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Section 1.1 Warm-Up

We begin by refreshing some basic skills for working with equations and graphs.

Subsection 1. Solve a linear equation

Recall that to solve an equation we want to "isolate" the variable on one side of the equals sign. We "undo" each operation performed on the variable by performing the opposite operation on both sides of the equation.

Subsubsection Examples

Example 1.1.

Solve the equation \(~~\dfrac{2}{3}x-5=7\)

Solution
\begin{equation*} \begin{aligned}[t] \dfrac{2}{3}x-5 \amp = 7 \amp \amp \blert{\text{Add 5 to both sides.}}\\ \dfrac{2}{3}x \amp = 12 \amp \amp \blert{\text{To divide both sides by } \dfrac{2}{3}, \text{we:}}\\ \dfrac{3}{2}\left(\dfrac{2}{3}x\right) \amp =\dfrac{3}{2}(12) \amp \amp \blert{\text{Multiply by the reciprocal of }\dfrac{2}{3}.}\\ x \amp = 18 \amp \amp \blert{\text{The solution is 18.}} \end{aligned} \end{equation*}
Example 1.2.

Solve the equation \(~~2x+7=4x-3\)

Solution
To begin, we must get both variable terms on the same side of the equation.
\begin{equation*} \begin{aligned}[t] 2x+7 \amp = 4x-3 \amp \amp \blert{\text{Subtract}~2x~\text{from both sides.}}\\ 7 \amp = 2x-3 \amp \amp \blert{\text{Add 3 to both sides.}}\\ 10 \amp =2x \amp \amp \blert{\text{Divide both sides by 2.}}\\ 5 \amp = x \amp \amp \blert{\text{The solution is 5.}} \end{aligned} \end{equation*}

Subsubsection Exercises

Solve the equation \(~~10=1-\dfrac{3x}{7}\)

Answer
\(-21\)

Solve the equation \(~~6p-8=-3p-26\)

Answer
\(-2\)

Solve the equation \(~~12=\dfrac{7u+4}{5}\)

Hint: Start by clearing the fraction: multiply both sides by 5.

Answer
\(8\)

Solve the equation \(~~0=13q+25-17q+7\)

Hint: Start by combining like terms.

Answer
\(8\)

Subsection 2. Solve a linear inequality

The rules for solving an inequality are the same as those for solving an equation, with one important difference:

Solving a Linear Inequality.

If we multiply or divide both sides by a negative number, we must reverse the direction of the inequality.

Subsubsection Examples

Example 1.7.

Solve \(~~-3x+1 \gt 7~~\) and graph the solutions on a number line.

Solution
\begin{align*} -3x+1 \amp \gt 7 \amp \amp ~\blert{\text{ Subtract 1 from both sides.}}\\ -3x \amp \gt 6 \amp \amp\ \begin{array}{l} \blert{\text{Divide both sides by }{-3}\text{, and reverse }}\\ \blert{\text{the direction of the inequality.}} \end{array}\\ x \amp \lt -2 \end{align*}

The solutions are all the numbers less than \(-2\text{.}\) The graph of the solutions is shown below.

line graph
Example 1.8.

Solve \(~~-3 \lt 2x-5 \le 6~~\) and graph the solutions on a number line.

Solution
\begin{align*} -3 \amp \lt 2x-5 \le 6 \amp \amp \blert{\text{Add 5 on all three sides of the inequality.}}\\ 2 \amp \lt 2x \le 11 \amp \amp \blert{\text{Divide each side by 2.}}\\ 1 \amp \lt x \le \dfrac{11}{2} \amp \amp \blert{\text{Notice that we did not reverse the inequality.}} \end{align*}

The solutions are all the numbers greater than 1 but less than 5.5. The graph of the solutions is shown below.

number line

Recall that a solid dot on a number line indicates that the number is part of the solution; an open dot means that the number is not part of the solution.

Subsubsection Exercises

Solve the inequality \(~~8-4x \gt -2~~\) and graph the solutions on a number line.

Answer

\(x \lt \dfrac{5}{2}\)

number line

Solve the inequality \(~~-6 \le \dfrac{4-x}{3} \lt 2~~\) and graph the solutions on a number line.

Answer

\(22 \ge x \gt -2\)

number line

Solve the inequality \(~~3x-5 \lt -6x+7~~\) and graph the solutions on a number line.

