5.2 Writing Linear Equations From a Table Reading Strategies

Learning Objectives

In this section, y'all volition:

• Solve equations in 1 variable algebraically.
• Solve a rational equation.
• Find a linear equation.
• Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
• Write the equation of a line parallel or perpendicular to a given line.

Caroline is a full-time higher student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $xv.00/hr, and she opened a savings account with an initial deposit of $400 on January fifteen. She arranged for directly eolith of her payroll checks. If spring suspension begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she tin can only work four hours per 24-hour interval, how many days per week volition she take to work? How many weeks volition it have? In this department, we will investigate problems like this and others, which generate graphs like the line in Effigy 1.

Effigy 1

Solving Linear Equations in One Variable

A linear equation is an equation of a direct line, written in one variable. The merely power of the variable is ane. Linear equations in one variable may take the class $a\mathrm{ten}+b=0$ and are solved using bones algebraic operations.

We begin by classifying linear equations in one variable every bit i of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.

$$\mathrm{three}x=\mathrm{ii}x+x$$

The solution set up consists of all values that make the equation truthful. For this equation, the solution gear up is all real numbers because any real number substituted for $\mathrm{10}$ will make the equation true.

A conditional equation is true for only some values of the variable. For example, if we are to solve the equation $5\mathrm{ten}+2=\mathrm{iii}x-\mathrm{half\; dozen},$ we accept the following:

$$\begin{array}{ccc}\hfill 5\mathrm{10}+2& =& \mathrm{iii}\mathrm{ten}-\mathrm{six}\hfill \\ \hfill 2\mathrm{10}& =& \mathrm{-8}\hfill \\ \hfill x& =& \mathrm{-4}\hfill \end{array}$$

The solution prepare consists of one number: $\left\{-4\right\}.$ It is the only solution and, therefore, nosotros have solved a conditional equation.

An inconsistent equation results in a simulated statement. For example, if we are to solve $5x-15=5\left(\mathrm{ten}-4\right),$ we take the following:

$$\begin{array}{cccc}\hfill \mathrm{five}x-\mathrm{fifteen}& =& \mathrm{five}\mathrm{10}-\mathrm{xx}\hfill & \\ \hfill 5x-15-\mathrm{five}x& =& 5\mathrm{10}-20-\mathrm{v}x\hfill & \text{Subtract}\mathrm{five}x\text{from both sides}.\hfill \\ \hfill \mathrm{-xv}& \ne & \mathrm{-xx}\hfill & \text{False argument}\hfill \end{array}$$

Indeed, $\mathrm{-fifteen}\ne \mathrm{-20.}$ At that place is no solution because this is an inconsistent equation.

Solving linear equations in one variable involves the central properties of equality and basic algebraic operations. A brief review of those operations follows.

Linear Equation in 1 Variable

A linear equation in one variable tin be written in the form

$$a\mathrm{ten}+b=0$$

where a and b are real numbers, $a\ne 0.$

How To

Given a linear equation in i variable, utilise algebra to solve it.

The post-obit steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads $\mathrm{ten}=\_\_\_\_\_\_\_\_\_,$ if ten is the unknown. There is no gear up order, as the steps used depend on what is given:

1. We may add together, subtract, multiply, or divide an equation by a number or an expression equally long as we do the same thing to both sides of the equal sign. Note that nosotros cannot split by zero.
2. Use the distributive property every bit needed: $a\left(b+c\right)=ab+ac.$
3. Isolate the variable on one side of the equation.
4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

Example 1

Solving an Equation in Ane Variable

Solve the post-obit equation: $2\mathrm{ten}+7=19.$

Try It #one

Solve the linear equation in one variable: $\mathrm{two}x+\mathrm{i}=\mathrm{-ix.}$

Example 2

Solving an Equation Algebraically When the Variable Appears on Both Sides

Solve the following equation: $4\left(\mathrm{ten}\mathrm{-3}\right)+12=15\mathrm{-5}\left(x+6\right).$

Analysis

This problem requires the distributive belongings to be applied twice, and so the backdrop of algebra are used to reach the final line, $\mathrm{ten}=-\frac{5}{3}.$

Try It #2

Solve the equation in ane variable: $\mathrm{-ii}\left(3\mathrm{10}-1\right)+x=\mathrm{xiv}-x.$

Solving a Rational Equation

In this section, we look at rational equations that, after some manipulation, upshot in a linear equation. If an equation contains at least 1 rational expression, it is a considered a rational equation.

