Skills Refresher: Solving Linear Equations, Inequalities, & Geometric Word Problems (2.1 – 2.3; 4.1; 4.5)) — Intermediate Algebra

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2 Skills Refresher: Solving Linear Equations, Inequalities, & Geometric Word Problems (2.1 – 2.3; 4.1; 4.5)) — Intermediate Algebra

2.1 Solving Elementary Linear Equations

Solving linear equations is an important and fundamental skill in algebra. In algebra, there are often problems in which the answer is known, but the variable part of the problem is missing. To find this missing variable, it is necessary to follow a series of steps that result in the variable equaling some solution.

Addition and Subtraction Problems

To solve equations, the general rule is to do the opposite of the order of operations. Consider the following.

Example 2.1.1

Solve for $x.$

$x-7=5$
$\phantom{1}$
$\begin{array}{rrrrr} x&-&7&=&-5\\ &+&7&&+7\\ \midrule &&x&=&2 \end{array}$
$4+x=8$
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$\begin{array}{rrrrr} 4&+&x&=&8\\ -4&&&&-4\\ \midrule &&x&=&4 \end{array}$
$7=x-9$
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$\begin{array}{rrrrr} 7&=&x&-&9\\ +9&&&+&9\\ \midrule 16&=&x&& \end{array}$
$5=8+x$
$\phantom{1}$
$\begin{array}{rrrrr} 5&=&8&+&x\\ -8&&-8&&\\ \midrule -3&=&x&& \end{array}$

Multiplication Problems

In a multiplication problem, get rid of the coefficient in front of the variable by dividing both sides of the equation by that number. Consider the following examples.

Example 2.1.2

Solve for $x.$

$\begin{array}{rrl} \\ \\ \\ \\ \\ 4x&=&20\\ \\ \dfrac {4x}{4}&=&\dfrac{20}{4}\\ \\ x&=&5 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ 8x&=&-24\\ \\ \dfrac {8x}{8}&=&\dfrac{-24}{8}\\ \\ x&=&-3 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ -4x&=&-20\\ \\ \dfrac {-4x}{-4}&=&\dfrac{-20}{-4}\\ \\ x&=&5 \end{array}$

Division Problems

In division problems, remove the denominator by multiplying both sides by it. Consider the following examples.

Example 2.1.3

Solve for $x.$

$\phantom{1}$
$\begin{array}{rrl}\\ \dfrac{x}{-7}&=&-2\\ \\ -7\left(\dfrac{x}{-7}\right)&=&(-2)-7 \\ \\ x&=&14\end{array}$
$\phantom{1}$
$\begin{array}{rrl}\\ \dfrac{x}{8}&=&5\\ \\ 8\left(\dfrac{x}{8}\right)&=&(5)8\\ \\ x&=&40\end{array}$
$\phantom{1}$
$\begin{array}{rrl}\\ \dfrac{x}{-4}&=&9\\ \\ -4\left(\dfrac{x}{-4}\right)&=&(9) -4\\ \\ x&=&-36\end{array}$

2.2 Solving Linear Equations

When working with questions that require two or more steps to solve, do the reverse of the order of operations to solve for the variable.

Example 2.2.1

Solve $4x+16=-4$ for $x.$

$\begin{array}{rrrrrl} 4x& +& 16 &=&-4& \\ &&-16&& -16&\text{Subtract 16 from each side} \\ \midrule &&\dfrac{4x}{4}& =& \dfrac{-20}{4}&\text{Divide each side by 4}\\ \\ &&x& =& -5 & \text{Solution} \end{array}$

In solving the above equation, notice that the general pattern followed was to do the opposite of the equation. $4x$ was added to 16, so 16 was then subtracted from both sides. The variable $x$ was multiplied by 4, so both sides were divided by 4.

