundary to shade.

For example, the inequality \(y > x + 2\) means all points whose \(y\)-value is greater than the \(y\)-value on the line \(y = x + 2\). So the solution is the region above that line.

2. Boundary lines: solid or dashed

Every linear inequality has a related linear equation called the boundary line. To get the boundary line, replace the inequality sign with an equal sign.

For example:

• For \(y < 2x - 1\), the boundary line is \(y = 2x - 1\).
• For \(3x + y \ge 6\), the boundary line is \(3x + y = 6\).

Then decide whether to draw the line as solid or dashed.

• Use a solid line for \(\le\) or \(\ge\), because points on the line are included in the solution.
• Use a dashed line for \(<\) or \(>\), because points on the line are not included.

So:

• \(y \le x + 4\) uses a solid line.
• \(y > -2x + 3\) uses a dashed line.

3. What is a half-plane?

A line divides the coordinate plane into two large regions. Each region is called a half-plane. When you graph a linear inequality, one of these half-planes represents all the solutions.

If the inequality is written with \(y\) by itself, the shading rule is often easy:

• \(y > mx + b\) or \(y \ge mx + b\): shade above the line.
• \(y < mx + b\) or \(y \le mx + b\): shade below the line.

This works because the inequality compares the actual \(y\)-value of a point to the \(y\)-value on the line.

4. How to graph a linear inequality

Here is a step-by-step method you can use every time.

1. Write the inequality clearly. If possible, solve for \(y\).
2. Graph the boundary line by changing the inequality sign to \(=\).
3. Decide whether the line should be solid or dashed.
4. Choose which side to shade.
5. If needed, test a point such as \((0,0)\) to check the correct side.

5. Using a test point

If it is not obvious which side to shade, use a test point. A test point is a point not on the boundary line. The point \((0,0)\) is often the easiest choice, as long as the boundary line does not pass through the origin.

Substitute the test point into the inequality:

• If the inequality is true, shade the side containing that point.
• If the inequality is false, shade the opposite side.

For example, for \(x + y > 4\), test \((0,0)\):

$$0 + 0 > 4$$

This is false, so do not shade the side containing \((0,0)\). Shade the other side instead.

6. Worked Example 1: Inequality in slope-intercept form

Graph \(y < 2x + 1\).

Step 1: Graph the boundary line.

Replace \(<\) with \(=\):

$$y = 2x + 1$$

This line has slope \(2\) and \(y\)-intercept \(1\).

Step 2: Decide solid or dashed.

The inequality is \(<\), not \(\le\), so the boundary line is dashed.

Step 3: Decide where to shade.

Since the inequality is \(y < 2x + 1\), shade below the line.

Answer: Draw a dashed line \(y = 2x + 1\) and shade the region below it.

7. Worked Example 2: Inequality in standard form

Graph \(2x + y \ge 3\).

Step 1: Solve for \(y\).

$$2x + y \ge 3$$ $$y \ge -2x + 3$$

Step 2: Graph the boundary line.

The boundary line is:

$$y = -2x + 3$$

Step 3: Decide solid or dashed.

The symbol is \(\ge\), so use a solid line.

Step 4: Shade.

Because \(y \ge -2x + 3\), shade above the line.

Check with a test point: Try \((0,0)\).

$$2(0) + 0 \ge 3$$

$$0 \ge 3$$

This is false, so \((0,0)\) is not in the solution. Shade the side not containing the origin.

Answer: Draw the solid line \(y = -2x + 3\) and shade above it.

8. Worked Example 3: Vertical boundary line

Graph \(x < -1\).

This inequality does not involve \(y\). Its boundary is the vertical line:

$$x = -1$$

Because the symbol is \(<\), use a dashed line.

The inequality \(x < -1\) means all points with \(x\)-coordinates less than \(-1\). These points are to the left of the line.

Answer: Draw a dashed vertical line at \(x = -1\) and shade the region to the left.

9. Worked Example 4: Graphing a system of inequalities

Graph the system:

$$y \le x + 2$$ $$y > -x$$

A system of inequalities means we need points that satisfy both inequalities at the same time.

