ORCCA Equations and Inequalities as True/False Statements

Section2.2Equations and Inequalities as True/False Statements

ObjectivesPCC Course Content and Outcome Guide

Item C.1.1:4.a
Item C.1.1:4.b
Item C.1.1:3.a

This section introduces the concepts of algebraic equations and inequalities, and what it means for a number to be a solution to an equation or inequality.

Figure2.2.1Alternative Video Lesson

Subsection2.2.1Equations, Inequalities, and Solutions

An equation is two mathematical expressions with an equals sign between them. The two expressions can be relatively simple or more complicated:

A relatively simple equation:

\begin{equation*} x+1=2 \end{equation*}

A more complicated equation:

\begin{equation*} \left(x^2+y^2-1\right)^3=x^2y^3 \end{equation*}

An inequality is quite similar, but the sign between the expressions is one of these: $\lt\text{,}$ $\leq\text{,}$ $\gt\text{,}$ $\geq\text{,}$ or $\neq\text{.}$

A relatively simple inequality:

\begin{equation*} x\geq15 \end{equation*}

A more complicated inequality:

\begin{equation*} x^2+y^2\lt1 \end{equation*}

A linear equation in one variable can be written in the form $ax+b=0\text{,}$ where $a, b$ are real numbers, and $a\ne0\text{.}$ The variable could be any letter other than $x\text{.}$ The variable cannot have other exponent than $1$ ($x=x^1$), and the variable cannot be inside a root symbol (square root, cube root, etc.) or in a denominator.

The following are some linear equations in one variable:

\begin{align*} 4-y\amp=5 \amp 4-z\amp=5z \amp 0\amp=\frac{1}{2}p \end{align*} \begin{align*} 3-2(q+2)\amp=10 \amp \sqrt{2}\cdot r+3\amp=10 \amp \frac{s}{2}+3\amp=5 \end{align*}

(Note that $r$ is outside the square root symbol.) We will see in later sections that all equations above can be converted into the form $ax+b=0\text{.}$

The following are some non-linear equations:

\begin{align*} 1+2=3 \amp\amp\amp \text{(There is no variable.)}\\ 4-2y^2=5 \amp\amp\amp \text{(The exponent of $y$ is not $1$.)}\\ \sqrt{2r}+3=10 \amp\amp\amp \text{($r$ is inside the square root.)}\\ \frac{2}{s}+3=5 \amp\amp\amp \text{($s$ is in a denominator.)} \end{align*}

This chapter focuses on linear equations in one variable. We will study other types of equations in later chapters.

The simplest equations and inequalities have numbers and no variables. When this happens, the equation is either true or false.The following equations and inequalities are true statements:

\begin{align*} 2\amp=2\amp-4\amp=-4\amp2\amp\gt1\amp-2\amp\lt-1\amp3\amp\ge3 \end{align*}

The following equations and inequalities are false statements:

\begin{align*} 2\amp=1\amp-4\amp=4\amp2\amp\lt1\amp-2\amp\ge-1\amp0\neq0 \end{align*}

When equations and inequalities have variables, we can consider substituting values in for the variables. If replacing a variable with a number makes an equation or inequality true, then that number is called a solution to the equation.

Example2.2.2A Solution

Consider the equation $y+2=3\text{,}$ which has only one variable, $y\text{.}$ If we substitute in $1$ for $y$ and then simplify:

\begin{align*} y+2\amp=3\\ \substitute{1}+2\amp\stackrel{?}{=}3\\ 3\amp\stackrel{\checkmark}{=}3 \end{align*}

we get a true equation. So we say that $1$ is a solution to $y+2=3\text{.}$ Notice that we used a question mark at first because we are unsure if the equation is true or false until the very end.

If replacing a variable with a number makes an equation or inequality false, then that number is not a solution.

Example2.2.3Not a Solution

Consider the inequality $x+4\gt 5\text{,}$ which has only one variable, $x\text{.}$ If we substitute in $0$ for $x$ and then simplify:

\begin{align*} x+4\amp\gt 5\\ \substitute{0}+4\amp\stackrel{?}{\gt}5\\ 4\amp\stackrel{\text{no}}{\gt}5 \end{align*}

we get a false equation. So we say that $0$ is not a solution to $x+4\gt 5\text{.}$

Subsection2.2.2Checking Possible Solutions

Given an equation or an inequality (with one variable), checking if some particular number is a solution is just a matter of replacing the value of the variable with the specified number and determining if the resulting equation/inequality is true or false. This may involve some amount of arithmetic simplification.

