ORCCA Chapter Review

Section3.6Chapter Review

Subsection3.6.1Solving Multistep Linear Equations and Inequalities Review

Steps to Solving Linear Equations

Simplify: Simplify the expressions on each side of the equation by distributing and combining like terms.
Isolate: Use addition or subtraction to separate the variable terms and constant terms (numbers) so that they are on different sides of the equation.
Eliminate: Use multiplication or division to eliminate the variable term's coefficient.
Check: Check the solution.
Summarize: State the solution set or (in the case of an application problem) summarize the result in a complete sentence using appropriate units.

Example3.6.1

Solve for $a$ in $4-(3-a)=-2-2(2a+1)\text{.}$

To solve this equation, we will simplify each side of the equation, manipulate it so that all variable terms are on one side and all constant terms are on the other, and then solve for $a\text{:}$

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-3+a\amp=-2-4a-2\\ 1+a\amp=-4-4a\\ 1+a\addright{4a}\amp=-4-4a\addright{4a}\\ 1+5a\amp=-4\\ 1+5a\subtractright{1}\amp=-4\subtractright{1}\\ 5a\amp=-5\\ \divideunder{5a}{5}\amp=\divideunder{-5}{5}\\ a\amp=-1 \end{align*}

Checking the solution $-1$ in the original equation, we get:

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-(3-(\substitute{-1}))\amp\stackrel{?}{=}-2-2(2(\substitute{-1})+1)\\ 4-(4)\amp\stackrel{?}{=}-2-2(-1)\\ 0\amp\stackrel{\checkmark}{=}0 \end{align*}

Therefore the solution to the equation is $-1$ and the solution set is $\{-1\}\text{.}$

Differentiating among Simplifying Expressions, Evaluating Expressions and Solving Equations

Let's summarize the differences among simplifying expressions, evaluating expressions and solving equations:

An expression like $10-3(x+2)$ can be simplified to $-3x+4$ (as in Example 3.1.15), but we cannot solve for $x$ in an expression.
As $x$ takes different values, an expression has different values. In Example 3.1.16, when $x=2\text{,}$ $10-3(x+2)=-2\text{;}$ but when $x=3\text{,}$ $10-3(x+2)=-5\text{.}$
An equation connects two expressions with an equals sign. In Example 3.1.17, $10-3(x+2)=x-16$ has the expression $10-3(x+2)$ on the left side of equals sign, and the expression $x-16$ on the right side.
When we solve the equation $10-3(x+2)=x-16\text{,}$ we are looking for a number which makes those two expressions have the same value. In Example 3.1.17, we found the solution to be $x=5\text{,}$ which makes both $10-3(x+2)=-11$ and $x-16=-11\text{,}$ as shown in the checking part.

Solving Multistep Inequalities

When solving a linear inequality, we follow the same steps in List 3.1.4. The only difference in our steps to solving is that when we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol must switch.

Example

Solve for $x$ in $-2-2(2x+1)\gt4-(3-x)\text{.}$ Write the solution set in both set-builder notation and interval notation.

\begin{align*} -2-2(2x+1)\amp\gt4-(3-x)\\ -2-4x-2\amp\gt4-3+x\\ -4x-4\amp\gt x+1\\ -4x-4\subtractright{x}\amp\gt x+1\subtractright{x}\\ -5x-4\amp\gt1\\ -5x-4\addright{4}\amp\gt1\addright{4}\\ -5x\amp\gt5\\ \divideunder{-5x}{-5}\amp\lt\divideunder{5}{-5}\\ x\amp\lt-1 \end{align*}

Note that when we divided both sides of the inequality by $-5\text{,}$ we had to switch the direction of the inequality symbol.

The solution set in set-builder notation is $\{x\mid x\lt-1\}\text{.}$

The solution set in interval notation is $(-\infty,-1)\text{.}$

Subsection3.6.2Solving Linear Equations and Inequalities with Fractions Review

Clearing Fractions

When there are fractions in a linear equation or inequality, we will multiply each side of the equation by the least common denominator (LCD) of all fractions.