First inequality: \(y \le x + 2\)

• Boundary line: \(y = x + 2\)
• Use a solid line because of \(\le\)
• Shade below the line

Second inequality: \(y > -x\)

• Boundary line: \(y = -x\)
• Use a dashed line because of \(>\)
• Shade above the line

The solution to the system is the region where the two shaded areas overlap.

Answer: The solution is the set of points above \(y = -x\) and at or below \(y = x + 2\).

10. Common mistakes to avoid

• Using the wrong type of line: Remember, \(\le\) and \(\ge\) mean solid; \(<\) and \(>\) mean dashed.
• Shading the wrong side: If you are unsure, use a test point.
• Forgetting to solve for \(y\): This is not always required, but it often makes the graph easier to understand.
• Shading the whole graph in a system: For systems, the answer is only the overlapping region.
• Including points on a dashed line: Dashed means the boundary itself is not part of the solution.

11. Quick guide for shading

• If the inequality is \(y > \text{line}\), shade above.
• If the inequality is \(y \ge \text{line}\), shade above and include the line.
• If the inequality is \(y < \text{line}\), shade below.
• If the inequality is \(y \le \text{line}\), shade below and include the line.
• If the inequality is \(x > a\), shade to the right of the vertical line.
• If the inequality is \(x < a\), shade to the left of the vertical line.

12. Why half-planes matter

Linear inequalities are important because they model constraints. In real situations, you may need values that stay under a budget, above a minimum score, or within a safe region. Each inequality gives one condition, and graphing shows all possible solutions.

Later, when working with optimization problems, systems of inequalities help define the region of possible answers. That is why understanding half-planes is such an important algebra skill.

Summary

A linear inequality in two variables graphs as a boundary line plus a shaded half-plane. First graph the related equation, then use a solid line for \(\le\) or \(\ge\) and a dashed line for \(<\) or \(>\). Next, shade above, below, left, or right depending on the inequality, or use a test point to confirm the correct side. For systems of inequalities, the solution is the overlapping shaded region.

Systems of Linear Inequalities help us describe all the points that satisfy more than one condition at the same time. Instead of finding just one line, we are looking for a region on the coordinate plane.

This idea is important in algebra, graphing, and real-world problems. For example, a business might need to stay under a spending limit and also produce at least a certain number of items. Each condition can be written as an inequality, and the solution is the set of all points that satisfy every condition together.

In this lesson, you will learn how to graph linear inequalities, combine them into a system, and identify whether the solution region is bounded or unbounded.

1. What is a linear inequality?

A linear inequality in two variables looks like one of these:

• \(y < 2x + 1\)
• \(y \geq -x + 4\)
• \(2x + 3y \leq 12\)

It is similar to a linear equation, but instead of using only \(=\), it uses inequality symbols such as \(<\), \(>\), \(\leq\), or \(\geq\).

The graph of a linear inequality includes:

• a boundary line, which comes from replacing the inequality sign with \(=\)
• a shaded region, showing all points that make the inequality true

2. Solid line or dashed line?

The boundary line tells where the inequality is exactly equal.

• Use a solid line for \(\leq\) or \(\geq\), because points on the line are included.
• Use a dashed line for \(<\) or \(>\), because points on the line are not included.

For example:

• \(y \leq 3x - 2\) uses a solid line
• \(y > -x + 5\) uses a dashed line

3. How do you know which side to shade?

After drawing the boundary line, choose a test point that is not on the line. A common choice is \((0,0)\), unless the line passes through the origin.

Substitute the test point into the inequality.

• If the inequality is true, shade the side containing the test point.
• If the inequality is false, shade the opposite side.

When the inequality is written as \(y > mx+b\) or \(y \geq mx+b\), you can also use a shortcut:

• shade above the line for \(y >\) or \(y \geq\)
• shade below the line for \(y <\) or \(y \leq\)

4. What is a system of linear inequalities?

A system of linear inequalities is a set of two or more inequalities that must all be true at the same time.

For example:

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

To solve the system, graph each inequality on the same coordinate plane. The solution to the system is the region where the shaded parts overlap.

This overlapping region is often called the feasible region. Every point in that region satisfies all the inequalities in the system.