Recall that a rational number is the ratio of two numbers, such as $\frac{\mathrm{two}}{3}$ or $\frac{\mathrm{seven}}{2}.$ A rational expression is the ratio, or caliber, of ii polynomials. Hither are iii examples.

$$\frac{\mathrm{ten}+\mathrm{i}}{x^{\mathrm{ii}}-4},\frac{1}{\mathrm{10}-3},\text{or}\frac{4}{{\mathrm{ten}}^2+x-2}$$

Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation past the to the lowest degree common denominator (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We practise this considering when the equation is multiplied past the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

Instance three

Solving a Rational Equation

Solve the rational equation: $\frac{7}{\mathrm{two}\mathrm{10}}-\frac{5}{3x}=\frac{22}{3}.$

A common mistake made when solving rational equations involves finding the LCD when ane of the denominators is a binomial—two terms added or subtracted—such as $\left(x+\mathrm{i}\right).$ E'er consider a binomial as an private gene—the terms cannot be separated. For instance, suppose a problem has three terms and the denominators are $x,$ $\mathrm{ten}-1,$ and $3x-3.$ First, gene all denominators. We then have $x,$ $(\mathrm{ten}-\mathrm{one}),$ and $3(x-1)$ equally the denominators. (Note the parentheses placed around the second denominator.) Simply the final ii denominators have a common factor of $(x-\mathrm{one}).$ The $x$ in the starting time denominator is separate from the $x$ in the $(x-1)$ denominators. An effective way to call back this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a carve up factor. The LCD in this case is plant by multiplying together the $x,$ ane factor of $\left(x-\mathrm{i}\right),$ and the 3. Thus, the LCD is the following:

$$x\left(x-1\right)3=3\mathrm{10}\left(x-\mathrm{i}\right)$$

And so, both sides of the equation would exist multiplied by $\mathrm{iii}x\left(\mathrm{ten}-1\right).$ Leave the LCD in factored class, as this makes it easier to see how each denominator in the trouble cancels out.

Another example is a problem with two denominators, such as $\mathrm{ten}$ and $x^{\mathrm{two}}+2\mathrm{ten}.$ Once the second denominator is factored every bit ${\mathrm{10}}^{\mathrm{ii}}+\mathrm{ii}\mathrm{ten}=x\left(x+\mathrm{ii}\right),$ there is a mutual factor of x in both denominators and the LCD is $\mathrm{10}\left(x+2\right).$

Sometimes we accept a rational equation in the grade of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.

$$\frac{a}{b}=\frac{c}{d}$$

We can use some other method of solving the equation without finding the LCD: cross-multiplication. We multiply terms past crossing over the equal sign.

Multiply $a\left(d\right)$ and $b\left(c\right),$ which results in $ad=bc.$

Whatever solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.

Rational Equations

A rational equation contains at least ane rational expression where the variable appears in at to the lowest degree 1 of the denominators.

How To

Given a rational equation, solve it.

1. Gene all denominators in the equation.
2. Find and exclude values that set each denominator equal to naught.
3. Find the LCD.
4. Multiply the whole equation by the LCD. If the LCD is correct, there volition be no denominators left.
5. Solve the remaining equation.
6. Brand certain to bank check solutions back in the original equations to avoid a solution producing zero in a denominator.