Example 2.2.2

$\begin{array}{rrrrr} \\ \\ \\ \\ \\ 5x &+& 7 &=& 7 \\ &-&7&&-7 \\ \midrule &&\dfrac{5x}{5} &=& \dfrac{0}{5}\\ \\ &&x& =& 0 \end{array}$
$\begin{array}{rrrrr} \\ \\ \\ \\ \\ 4 &- &2x& =& 10 \\ -4&&&&-4\\ \midrule &&\dfrac{-2x}{-2}& = &\dfrac{6}{-2}\\ \\ &&x& =& -3 \end{array}$
$\begin{array}{rrrrr} \\ \\ \\ \\ \\ -3x& -& 7& =& 8 \\ &+&7& = & + 7 \\ \midrule &&\dfrac{-3x}{-3}& =& \dfrac{15}{-3} \\ \\ &&x &= &-5\\ \end{array}$

2.3 Solving Intermediate Linear Equations

When working with linear equations with parentheses, the first objective is to isolate the parentheses. Once isolated, the parentheses can be removed and then the variable solved.

Example 2.3.1

Solve for $x$ in the equation $4(2x-6) = 16.$

$\begin{array}{rrrl} \dfrac{4(2x-6)}{4}&=&\dfrac{16}{4}&\text{Divide both sides by 4} \\ \\ (2x - 6)&=&4&\text{Remove the parentheses} \\ 2x -6&=&4&\text{Add 6 to both sides to remove }-6 \\ +6&&+6& \\ \midrule \dfrac{2x}{2}&=&\dfrac{10}{2}&\text{Divide both sides by 2} \\ \\ x&=&5&\text{Solution} \end{array}$

Example 2.3.2

Solve for $x$ in the equation $3(2x - 4) + 9 = 15.$

$\begin{array}{rrrl} 3(2x - 4) + 9&=&15&\text{Subtract 9 from both sides} \\ -9&&-9& \\ \midrule \dfrac{3(2x - 4)}{3}&=& \dfrac{6}{3}&\text{Divide both sides by 3 and remove parentheses} \\ \\ 2x - 4&=&2&\text{Add 4 to both sides} \\ +4&&+4& \\ \midrule \dfrac{2x}{2}&=&\dfrac{6}{2}&\text{Divide both sides by 2} \\ \\ x&=&3&\text{Solution} \end{array}$

For some problems, it is too difficult to isolate the parentheses. In these problems, it is necessary to multiply or divide throughout the parentheses by whatever coefficient is in front of it.

Example 2.3.3

Solve for $x$ in the equation $3(4x - 5) - 4(2x + 1) = 5.$

$\begin{array}{rrrl} 3(4x - 5) - 4(2x + 1)&=&5&\text{Distribute} \\ 12x - 15 - 8x - 4&=&5&\text{Combine similar terms} \\ 4x-19&=&5&\text{Add 19 to both sides} \\ +19&&+19& \\ \midrule \dfrac{4x}{4}&=&\dfrac{24}{4}&\text{Divide both sides by 4} \\ \\ x&=&6& \end{array}$

4.1 Solving and Graphing Linear Inequalities

When given an equation, such as $x = 4$ or $x = -5,$ there are specific values for the variable. However, with inequalities, there is a range of values for the variable rather than a defined value. To write the inequality, use the following notation and symbols:

Symbol	Meaning
	> Greater than
	≤ Greater than or equal to
	< Less than
	≥ Less than or equal to

Example 4.1.1

Given a variable $x$ such that $x$ > $4$ , this means that $x$ can be as close to 4 as possible but always larger. For $x$ > $4$ , $x$ can equal 5, 6, 7, 199. Even $x =$ 4.000000000000001 is true, since $x$ is larger than 4, so all of these are solutions to the inequality. The line graph of this inequality is shown below:

Written in interval notation, $x$ > $4$ is shown as $(4, \infty)$ .

Example 4.1.2

Likewise, if $x < 3$ , then $x$ can be any value less than 3, such as 2, 1, −102, even 2.99999999999. The line graph of this inequality is shown below:

Written in interval notation, $x < 3$ is shown as $(-\infty, 3)$ .