Once the LCD is distributed and all fractions are reduced, we will have an equation that is equivalent to the original, yet without any fractions

Example

Solve for $x$ in $\frac{1}{4}x+\frac{2}{3}=\frac{1}{6}\text{.}$

We'll solve by multiplying each side of the equation by $12\text{:}$

\begin{align*} \frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\ \multiplyleft{12}\left(\frac{1}{4}x+\frac{2}{3}\right)\amp=\multiplyleft{12}\frac{1}{6}\\ 12\cdot\left(\frac{1}{4}x\right)+12\cdot\left(\frac{2}{3}\right)\amp=12\cdot\frac{1}{6}\\ 3x+8\amp=2\\ 3x\amp=-6\\ \divideunder{3x}{3}\amp=\divideunder{-6}{3}\\ x\amp=-2 \end{align*}

Checking the solution:

\begin{align*} \frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\ \frac{1}{4}(-2)+\frac{2}{3}\amp\stackrel{?}{=}\frac{1}{6}\\ -\frac{2}{4}+\frac{2}{3}\amp\stackrel{?}{=}\frac{1}{6}\\ -\frac{6}{12}+\frac{8}{12}\amp\stackrel{?}{=}\frac{1}{6}\\ \frac{2}{12}\amp\stackrel{?}{=}\frac{1}{6}\\ \frac{1}{6}\amp\stackrel{\checkmark}{=}\frac{1}{6} \end{align*}

The solution is therefore $-2\text{.}$ We write the solution set s $\{-2\}\text{.}$

Subsection3.6.3Isolating a Linear Variable Review

Isolating a Linear Variable

When we isolate a linear variable, we follow the same steps of solving linear equations, except we treat all the other variables as numbers. For example, when we solve for $x$ in $y=mx+b\text{,}$ we follow the same steps as solving for $x$ in $1=2x+3\text{.}$

Example

Solve for $x$ in $y=mx+b\text{.}$

\begin{align*} y\amp=mx+b\\ y\subtractright{b}\amp=mx+b\subtractright{b}\\ y-b\amp=mx\\ \divideunder{y-b}{m}\amp=\divideunder{mx}{m}\\ \frac{y-b}{m}\amp=x \end{align*}

Subsection3.6.4Ratios and Proportions Review

Ratios and Proportions

A ratio is a comparison of two values, usually written as a fraction.

A proportion is a statement that two ratios are equal. Many applications require solving for a variable in a proportion.

Example

Property taxes for a residential property are proportional to the assessed value of the property. Assume that a certain property in a given neighborhood is assessed at $\$234{,}100$ and its annual property taxes are $\$2{,}518.92\text{.}$ What are the annual property taxes for a house that is assessed at $\$287{,}500\text{?}$

Let $T$ be the annual property taxes (in dollars) for a property assessed at $\$287{,}500\text{.}$ We can write and solve this proportion:

\begin{align*} \frac{\text{tax}}{\text{property value}}\amp=\frac{\text{tax}}{\text{property value}}\\ \frac{2518.92}{234100}\amp=\frac{T}{287500}\\ \multiplyleft{234100\cdot287500}\frac{2518.92}{234100}\amp=\frac{T}{287500}\multiplyright{234100\cdot287500}\\ 287500\cdot2518.92 \amp=T\cdot 234100\\ \frac{287500\cdot 2518.92}{234100} \amp=\frac{234100T}{234100}\\ T\amp\approx3093.50 \end{align*}

The property taxes for a property assessed at $\$287{,}500$ are $\$3{,}093.50\text{.}$

Subsection3.6.5Special Solution Sets Review

Special Solution Sets

Most of the time, a linear equation has only one solution. It's possible that the equation has no solution and it's also possible that every real number is a solution.

When solving linear inequalities, it's also possible that no solution exists or that all real numbers are solutions.

Examples

Solve for $x$ in $3x=3x+4\text{.}$

To solve this equation, we need to move all terms containing $x$ to one side of the equals sign:

\begin{align*} 3x\amp=3x+4\\ 3x\subtractright{3x}\amp=3x+4\subtractright{3x}\\ 0\amp=4 \end{align*}

This equation has no solution. We write the solution set as $\emptyset\text{,}$ which is the symbol for the empty set.

Solve for $t$ in the inequality $4t+5\gt 4t+2\text{.}$

To solve for $t\text{,}$ we will first subtract $4t$ from each side to get all terms containing $t$ on one side:

\begin{align*} 4t+5\amp\gt 4t+2\\ 4t+5\subtractright{4t}\amp\gt 4t+2\subtractright{4t}\\ 5\amp\gt 2 \end{align*}

All values of the variable $t$ make the inequality true. The solution set is all real numbers, which we can write as $\{t\mid t\text{ is a real number}\}$ in set notation, or $(-\infty,\infty)$ in interval notation.

Subsection3.6.6Exercises

Steps to Solving Linear Equations

Example3.6.1

Differentiating among Simplifying Expressions, Evaluating Expressions and Solving Equations

Solving Multistep Inequalities

Example

Clearing Fractions

Example

Isolating a Linear Variable

Example

Ratios and Proportions

Example

Special Solution Sets

Examples

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