Instance 4

Solving a Rational Equation without Factoring

Solve the post-obit rational equation:

$$\frac{\mathrm{two}}{x}-\frac{\mathrm{iii}}{\mathrm{two}}=\frac{7}{2x}$$

Attempt Information technology #3

Solve the rational equation: $\frac{2}{\mathrm{iii}x}=\frac{1}{4}-\frac{\mathrm{ane}}{6\mathrm{ten}}.$

Example 5

Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: $\frac{\mathrm{ane}}{\mathrm{10}}=\frac{1}{10}-\frac{\mathrm{iii}}{4x}.$

Try Information technology #4

Solve the rational equation: $-\frac{\mathrm{five}}{2x}+\frac{3}{4\mathrm{10}}=-\frac{7}{4}.$

Case six

Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

1. ⓐ $\frac{\mathrm{iii}}{x-6}=\frac{5}{\mathrm{ten}}$
2. ⓑ $\frac{\mathrm{ten}}{x-3}=\frac{5}{\mathrm{ten}-3}-\frac{1}{\mathrm{ii}}$
3. ⓒ $\frac{x}{x-2}=\frac{5}{x-2}-\frac{1}{2}$

Try It #5

Solve $\frac{-\mathrm{iii}}{2x+\mathrm{ane}}=\frac{4}{\mathrm{three}\mathrm{10}+1}.$ State the excluded values.

Instance 7

Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: $\frac{2}{x+1}-\frac{1}{x-\mathrm{ane}}=\frac{2x}{{\mathrm{ten}}^2-\mathrm{ane}}.$ State the excluded values.

Try It #half dozen

Solve the rational equation: $\frac{2}{x-2}+\frac{\mathrm{one}}{x+1}=\frac{1}{{\mathrm{10}}^2-\mathrm{10}-2}.$

Finding a Linear Equation

Mayhap the most familiar course of a linear equation is the slope-intercept class, written every bit $y=mx+b,$ where $\mathrm{chiliad}=\text{slope}$ and $b=y\text{-intercept}\text{.}$ Let united states of america begin with the gradient.

The Slope of a Line

The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between whatever two points on a line. It indicates the direction in which a line slants likewise as its steepness. Slope is sometimes described every bit rising over run.

$$m=\frac{y_{\mathrm{two}}-y_{\mathrm{ane}}}{{\mathrm{ten}}_{\mathrm{two}}-x_{\mathrm{ane}}}$$

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. Equally the gradient increases, the line becomes steeper. Some examples are shown in Figure 2. The lines indicate the following slopes: $m=\mathrm{-3},$ $\mathrm{thou}=2,$ and $m=\frac{1}{3}.$

Figure 2

The Gradient of a Line

The slope of a line, chiliad, represents the change in y over the change in x. Given two points, $\left(x_1,y_1\right)$ and $\left(x_{\mathrm{ii}},y_{\mathrm{ii}}\right),$ the following formula determines the gradient of a line containing these points:

$$m=\frac{y_2-y_{\mathrm{one}}}{x_{\mathrm{two}}-{\mathrm{ten}}_1}$$

Example viii

Finding the Slope of a Line Given Two Points

Find the gradient of a line that passes through the points $\left(\mathrm{ii},\mathrm{-1}\right)$ and $\left(\mathrm{-5},3\right).$

Assay

Information technology does non thing which point is called $\left(x_{\mathrm{i}},y_1\right)$ or $\left(x_2,y_2\right).$ Every bit long as we are consequent with the order of the y terms and the social club of the x terms in the numerator and denominator, the calculation will yield the same result.

Attempt It #7

Detect the slope of the line that passes through the points $\left(\mathrm{-2},\mathrm{vi}\right)$ and $\left(1,4\right).$

Example 9

Identifying the Gradient and y-intercept of a Line Given an Equation

Place the gradient and y-intercept, given the equation $y=-\frac{3}{4}x-4.$

Analysis

The y-intercept is the point at which the line crosses the y-axis. On the y-centrality, $x=0.$ We can always identify the y-intercept when the line is in slope-intercept course, every bit it volition always equal b. Or, merely substitute $\mathrm{ten}=0$ and solve for y.

The Bespeak-Slope Formula

Given the gradient and ane bespeak on a line, we tin find the equation of the line using the indicate-gradient formula.

$$y-y_{\mathrm{i}}=\mathrm{1000}\left(x-x_{\mathrm{one}}\right)$$

This is an important formula, as information technology volition exist used in other areas of college algebra and ofttimes in calculus to observe the equation of a tangent line. We need only i point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

The Bespeak-Slope Formula

Given i point and the slope, the point-slope formula volition lead to the equation of a line:

$$y-y_{\mathrm{ane}}=m\left(x-{\mathrm{10}}_{\mathrm{ane}}\right)$$

Example ten

Finding the Equation of a Line Given the Slope and I Point

Write the equation of the line with slope $m=\mathrm{-three}$ and passing through the signal $\left(4,8\right).$ Write the final equation in slope-intercept grade.