5. Bounded and unbounded regions

Once you graph a system, the feasible region may be:

• bounded: the region is closed in and has a limited area
• unbounded: the region continues forever in at least one direction

A bounded region often looks like a polygonal shape, such as a triangle or quadrilateral. An unbounded region is open on one or more sides.

6. Steps for graphing a system of linear inequalities

1. Rewrite each inequality in a graph-friendly form if needed, often slope-intercept form \(y = mx+b\).
2. Graph the boundary line for each inequality.
3. Use a solid or dashed line based on the inequality symbol.
4. Shade the correct side of each line.
5. Find the overlap of all shaded regions.
6. Decide whether the feasible region is bounded or unbounded.

Worked Example 1: One inequality review

Graph the inequality

$$y \leq 2x + 1$$

Step 1: The boundary line is

$$y = 2x + 1$$

This line has slope \(2\) and \(y\)-intercept \(1\).

Step 2: Because the inequality is \(\leq\), draw a solid line.

Step 3: Since \(y\) is less than or equal to the line, shade below the line.

Answer: The solution is all points on or below the line \(y=2x+1\).

Worked Example 2: A system with an unbounded region

Graph the system

$$ \begin{cases} y \geq x + 1 \\ y < 4 \end{cases} $$

First inequality: \(y \geq x+1\)

• Boundary line: \(y=x+1\)
• Use a solid line
• Shade above the line

Second inequality: \(y < 4\)

• Boundary line: \(y=4\)
• Use a dashed line
• Shade below the line

The solution is the overlap: points that are above or on \(y=x+1\) and also below \(y=4\).

These two conditions create a region between the two lines. The region extends forever to the left, so it is unbounded.

Worked Example 3: A bounded feasible region

Graph the system

$$ \begin{cases} y \geq 0 \\ x \geq 0 \\ y \leq -x + 6 \end{cases} $$

This system includes the coordinate axes and one diagonal line.

Step 1: Graph \(y \geq 0\).

• Boundary line: \(y=0\), which is the \(x\)-axis
• Solid line
• Shade above the axis

Step 2: Graph \(x \geq 0\).

• Boundary line: \(x=0\), which is the \(y\)-axis
• Solid line
• Shade to the right of the axis

Step 3: Graph \(y \leq -x+6\).

• Boundary line: \(y=-x+6\)
• Solid line
• Shade below the line

The overlap forms a triangle in the first quadrant.

To identify its corners, find the intercepts of \(y=-x+6\):

• if \(x=0\), then \(y=6\), so one point is \((0,6)\)
• if \(y=0\), then \(x=6\), so one point is \((6,0)\)

Along with the origin \((0,0)\), the vertices are:

$$ (0,0), \quad (0,6), \quad (6,0) $$

This feasible region is bounded because it is enclosed on all sides.

Worked Example 4: Solve and describe the feasible region

Graph the system

$$ \begin{cases} 2x + y \leq 8 \\ x - y < 2 \\ x \geq 1 \end{cases} $$

Step 1: Rewrite in forms that are easier to graph.

For the first inequality:

$$2x + y \leq 8$$ $$y \leq -2x + 8$$

For the second inequality:

$$x - y < 2$$ $$-y < 2 - x$$

When multiplying by \(-1\), reverse the inequality sign:

$$y > x - 2$$

So the system becomes

$$ \begin{cases} y \leq -2x + 8 \\ y > x - 2 \\ x \geq 1 \end{cases} $$

Step 2: Graph each boundary.

• \(y=-2x+8\): solid line, shade below
• \(y=x-2\): dashed line, shade above
• \(x=1\): solid vertical line, shade to the right

Step 3: Find the overlap.

The solution is the set of points to the right of \(x=1\), above the line \(y=x-2\), and on or below the line \(y=-2x+8\).

To see where the two slanted lines meet, solve

$$x - 2 = -2x + 8$$ $$3x = 10$$ $$x = \frac{10}{3}$$

Then

$$y = x - 2 = \frac{10}{3} - 2 = \frac{4}{3}$$

So the lines intersect at

$$\left(\frac{10}{3}, \frac{4}{3}\right)$$

The vertical line \(x=1\) also helps form the region. Because the region is enclosed between these boundaries, it is bounded on the graph shown by the three constraints.

