# 2.2 Linear equations in one variable  (Page 7/15)

 Page 7 / 15

Find the equation of the line parallel to $\text{\hspace{0.17em}}5x=7+y\text{\hspace{0.17em}}$ and passing through the point $\text{\hspace{0.17em}}\left(-1,-2\right).$

$y=5x+3$

## Finding the equation of a line perpendicular to a given line passing through a given point

Find the equation of the line perpendicular to $\text{\hspace{0.17em}}5x-3y+4=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-4,1\right).$

The first step is to write the equation in slope-intercept form.

$\begin{array}{ccc}\hfill 5x-3y+4& =& 0\hfill \\ \hfill -3y& =& -5x-4\hfill \\ \hfill y& =& \frac{5}{3}x+\frac{4}{3}\hfill \end{array}$

We see that the slope is $\text{\hspace{0.17em}}m=\frac{5}{3}.\text{\hspace{0.17em}}$ This means that the slope of the line perpendicular to the given line is the negative reciprocal, or $-\frac{3}{5}.\text{\hspace{0.17em}}$ Next, we use the point-slope formula with this new slope and the given point.

$\begin{array}{ccc}\hfill y-1& =& -\frac{3}{5}\left(x-\left(-4\right)\right)\hfill \\ \hfill y-1& =& -\frac{3}{5}x-\frac{12}{5}\hfill \\ \hfill y& =& -\frac{3}{5}x-\frac{12}{5}+\frac{5}{5}\hfill \\ \hfill y& =& -\frac{3}{5}x-\frac{7}{5}\hfill \end{array}$

Access these online resources for additional instruction and practice with linear equations.

## Key concepts

• We can solve linear equations in one variable in the form $\text{\hspace{0.17em}}ax+b=0\text{\hspace{0.17em}}$ using standard algebraic properties. See [link] and [link] .
• A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See [link] and [link] .
• All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See [link] and [link] .
• Given two points, we can find the slope of a line using the slope formula. See [link] .
• We can identify the slope and y -intercept of an equation in slope-intercept form. See [link] .
• We can find the equation of a line given the slope and a point. See [link] .
• We can also find the equation of a line given two points. Find the slope and use the point-slope formula. See [link] .
• The standard form of a line has no fractions. See [link] .
• Horizontal lines have a slope of zero and are defined as $\text{\hspace{0.17em}}y=c,$ where c is a constant.
• Vertical lines have an undefined slope (zero in the denominator), and are defined as $\text{\hspace{0.17em}}x=c,$ where c is a constant. See [link] .
• Parallel lines have the same slope and different y- intercepts. See [link] .
• Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical. See [link] .

## Verbal

What does it mean when we say that two lines are parallel?

It means they have the same slope.

What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?

How do we recognize when an equation, for example $\text{\hspace{0.17em}}y=4x+3,$ will be a straight line (linear) when graphed?

The exponent of the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ variable is 1. It is called a first-degree equation.

What does it mean when we say that a linear equation is inconsistent?

When solving the following equation:

$\frac{2}{x-5}=\frac{4}{x+1}$

explain why we must exclude $\text{\hspace{0.17em}}x=5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-1\text{\hspace{0.17em}}$ as possible solutions from the solution set.

If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).

## Algebraic

For the following exercises, solve the equation for $\text{\hspace{0.17em}}x.$

$7x+2=3x-9$

$4x-3=5$

$x=2$

$3\left(x+2\right)-12=5\left(x+1\right)$

$12-5\left(x+3\right)=2x-5$

$x=\frac{2}{7}$

$\frac{1}{2}-\frac{1}{3}x=\frac{4}{3}$

$\frac{x}{3}-\frac{3}{4}=\frac{2x+3}{12}$

$x=6$

$\frac{2}{3}x+\frac{1}{2}=\frac{31}{6}$

$3\left(2x-1\right)+x=5x+3$

$x=3$

$\frac{2x}{3}-\frac{3}{4}=\frac{x}{6}+\frac{21}{4}$

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