Example2.2.4

Is $8$ a solution to $x^2-5x=\sqrt{2x}+20\text{?}$

To find out, substitute in $8$ for $x$ and see what happens.

\begin{align*} x^2-5x\amp=\sqrt{2x}+20\\ \substitute{8}^2-5(\substitute{8})\amp\stackrel{?}{=}\sqrt{2(\substitute{8})}+20\\ \highlight{64}-5(8)\amp\stackrel{?}{=}\sqrt{\highlight{16}}+20\\ 64-\highlight{40}\amp\stackrel{?}{=}\highlight{4}+20\\ \highlight{24}\amp\stackrel{\checkmark}{=}\highlight{24} \end{align*}

So yes, $8$ is a solution to $x^2-5x=\sqrt{2x}+20\text{.}$

Example2.2.5

Is $-5$ a solution to $\sqrt{169-y^2}=y^2-2y\text{?}$

To find out, substitute in $-5$ for $y$ and see what happens.

\begin{align*} \sqrt{169-y^2}\amp=y^2-2y\\ \sqrt{169-\substitute{(-5)}^2}\amp\stackrel{?}{=}\substitute{(-5)}^2-2(\substitute{-5})\\ \sqrt{169-\highlight{25}}\amp\stackrel{?}{=}\highlight{25}-2(-5)\\ \sqrt{\highlight{144}}\amp\stackrel{?}{=}25-(\highlight{-10})\\ \highlight{12}\amp\stackrel{\text{no}}{=}\highlight{35} \end{align*}

So no, $-5$ is not a solution to $\sqrt{169-y^2}=y^2-2y\text{.}$

But is $-5$ a solution to the inequality $\sqrt{169-y^2}\leq y^2-2y\text{?}$ Yes, because substituting $-5$ in for $y$ would give you

\begin{equation*} 12\leq35\text{,} \end{equation*}

which is true.

Exercise2.2.6

Exercise2.2.7

Exercise2.2.8

Exercise2.2.9

Exercise2.2.10

Example2.2.11Cylinder Volume

A cylinder's volume is related to its radius and its height by the equation

\begin{equation*} V=\pi r^2h\text{,} \end{equation*}

where $V$ stands for volume, $r$ stands for the base's radius, and $h$ stands for the cylinder's height. If we have a cylinder where we know the volume is 96$\pi$ cm³ and the radius is 4 cm, then this equation simplifies to

\begin{equation*} 96\pi=16\pi h \end{equation*}

Figure2.2.12A Cylinder with volume 96$\pi$ cm³, radius 4 cm, and unknown height

Is 4 cm the height of the cylinder? In other words, is $4$ a solution to $96\pi=16\pi h\text{?}$ We will substitute $h$ in the equation with $4$ to check:

\begin{align*} 96\pi\amp=16\pi h\\ 96\pi\amp\stackrel{?}{=}16\pi \cdot\substitute{4}\\ 96\pi\amp\stackrel{\text{no}}{=}64\pi \end{align*}

Since $96\pi=64\pi$ is false, $h=4$ does not satisfy the equation $96\pi=16\pi h\text{.}$

Next, we will try $h=6\text{:}$

\begin{align*} 96\pi\amp=16\pi h\\ 96\pi\amp\stackrel{?}{=}16\pi \cdot\substitute{6}\\ 96\pi\amp\stackrel{\checkmark}{=}96\pi \end{align*}

When $h=6\text{,}$ the equation $96\pi=16\pi h$ is true. This tells us that $6$ is a solution for $h$ in the equation $96\pi=16\pi h\text{.}$

Remark2.2.13

Note that we did not approximate $\pi$ with $3.14$ or any other approximation. We often leave $\pi$ as $\pi$ throughout our calculations. If we need to round, we do so as a final step.

Example2.2.14

Ann has budgeted a maximum of $\$300$ for an appliance repair. The total cost of the repair can be modeled by $89+110(h-0.25)\text{,}$ where $\$89$ is the initial cost and $\$110$ is the hourly labor charge after the first quarter hour. Is $2$ a solution for $h$ in the inequality $89+110(h-0.25)\le 300\text{?}$

To determine if $h=2$ satisfies the inequality $89+110(h-0.25)\le 300\text{,}$ we will replace $h$ with $2$ and check if the statement is true or false:

\begin{align*} 89+110(h-0.25)\amp\le 300\\ 89+110(\substitute{2}-0.25)\amp\stackrel{?}{\le} 300\\ 89+110(1.75)\amp\stackrel{?}{\le} 300\\ 89+192.5\amp\stackrel{?}{\le} 300\\ 281.5\amp\stackrel{\checkmark}{\le} 300 \end{align*}

Thus $2$ is a solution for $h$ in the inequality $89+110(h-0.25)\le 300\text{.}$ In context, this means that Ann would stay within her $\$300$ budget if $2$ hours of labor were performed.