Example 4.1.3

For greater than or equal (≥) and less than or equal (≤), the inequality starts at a defined number and then grows larger or smaller. For $x \ge 4,$ $x$ can equal 5, 6, 7, 199, or 4. The line graph of this inequality is shown below:

Written in interval notation, $x \ge 4$ is shown as $[4, \infty)$ .

Example 4.1.4

If $x \le 3$ , then $x$ can be any value less than or equal to 3, such as 2, 1, −102, or 3. The line graph of this inequality is shown below:

Written in interval notation, $x \le 3$ is shown as $(-\infty, 3].$

When solving inequalities, the direction of the inequality sign (called the sense) can flip over. The sense will flip under two conditions:

First, the sense flips when the inequality is divided or multiplied by a negative. For instance, in reducing $-3x < 12$ , it is necessary to divide both sides by −3. This leaves $x$ > $-4.$

Second, the sense will flip over if the entire equation is flipped over. For instance, $x$ > $2$ , when flipped over, would look like $2 < x.$ In both cases, the 2 must be shown to be smaller than the $x$ , or the $x$ is always greater than 2, no matter which side each term is on.

Example 4.1.5

Solve the inequality $5-2x$ > $11$ and show the solution on both a number line and in interval notation.

First, subtract 5 from both sides:

$\begin{array}{rrrrr} 5&-&2x&\ge &11 \\ -5&&&&-5 \\ \midrule &&-2x&\ge &6 \end{array}$

Divide both sides by −2:

$\begin{array}{rrr} \dfrac{-2x}{-2} &\ge &\dfrac{6}{-2} \\ \end{array}$

Since the inequality is divided by a negative, it is necessary to flip the direction of the sense.

This leaves:

$x \le -3$

In interval notation, the solution is written as $(-\infty, -3]$ .

On a number line, the solution looks like:

Inequalities can get as complex as the linear equations previously solved in this textbook. All the same patterns for solving inequalities are used for solving linear equations.

Example 4.1.6

Solve and give interval notation of $3 (2x - 4) + 4x < 4 (3x - 7) + 8$ .

Multiply out the parentheses:

$6x - 12 + 4x < 12x - 28 + 8$

Simplify both sides:

$10x - 12 < 12x - 20$

Combine like terms:

$\begin{array}{rrrrrrr} 10x&-&12&<&12x&-&20 \\ -12x&+&12&&-12x&+&12 \\ \midrule &&-2x&<&-8&& \end{array}$

The last thing to do is to isolate $x$ from the −2. This is done by dividing both sides by −2. Because both sides are divided by a negative, the direction of the sense must be flipped.

This means:

$\dfrac{-2x}{-2}< \dfrac{-8}{-2}$

Will end up looking like:

The solution written on a number line is:

Written in interval notation, $x$ > $4$ is shown as $(4, \infty)$ .

4.5 Geometric Word Problems

It is common to run into geometry-based word problems that look at either the interior angles, perimeter, or area of shapes. When looking at interior angles, the sum of the angles of any polygon can be found by taking the number of sides, subtracting 2, and then multiplying the result by 180°. In other words:

$\text{sum of interior angles} = 180^{\circ} \times (\text{number of sides} - 2)$

This means the interior angles of a triangle add up to 180° × (3 − 2), or 180°. Any four-sided polygon will have interior angles adding to 180° × (4 − 2), or 360°. A chart can be made of these:

$\begin{array}{rrrrrr} \text{3 sides:}&180^{\circ}&\times&(3-2)&=&180^{\circ} \\ \text{4 sides:}&180^{\circ}&\times&(4-2)&=&360^{\circ} \\ \text{5 sides:}&180^{\circ}&\times&(5-2)&=&540^{\circ} \\ \text{6 sides:}&180^{\circ}&\times&(6-2)&=&720^{\circ} \\ \text{7 sides:}&180^{\circ}&\times&(7-2)&=&900^{\circ} \\ \text{8 sides:}&180^{\circ}&\times&(8-2)&=&1080^{\circ} \\ \end{array}$

Example 4.5.1

The second angle $(A_2)$ of a triangle is double the first $(A_1).$ The third angle $(A_3)$ is 40° less than the first $(A_1).$ Find the three angles.