Analysis

Note that whatsoever point on the line can be used to find the equation. If washed correctly, the same final equation will be obtained.

Try It #8

Given $\mathrm{grand}=4,$ find the equation of the line in gradient-intercept grade passing through the point $\left(\mathrm{ii},\mathrm{v}\right).$

Example 11

Finding the Equation of a Line Passing Through Ii Given Points

Find the equation of the line passing through the points $\left(3,\mathrm{iv}\right)$ and $\left(0,\mathrm{-3}\right).$ Write the final equation in slope-intercept form.

Analysis

To prove that either point can exist used, permit u.s. use the second signal $\left(0,\mathrm{-3}\right)$ and run into if we get the same equation.

$$\begin{array}{ccc}\hfill y-(-\mathrm{three})& =& \frac{7}{\mathrm{iii}}(x-0)\hfill \\ \hfill y+3& =& \frac{\mathrm{seven}}{3}\mathrm{ten}\hfill \\ \hfill y& =& \frac{7}{\mathrm{three}}x-3\hfill \end{array}$$

We encounter that the same line volition be obtained using either point. This makes sense because nosotros used both points to summate the slope.

Standard Form of a Line

Another way that we can correspond the equation of a line is in standard form. Standard form is given as

$$A\mathrm{10}+By=C$$

where $A,$ $B,$ and $C$ are integers. The x- and y-terms are on one side of the equal sign and the constant term is on the other side.

Example 12

Finding the Equation of a Line and Writing It in Standard Form

Discover the equation of the line with $\mathrm{thou}=\mathrm{-half-dozen}$ and passing through the point $\left(\frac{1}{4},\mathrm{-2}\right).$ Write the equation in standard class.

Endeavor It #nine

Find the equation of the line in standard form with slope $\mathrm{thousand}=-\frac{1}{\mathrm{three}}$ and passing through the point $\left(\mathrm{ane},\frac{1}{3}\right).$

Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not crave whatever of the preceding formulas, although we can utilize the formulas to prove that the equations are correct. The equation of a vertical line is given as

$$x=c$$

where c is a abiding. The slope of a vertical line is undefined, and regardless of the y-value of any bespeak on the line, the x-coordinate of the point volition exist c.

Suppose that nosotros desire to discover the equation of a line containing the following points: $\left(\mathrm{-3},\mathrm{-5}\right),\left(\mathrm{-3},1\right),\left(\mathrm{-iii},3\right),$ and $\left(\mathrm{-three},5\right).$ Beginning, we volition find the slope.

$$m=\frac{5-3}{-\mathrm{three}-\left(\mathrm{-3}\right)}=\frac{\mathrm{two}}{0}$$

Zero in the denominator ways that the slope is undefined and, therefore, we cannot employ the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and nosotros notice a vertical line through $\mathrm{10}=\mathrm{-3.}$ See Figure iii.

The equation of a horizontal line is given as

$$y=c$$

where c is a constant. The slope of a horizontal line is nix, and for whatsoever 10-value of a point on the line, the y-coordinate will be c.

Suppose we want to find the equation of a line that contains the post-obit set of points: $\left(\mathrm{-ii},\mathrm{-two}\right),\left(0,\mathrm{-2}\right),\left(3,\mathrm{-two}\right),$ and $\left(\mathrm{five},\mathrm{-two}\right).$ Nosotros tin use the bespeak-slope formula. Beginning, we find the slope using whatever 2 points on the line.

$$\begin{array}{ccc}\hfill m& =& \frac{\mathrm{-2}-(\mathrm{-2})}{0-(\mathrm{-ii})}\hfill \\ & =& \frac{0}{\mathrm{two}}\hfill \\ & =& 0\hfill \end{array}$$

Use any signal for $\left({\mathrm{ten}}_1,y_1\right)$ in the formula, or use the y-intercept.

$$\begin{array}{ccc}\hfill y-(\mathrm{-2})& =& 0(x-3)\hfill \\ \hfill y+\mathrm{ii}& =& 0\hfill \\ \hfill y& =& \mathrm{-2}\hfill \end{array}$$

The graph is a horizontal line through $y=\mathrm{-2.}$ Notice that all of the y-coordinates are the same. Encounter Figure three.

