Intermediate Algebra: Functions and Graphs

Subsection 1.2.1 Equations and Solutions

In this section we review some techniques and terminology related to equations, inequalities, and their graphs.

For example, \(x = \alert{−2}\) is a solution of the equation

because \(4(\alert{−2})^3 + 5(\alert{−2})^2 - 8(\alert{−2}) = -2 + 20 + 16 = 4\text{.}\)

A linear equation, such as the models we studied in the last section, has at most one solution. We find the solution by transforming the equation into a simpler equivalent equation whose solution is obvious.

Subsection 1.2.2 Linear Inequalities

Although a linear equation can have at most one solution, a linear inequality can have many solutions. For example, complete the table of values for the expression \(5 − 2x\text{:}\)

Now use your table to list at least three solutions of the inequality \(5 - 2x \lt 2\text{.}\)

Use algebra to solve the inequality \(~~5 -2x \lt 2\text{.}\)

Solution.

We begin by isolating the term containing the variable, just as we do when solving a linear equation. We subtract 5 from both sides to obtain

\begin{gather*} -2x \lt -3 \end{gather*}

Then we divide both sides by \(−2\) to find

\begin{align*} x\gt \frac{-3}{-2}\amp =\frac{3}{2}\amp\amp\blert{\text{Reverse the direction of the inequality.}} \end{align*}

Any value of \(x\) greater than \(\frac{3}{2}\) is a solution of the inequality. We write the solutions as \(x \gt \frac{3}{2}\text{.}\) Because we cannot list all of these solutions, we often illustrate them as a graph on a number line, as shown below.

In the Example above, we used the following rule for solving linear inequalities.

Other than the rule stated in the box above, the rules for solving a linear inequality are the same as the rules for solving a linear equation.

Solve the inequality \(2x + 1 \lt 7\text{,}\) and graph the solutions on a number line.

Solution:

Answer.

\(x < 3\)

Solution.

\(x \lt 3\)

A number line is below.

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Subsection 1.2.3 Equations in Two Variables

An equation in two variables, such as

has many solutions. Each solution consists of an ordered pair of values, one for \(x\) and one for \(y\text{,}\) that together satisfy the equation (make the equation true.)

Subsection 1.2.4 What is the Graph of an Equation?

You might think it would be difficult to find all the solutions of an equation, but for a linear equation \(Ax + By = C\text{,}\) we can at least illustrate the solutions: all the solutions lie on a straight line. (Later on we can prove that this is true.)

The figure shows a graph of the equation \(~Ax + By = C\text{.}\) Which of the following equations are true?

1. \(\displaystyle 24B=C\)

2. \(\displaystyle Ap+Bq=C\)

3. \(\displaystyle Ag+Bh=C\)

Solution.

1. The point \((0, 24)\) lies on the graph, so \(x = 0\text{,}\) \(y = 24\) is a solution of the equation. Thus, \(A \cdot 0 + B \cdot 24 = C\text{,}\) or \(24B = C\) is a true statement.

2. The point \((p, q)\) does not lie on the graph, so \(x = p\text{,}\) \(y = q\) does not satisfy the equation, and \(A p+ B q = C\) is a not true.

3. The point \((g,h )\) does lie on the graph, so \(x = g\text{,}\) \(y = h\) does satisfy the equation, and \(A g+ B h = C\) is a true statement.

On the grid above, plot the points you found in Practice 3. All the points should lie on a straight line; draw the line with a ruler or straightedge. Which of the following points lie on the graph?

• \(\displaystyle (-13,4) \)

• \(\displaystyle (-1.6,4.8) \)

• \(\displaystyle (1.125,7) \)

Which of the points above satisfy the equation in Practice 3?

• \(\displaystyle (-13,4) \)

• \(\displaystyle (-1.6,4.8) \)

• \(\displaystyle (1.125,7) \)

Answer 1.

\(\text{Choice 2}\)

Answer 2.

\(\text{Choice 2}\)

Solution.

\((-1.6,4.8)\) lies on the graph and satisfies the equation.

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Subsection 1.2.5 Graphical Solution of Equations

We can use graphs to find solutions to equations in one variable.

Use the graph of \(~y = 285 - 15x~\) to solve the equation

\begin{equation*} 150 = 285 - 15x \end{equation*}

Solution.

Compare the two equations in the problem. In the equation we want to solve, \(y\) has been replaced by \(\alert{150}\text{.}\) We begin by locating the point \(P\) on the graph for which \(y = 150\text{.}\)

Next we find the \(x\)-coordinate of point \(P\) by drawing an imaginary line from \(P\) straight down to the \(x\)-axis. The \(x\)-coordinate of \(P\) is \(x = 9\text{.}\)

Thus, \(P\) is the point \((9,150)\text{,}\) and \(x = 9\) when \(y = 150\text{.}\) The solution we seek is \(x = 9\text{.}\)

You can verify the solution algebraically by substituting \(x = \alert{9}\) into the equation:

Does \(150 = 285 - 15(\alert{9})\text{?}\)

\begin{equation*} 285 - 15(\alert{9}) = 285 - 135 = 150 ~~~~~~\blert{\text{Yes}} \end{equation*}

1. Use the graph of \(y = 30 - 8x\) to solve the equation \(30 - 8x = 50\text{.}\)Follow the steps:

   • Step 1: Locate the point \(P\) on the graph with \(y = 50\text{.}\)

   • Step 2: Find the \(x\)-coordinate of your point \(P\text{.}\)

2. Verify your solution algebraically.

\(x=\)

Answer.