The relationships described in equation form are as follows:

$A_2 = 2A_1 \text{ and } A_3 = A_1 - 40^{\circ}$

Because the shape in question is a triangle, the interior angles add up to 180°. Therefore:

$A_1 + A_2 + A_3 = 180^{\circ}$

Which can be simplified using substitutions:

$A_1 + (2A_1) + (A_1 - 40^{\circ}) = 180^{\circ}$

Which leaves:

$\begin{array}{rrrrrrrrrrr} 2A_1&+&A_1&+&A_1&-&40^{\circ}&=&180^{\circ}&&&\\ &&&&4A_1&-&40^{\circ}&=&180^{\circ}&&\\ \\ &&&&&&4A_1&=&180^{\circ}&+&40^{\circ}\\ \\ &&&&&&A_1&=&\dfrac{220^{\circ}}{4}&\text{or}&55^{\circ} \end{array}$

This means $A_2 = 2 (55^{\circ})$ or 110° and $A_3 = 55^{\circ}-40^{\circ}$ or 15°.

Common Geometric Shapes with Related Area and Perimeter Equations
Shape	Area	Perimeter
Circle	$\pi r^2$	$2\pi r$
Square	$s^2$	$4s$
Rectangle	$lw$	$2l+2w$
Triangle	$\dfrac{1}{2}bh$	$s_1+s_2+s_3$
Rhombus	$bh$	$4b$
Trapezoid	$\dfrac{1}{2}\left(l_1+l_2\right)h$	$l_1+l_2+h_1+h_2$
Parallelogram	$bh$	$2h_1+2b$
Regular polygon ( $n$ -gon)	$\left(\dfrac{1}{2}sh\right)(\text{number of sides})$	$s(\text{number of sides})$

Another common geometry word problem involves perimeter, or the distance around an object. For example, consider a rectangle, for which $\text{perimeter} = 2l + 2w.$

Example 4.5.2

If the length of a rectangle is 5 m less than twice the width, and the perimeter is 44 m long, find its length and width.

The relationships described in equation form are as follows:

$L = 2W - 5 \text{ and } P = 44$

For a rectangle, the perimeter is defined by:

$P = 2 W + 2 L$

Substituting for $L$ and the value for the perimeter yields:

$44 = 2W + 2 (2W - 5)$

Which simplifies to:

$44 = 2W + 4W - 10$

Further simplify to find the length and width:

$\begin{array}{rrrrlrrrr} 44&+&10&=&6W&&&& \\ \\ &&54&=&6W&&&& \\ \\ &&W&=&\dfrac{54}{6}&\text{or}&9&& \\ \\ &\text{So}&L&=&2(9)&-&5&\text{or}&13 \\ \end{array}$

The width is 9 m and the length is 13 m.

Extra Reading and Instructional Videos

Article to read: New theory finds ‘traffic jams’ in jet stream cause abnormal weather patterns.

The abstract reads:

A study offers an explanation for a mysterious and sometimes deadly weather pattern in which the jet stream, the global air currents that circle the Earth, stalls out over a region. Much like highways, the jet stream has a capacity, researchers said, and when it’s exceeded, blockages form that are remarkably similar to traffic jams — and climate forecasters can use the same math to model them both.

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2.1 Solving Elementary Linear Equations

Addition and Subtraction Problems

Multiplication Problems

Division Problems

2.2 Solving Linear Equations

2.3 Solving Intermediate Linear Equations

4.1 Solving and Graphing Linear Inequalities

4.5 Geometric Word Problems

Extra Reading and Instructional Videos

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