7. How to check if a point is a solution

A point is a solution to the system only if it makes every inequality true.

For example, consider the system

$$ \begin{cases} y \geq x - 1 \\ y \leq 3 \end{cases} $$

Check whether \((2,2)\) is a solution.

• First inequality: \(2 \geq 2-1\), so \(2 \geq 1\), true
• Second inequality: \(2 \leq 3\), true

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

Check whether \((4,2)\) is a solution.

• First inequality: \(2 \geq 4-1\), so \(2 \geq 3\), false

Since one inequality is false, \((4,2)\) is not a solution.

8. Common mistakes to avoid

• Using the wrong type of line: remember solid for \(\leq\) and \(\geq\), dashed for \(<\) and \(>\).
• Shading the wrong side: use a test point if you are unsure.
• Forgetting to reverse the sign: when multiplying or dividing an inequality by a negative number, flip the inequality symbol.
• Choosing only one inequality: the solution to a system must satisfy all inequalities, so look only at the overlap.
• Ignoring boundary inclusion: if the line is dashed, points on that line are not part of the solution.

9. Why bounded and unbounded matter

In many real-world problems, especially optimization, the feasible region shows all possible choices. If the region is bounded, the possibilities are limited to a closed area. If it is unbounded, the choices continue indefinitely in some direction.

This matters when deciding maximum or minimum values later in algebra and higher math.

10. Summary

A system of linear inequalities is graphed by drawing each boundary line, shading the correct side, and finding the overlap. The overlap is the feasible region, which represents all solutions to the system.

A feasible region can be bounded if it is enclosed, or unbounded if it continues forever. To graph correctly, pay attention to solid versus dashed lines, the direction of shading, and whether a point satisfies every inequality in the system.

Linear Programming Fundamentals

Linear programming is a method for finding the best possible value of a linear expression, called an objective function, when there are rules or limits called constraints.

In 11th Grade maths, linear programming is usually done in two variables, often written as \(x\) and \(y\). You graph the constraints, find the region that satisfies all of them, and then test the corner points to see where the maximum or minimum value occurs.

This topic connects algebra and graphing. You use inequalities to describe limits, graph those inequalities on a coordinate plane, and then use the shape formed by the overlap to solve an optimization problem.

Why it matters: linear programming is used in planning, budgeting, production, transportation, and scheduling. For example, a company may want to maximize profit while using limited time and materials.

Key Idea

A linear programming problem has three main parts:

• Variables: the unknown quantities, such as \(x\) and \(y\)
• Constraints: linear inequalities that limit the possible values of the variables
• Objective function: a linear expression that you want to maximize or minimize

For example, suppose we want to maximize

$$P = 3x + 2y$$

subject to the constraints

$$x + y \le 6$$ $$x \ge 0$$ $$y \ge 0$$

Here:

• \(x\) and \(y\) are the variables
• \(x + y \le 6\), \(x \ge 0\), and \(y \ge 0\) are the constraints
• \(P = 3x + 2y\) is the objective function

Step-by-Step Method

1. Define the variables.
2. Write the objective function.
3. Write the constraints as linear inequalities.
4. Graph each constraint line and shade the correct side.
5. Find the feasible region, which is the overlap of all the shaded regions.
6. Find the vertices (corner points) of the feasible region.
7. Substitute each vertex into the objective function.
8. Choose the greatest value for a maximum problem or the smallest value for a minimum problem.

Important Vocabulary

• Feasible region: the set of all points that satisfy every constraint
• Vertex: a corner point of the feasible region
• Maximum: the greatest value of the objective function
• Minimum: the smallest value of the objective function
• Nonnegative constraints: restrictions like \(x \ge 0\) and \(y \ge 0\), which keep the solution in the first quadrant

Why do we test only the vertices?

For linear programming problems in two variables, the maximum or minimum value of the objective function happens at a vertex of the feasible region, as long as a best value exists.

This is why the graph is so important. Instead of checking every point in the region, you only need to check the corner points.