Effigy 3 The line x = −3 is a vertical line. The line y = −ii is a horizontal line.

Example 13

Finding the Equation of a Line Passing Through the Given Points

Observe the equation of the line passing through the given points: $\left(\mathrm{one},\mathrm{-three}\right)$ and $\left(\mathrm{one},4\right).$

Try Information technology #10

Find the equation of the line passing through $\left(\mathrm{-5},2\right)$ and $\left(2,2\right).$

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines take the same gradient and different y-intercepts. Lines that are parallel to each other will never intersect. For example, Effigy iv shows the graphs of various lines with the same slope, $m=2.$

Effigy four Parallel lines

All of the lines shown in the graph are parallel considering they have the same gradient and dissimilar y-intercepts.

Lines that are perpendicular intersect to form a $\mathrm{90°}$ -angle. The slope of one line is the negative reciprocal of the other. Nosotros tin can bear witness that ii lines are perpendicular if the product of the ii slopes is $\mathrm{-1}:{\mathrm{thousand}}_1\cdot m_2=\mathrm{-1.}$ For case, Figure 5 shows the graph of ii perpendicular lines. 1 line has a slope of 3; the other line has a gradient of $-\frac{1}{3}.$

$$\begin{array}{ccc}\hfill {\mathrm{thousand}}_{\mathrm{i}}\cdot {\mathrm{grand}}_{\mathrm{two}}& =& \mathrm{-i}\hfill \\ \hfill 3\cdot \left(-\frac{1}{3}\right)& =& \mathrm{-i}\hfill \end{array}$$

Figure 5 Perpendicular lines

Example 14

Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: $3y=-\mathrm{four}x+\mathrm{three}$ and $\mathrm{three}x-4y=8.$

Try It #11

Graph the two lines and decide whether they are parallel, perpendicular, or neither: $2y-x=10$ and $2y=\mathrm{10}+4.$

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we accept learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the aforementioned principles equally we do for finding the equation of any line. After finding the gradient, use the point-slope formula to write the equation of the new line.

How To

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

1. Discover the gradient of the given line. The easiest way to do this is to write the equation in gradient-intercept form.
2. Utilize the gradient and the given betoken with the signal-gradient formula.
3. Simplify the line to slope-intercept form and compare the equation to the given line.

Example 15

Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a $5x+\mathrm{three}y=\mathrm{i}$ and passing through the point $\left(\mathrm{iii},5\right).$

Try Information technology #12

Observe the equation of the line parallel to $\mathrm{v}\mathrm{ten}=7+y$ and passing through the signal $\left(\mathrm{-1},\mathrm{-2}\right).$

Instance xvi

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Signal

Find the equation of the line perpendicular to $5x-3y+\mathrm{four}=0$ and passing through the betoken $(-\mathrm{four},\mathrm{ane}).$

ii.2 Section Exercises

Exact

1 .

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

2 .

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

three .

How practice nosotros recognize when an equation, for example $y=\mathrm{four}x+\mathrm{three},$ will be a straight line (linear) when graphed?

4 .

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

5 .

When solving the following equation:

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

explain why nosotros must exclude $x=5$ and $\mathrm{10}=\mathrm{-ane}$ every bit possible solutions from the solution ready.

Algebraic

For the following exercises, solve the equation for $x.$

8 .

$\mathrm{three}(\mathrm{ten}+2)-12=5(x+1)$

9 .

$12-\mathrm{v}(\mathrm{ten}+3)=2\mathrm{ten}-5$

10 .

$\frac{\mathrm{one}}{2}-\frac{\mathrm{one}}{\mathrm{three}}x=\frac{\mathrm{four}}{3}$

11 .