\(-2.5\)

Solution.

\(x=-2.5\)

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Subsection 1.2.6 Graphical Solution of Inequalities

We can also use graphs to solve inequalities. Consider the inequality

To solve this inequality means to find all values of \(x\) that make the expression \(285 -15x\) greater than or equal to 150. We could begin by trying some values of \(x\text{.}\) Here is a table obtained by evaluating \(285 − 15x\text{.}\)

From the table, we see that values of \(x\) less than or equal to 8 are solutions of the inequality, but we have not checked all possible \(x\)-values. We can get a more complete picture from a graph.

Use the graph of the equation \(~y = 285 - 15x~\) to solve the inequality

\begin{equation*} 285 - 15x \gt 150 \end{equation*}

Solution.

We look for points on the graph with \(y\)-coordinates greater than or equal to 150. These points are shown in color. Which \(x\)-values produced these points?

We can read the \(x\)-coordinates by dropping straight down to the \(x\)-axis, as shown by the arows. For example, the \(x\)-value corresponding to \(y=150\) is \(x=9\text{.}\) For larger values of \(285-15x\text{,}\) we must choose \(x\)-values less than 9. Thus, all values of \(x\) less than or equal to 9 are solutions, as shown on the \(x\)-axis.

We write the solutions as \(x \le 9\text{.}\)

1. Use the graph of \(y = 30 - 8x\) to solve the equation \(30 - 8x \lt 14\text{.}\)Follow the steps:

   • Step 1: Locate the point \(P\) on the graph with \(y = 14\text{.}\)

   • Step 2: Find the \(x\)-coordinate of the point \(P\text{.}\)

   • Step 3: Which points on the graph have \(y \lt 14\text{?}\) Mark them on the graph.

   • Step 4: Find the \(x\)-coordinates of the points in Step 3. Mark them all on the \(x\)-axis.

2. Verify your solution algebraically.

Solution:

Answer.

\(x > 2\)

Solution.

\(x \gt 2\)

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Subsection 1.2.7 Using a Graphing Utility

We can use a graphing utility to graph equations if they are written in the form \(~y = (\text{expression in }x)~\text{.}\) First, let's review how to solve an equation for \(y\) in terms of \(x\text{.}\)

Now we are ready to graph an equation with technology. On most calculators, we follow three steps.

Choosing a graphing window corresponds to drawing the \(x\)- and \(y\)-axes and marking a scale on each axis when we graph by hand. The standard graphing window displays values from \(-10\) to \(10\) on both axes. We can press the ZOOM key and then the number \(6\) to select this window.

Use a graphing utility to graph the equation \(3x + 2y = 16\text{.}\)

Solution.

First, we solve the equation for \(y\) in terms of \(x\text{.}\)

\begin{align*} 3x+2y \amp = 16\amp\amp\blert{\text{Subtract } 3x \text{ from both sides.}} \\ 2y\amp = -3x+16 \amp\amp\blert{\text{Divide both sides by 2.}} \\ y\amp = \frac{-3x}{2}+\frac{16}{2}\amp\amp \blert{\text{Simplify.}} \\ y\amp= -1.5x+8 \end{align*}

Now press Y= and enter \(-1.5x+8\) after \(Y_1=\text{.}\) (Use the negation key, (-), to enter \(-1.5x\text{;}\) do not use the subtraction key.) We choose the standard graphing window by pressing ZOOM 6. The graph is shown at right.

When you want to erase the graph, press Y= and then press CLEAR. This deletes the expression following \(Y_1\text{.}\)

1. Solve the equation \(-3x + 4y = 24\) for \(y\) in terms of \(x\text{.}\)

   \(y=\)

2. Graph the equation in the standard window.

Answer.

\(\frac{3}{4}x+6\)

Solution.

\(y=\dfrac{3}{4}x+6\)

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Graph for part (b):

Subsection 1.2.8 Finding Coordinates with a Graphing Calculator

We can use the TRACE feature to find the coordinates of points on a graph.

Use a graph to find a solution to the equation \(y = -2.6x - 5.4\) with \(y\)-coordinate \(-10.6\text{.}\)

Solution.

First we graph the equation \(y = -2.6x - 5.4\) in the window

Xmin\(=-5\)		Xmax\(=4.4\)
Ymin\(=-20\)		Ymax\(=15\)

We press TRACE, and a "bug" begins flashing on the display. The coordinates of the bug appear at the bottom of the display, as shown in the figure. We use the left and right arrows to move the bug along the graph. You can check that the coordinates of the point \((2, -10.6)\) do satisfy the equation \(y = -2.6x - 5.4\text{.}\)

1. Graph the equation \(y=32x-42\) in the window:

   Xmin\(=-4.7 \qquad\)	Xmax\(=4.7 \qquad\)	Xscl\(=1 \qquad\)
   Ymin\(=-250 \qquad\)	Ymax\(=50 \qquad\)	Yscl\(=25 \qquad\)

2. Use the Trace feature to find the point that has \(y\)-coordinate \(-122\text{.}\)

   Enter an ordered pair.

3. Verify your answer algebraically by substituting your \(x\)-value into the equation.

Answer.

\(\left(-2.5,-122\right)\)

Solution.

1. A graph is below.

2. \(\displaystyle (-2.5,-122)\)

3. \(\displaystyle \blert{-122} = 32(\alert{-2.5}) - 42\)

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Graph for part (a):