Graphing Constraints

Each constraint is a linear inequality. To graph it:

1. Replace the inequality sign with an equals sign to draw the boundary line.
2. Use a solid line for \(\le\) or \(\ge\), because points on the line are included.
3. Use a dashed line for \(<\) or \(>\), because points on the line are not included.
4. Shade the side that makes the inequality true.

For example, for \(y \le 2x + 1\):

• Draw the line \(y = 2x + 1\)
• Because the sign is \(\le\), use a solid line
• Shade the region below the line

Worked Example 1: A Basic Maximum Problem

Maximize

$$P = 4x + 3y$$

subject to

$$x + y \le 5$$ $$x \ge 0$$ $$y \ge 0$$

Step 1: Identify the feasible region.

The inequality \(x + y \le 5\) is the region on or below the line \(x + y = 5\). Since \(x \ge 0\) and \(y \ge 0\), we only look in the first quadrant.

The feasible region is a triangle with vertices:

• \((0,0)\)
• \((5,0)\)
• \((0,5)\)

Step 2: Test the vertices in the objective function.

At \((0,0)\):

$$P = 4(0) + 3(0) = 0$$

At \((5,0)\):

$$P = 4(5) + 3(0) = 20$$

At \((0,5)\):

$$P = 4(0) + 3(5) = 15$$

Conclusion: the maximum value is \(20\), and it occurs at \((5,0)\).

Worked Example 2: Maximum with Two Main Constraints

Maximize

$$Z = 2x + 5y$$

subject to

$$x + y \le 6$$ $$x + 3y \le 12$$ $$x \ge 0$$ $$y \ge 0$$

Step 1: Find the vertices of the feasible region.

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

Now find the intercepts and the intersection point of the two lines.

For \(x + y = 6\):

• If \(x=0\), then \(y=6\)
• If \(y=0\), then \(x=6\)

For \(x + 3y = 12\):

• If \(x=0\), then \(y=4\)
• If \(y=0\), then \(x=12\)

Now find the intersection of

$$x + y = 6$$ $$x + 3y = 12$$

Subtract the first equation from the second:

$$2y = 6$$

$$y = 3$$

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

$$x + 3 = 6$$

$$x = 3$$

So the intersection point is \((3,3)\).

The feasible region has vertices:

• \((0,0)\)
• \((6,0)\)
• \((3,3)\)
• \((0,4)\)

Step 2: Evaluate the objective function.

At \((0,0)\):

$$Z = 2(0) + 5(0) = 0$$

At \((6,0)\):

$$Z = 2(6) + 5(0) = 12$$

At \((3,3)\):

$$Z = 2(3) + 5(3) = 6 + 15 = 21$$

At \((0,4)\):

$$Z = 2(0) + 5(4) = 20$$

Conclusion: the maximum value is \(21\) at \((3,3)\).

Worked Example 3: A Minimum Problem

Minimize

$$C = 3x + 2y$$

subject to

$$x + y \ge 4$$ $$x + 2y \ge 6$$ $$x \ge 0$$ $$y \ge 0$$

This time, the inequalities use \(\ge\), so we shade the region above each line.

Step 1: Find the vertices of the feasible region.

We need the corner points formed by the boundary lines.

First, find the intersection of

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

Subtract the first equation from the second:

$$y = 2$$

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

$$x + 2 = 4$$

$$x = 2$$

So one vertex is \((2,2)\).

Now check the axis boundary points that satisfy all constraints:

• On the \(x\)-axis, \(y=0\). Then \(x+y\ge4\) gives \(x\ge4\), and \(x+2y\ge6\) gives \(x\ge6\). So the nearest point is \((6,0)\).
• On the \(y\)-axis, \(x=0\). Then \(x+y\ge4\) gives \(y\ge4\), and \(x+2y\ge6\) gives \(y\ge3\). So the nearest point is \((0,4)\).

The relevant vertices are:

• \((2,2)\)
• \((6,0)\)
• \((0,4)\)

Step 2: Test the objective function.