Answer

\(x \lt \dfrac{4}{3}\)

number line

Solve the inequality \(~~-6 \gt 4-5b \gt -21~~\) and graph the solutions on a number line.

Answer

\(2 \lt b \lt 5\)

number line

Subsection 3. Verify a solution

We can always check a solution to an equation by verifying that it makes the equation true.

Subsubsection Examples

Example 1.13.

Verify that \(x=-5\) is a solution of the equation

\begin{equation*} ~~x^2+2x-15=0 \end{equation*}
Solution

We show that substituting \(-5\) for \(x\) makes the equation true. When we substitute a negative number for a variable, we should enclose the number in parentheses.

\begin{equation*} \begin{aligned}[t] x^2+2x-15 \amp = (\alert{-5})^2+2(\alert{-5})-15\\ \amp = 25-10-15=0 \end{aligned} \end{equation*}

Because the expression does equal \(0\text{,}\) we see that \(x=-5\) is a solution.

Example 1.14.

Verify that \(x=-3\) is not a solution of the equation

\begin{equation*} ~~\sqrt{2x+10}-3x=8 \end{equation*}
Solution

We show that substituting \(-3\) for \(x\) does not make the equation true.

\begin{equation*} \begin{aligned}[t] \sqrt{2x+10}-3x \amp = \sqrt{2(\alert{-3})+10}-3(\alert{-3})\\ \amp = \sqrt{4}+9=2+9=11 \not= 8 \end{aligned} \end{equation*}

The left side of the equation does not equal 8 when \(x=-3\text{,}\) so \(x=-3\) is not a solution.

Subsubsection Exercises

Decide whether the given value is a solution of the equation.

\begin{equation*} x^3-3x^2-4x+2=10;~~~~x=-2 \end{equation*}
Answer

Yes

Decide whether the given value is a solution of the equation.

\begin{equation*} \sqrt{3x+5}=10+\sqrt{x+7};~~~~x=9 \end{equation*}
Answer

No

Decide whether the given value is a solution of the equation.

\begin{equation*} \dfrac{2x-1}{x+1}+2=\dfrac{x+1}{x-1};~~~~x=2 \end{equation*}
Answer

Yes

Decide whether the given value is a solution of the equation.

\begin{equation*} 9-4x=5\sqrt{x+3};~~~~x=6 \end{equation*}
Answer

No

Subsection 4. Solve an equation in two variables

A solution of an equation in two variables \(x\) and \(y\) is written as an ordered pair, \((x,y)\text{.}\) For example, the solution \((-2,5)\) means that \(x=-2\) and \(y=5\text{.}\)

Subsubsection Examples

Example 1.19.

Is \((-3,2)\) a solution of the equation \(~~x^2+4y^2=25\) ?

Solution

We substitute \(x=\alert{-3}\) and \(y=\alert{2}\) into the equation.

\begin{equation*} (\alert{-3})^2+4(\alert{2})^2=9+4(4)=9+16=25 \end{equation*}

The ordered pair \((-3,2)\) satisfies the equation, so it is a solution.

Example 1.20.

Which of the following ordered pairs are solutions of the equation whose graph is shown?

  1. \(\displaystyle (-3,-2)\)
  2. \(\displaystyle (-5,0)\)
  3. \(\displaystyle (1,-4)\)
  4. \(\displaystyle (-1,-6)\)
parabola
Solution

The graph of an equation is just a picture of its solutions, so points that lie on the graph are solutions of the equation.

The points \((-3,-2)\) and \((-1,-6)\) lie on the graph, so they represent solutions of the equation. The points \((-5,0)\) and \((1,-4)\) do not lie on the graph, so they are not solutions of the equation.

Subsubsection Exercises

Find a solution of the equation with the given coordinate.

\begin{equation*} 6x-5y=-3,~~~~(2,?) \end{equation*}
Answer
\((2,3)\)

Find a solution of the equation with the given coordinate.

\begin{equation*} y=\frac{3}{4}x+8,~~~~(?,-1) \end{equation*}
Answer
\((-12,-1)\)

Find a solution of the equation with the given coordinate.

graph
Answer
\((16,4)\)

Find a solution of the equation with the given coordinate.

graph
Answer
\((2,-5)\)