At \((2,2)\):

$$C = 3(2) + 2(2) = 6 + 4 = 10$$

At \((6,0)\):

$$C = 3(6) + 2(0) = 18$$

At \((0,4)\):

$$C = 3(0) + 2(4) = 8$$

Conclusion: the minimum value is \(8\), at \((0,4)\).

Worked Example 4: A Word Problem

A school club is making two types of posters, Type A and Type B.

• Each Type A poster takes 2 hours to design and 1 hour to print.
• Each Type B poster takes 1 hour to design and 2 hours to print.
• At most 8 hours are available for design.
• At most 8 hours are available for printing.
• The profit is $5 for each Type A poster and $4 for each Type B poster.

How many of each poster should be made to maximize profit?

Step 1: Define the variables.

Let

• \(x\) = number of Type A posters
• \(y\) = number of Type B posters

Step 2: Write the constraints.

Design time:

$$2x + y \le 8$$

Printing time:

$$x + 2y \le 8$$

Also,

$$x \ge 0, \quad y \ge 0$$

Step 3: Write the objective function.

Profit:

$$P = 5x + 4y$$

Step 4: Find the vertices.

First find the intercepts:

• For \(2x + y = 8\): intercepts are \((4,0)\) and \((0,8)\)
• For \(x + 2y = 8\): intercepts are \((8,0)\) and \((0,4)\)

Now find the intersection:

$$2x + y = 8$$ $$x + 2y = 8$$

Multiply the second equation by 2:

$$2x + 4y = 16$$

Subtract the first equation:

$$3y = 8$$

$$y = \frac{8}{3}$$

Substitute into \(2x + y = 8\):

$$2x + \frac{8}{3} = 8$$

$$2x = \frac{16}{3}$$

$$x = \frac{8}{3}$$

So the intersection is \(\left(\frac{8}{3}, \frac{8}{3}\right)\).

The feasible region vertices are:

• \((0,0)\)
• \((4,0)\)
• \(\left(\frac{8}{3}, \frac{8}{3}\right)\)
• \((0,4)\)

Step 5: Evaluate the profit.

At \((0,0)\):

$$P = 5(0) + 4(0) = 0$$

At \((4,0)\):

$$P = 5(4) + 4(0) = 20$$

At \(\left(\frac{8}{3}, \frac{8}{3}\right)\):

$$P = 5\left(\frac{8}{3}\right) + 4\left(\frac{8}{3}\right) = \frac{40}{3} + \frac{32}{3} = \frac{72}{3} = 24$$

At \((0,4)\):

$$P = 5(0) + 4(4) = 16$$

Conclusion: maximum profit is \(\$24\), when the club makes \(\frac{8}{3}\) Type A posters and \(\frac{8}{3}\) Type B posters.

In a real-life situation, the number of posters usually must be whole numbers. In that case, you would check nearby whole-number points. But in graph-based linear programming, the corner-point method gives the best value for the continuous model.

Common Mistakes to Avoid

• Forgetting nonnegative constraints: if \(x \ge 0\) and \(y \ge 0\), the solution must be in the first quadrant.
• Shading the wrong side of a line: test a point like \((0,0)\) if it is not on the line.
• Using the wrong vertices: only use corner points of the feasible region, not all intercepts from every line.
• Not checking all vertices: the best value may happen at any corner.
• Mixing up maximum and minimum: choose the greatest value for a maximum problem and the smallest for a minimum problem.

How to Check Your Work

• Make sure each chosen vertex satisfies all constraints.
• Recalculate intersections carefully by solving the system of equations.
• Substitute each vertex into the objective function one at a time.
• Read the final answer in context, especially in word problems.

Quick Strategy Guide

When solving a linear programming problem, remember this pattern:

1. Write inequalities from the problem.
2. Graph the boundary lines.
3. Shade to find the feasible region.
4. Locate the vertices.
5. Evaluate the objective function at each vertex.
6. Pick the maximum or minimum value.

Brief Summary

Linear programming helps you find the greatest or least value of a linear objective function under given linear constraints.

In two-variable problems, you graph the constraints to form a feasible region, identify its vertices, and test the objective function at each vertex.

The best solution occurs at a corner point of the feasible region. That is the main idea to remember.