$\frac{x}{\mathrm{three}}-\frac{3}{4}=\frac{\mathrm{ii}\mathrm{10}+\mathrm{iii}}{12}$

12 .

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

xiii .

$3(2x-\mathrm{i})+x=5x+3$

14 .

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

15 .

$\frac{x+2}{4}-\frac{\mathrm{ten}-1}{3}=\mathrm{two}$

For the following exercises, solve each rational equation for $x.$ State all x-values that are excluded from the solution set.

16 .

$\frac{\mathrm{three}}{x}-\frac{\mathrm{i}}{3}=\frac{\mathrm{ane}}{\mathrm{six}}$

17 .

$\mathrm{two}-\frac{3}{x+4}=\frac{\mathrm{10}+\mathrm{ii}}{\mathrm{10}+4}$

18 .

$\frac{3}{x-2}=\frac{1}{\mathrm{10}-\mathrm{one}}+\frac{7}{(\mathrm{ten}-\mathrm{one})(x-\mathrm{two})}$

19 .

$\frac{3\mathrm{10}}{x-1}+2=\frac{3}{x-\mathrm{i}}$

20 .

$\frac{\mathrm{five}}{x+\mathrm{i}}+\frac{\mathrm{one}}{\mathrm{10}-\mathrm{iii}}=\frac{-6}{{\mathrm{10}}^{\mathrm{two}}-2x-\mathrm{three}}$

21 .

$\frac{1}{\mathrm{10}}=\frac{1}{5}+\frac{\mathrm{three}}{\mathrm{ii}x}$

For the post-obit exercises, notice the equation of the line using the point-gradient formula. Write all the final equations using the slope-intercept grade.

22 .

$\left(0,\mathrm{three}\right)$ with a slope of $\frac{2}{3}$

23 .

$\left(\mathrm{one},2\right)$ with a slope of $-\frac{\mathrm{four}}{5}$

24 .

10-intercept is 1, and $\left(\mathrm{-2},6\right)$

25 .

y-intercept is 2, and $\left(\mathrm{iv},\mathrm{-one}\right)$

26 .

$(\mathrm{-3},\mathrm{ten})$ and $(5,\mathrm{-6})$

27 .

$\left(1,3\right)\text{ and}\left(5,5\right)$

28 .

parallel to $y=2x+5$ and passes through the bespeak $\left(\mathrm{four},\mathrm{iii}\right)$

29 .

perpendicular to $\text{3}y=x-4$ and passes through the signal $\left(\mathrm{-2},1\right)$ .

For the following exercises, notice the equation of the line using the given information.

30 .

$\left(-2,0\right)$ and $\left(\mathrm{-2},5\right)$

31 .

$\left(1,7\right)$ and $\left(3,7\right)$

32 .

The slope is undefined and it passes through the point $\left(2,3\right).$

33 .

The slope equals zero and it passes through the point $\left(\mathrm{ane},\mathrm{-4}\right).$

34 .

The slope is $\frac{3}{4}$ and it passes through the point $(1,\mathrm{iv})$ .

35 .

$(\mathrm{–1},3)$ and $(\mathrm{four},\mathrm{–v})$

Graphical

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

36 .

$\begin{array}{l}y=\mathrm{ii}x+\mathrm{seven}\hfill \\ y=-\frac{1}{2}x-4\hfill \end{array}$

37 .

$\begin{array}{l}\mathrm{iii}x-2y=\mathrm{five}\hfill \\ 6y-\mathrm{ix}x=6\hfill \end{array}$

38 .

$\begin{array}{l}y=\frac{3x+\mathrm{ane}}{4}\hfill \\ y=3\mathrm{10}+2\hfill \end{array}$

Numeric

For the following exercises, find the gradient of the line that passes through the given points.

40 .

$\left(5,\mathrm{four}\right)$ and $\left(7,9\right)$

41 .

$\left(\mathrm{-3},\mathrm{ii}\right)$ and $\left(4,\mathrm{-seven}\right)$

42 .

$\left(\mathrm{-5},\mathrm{four}\right)$ and $\left(2,\mathrm{four}\right)$

43 .

$\left(\mathrm{-1},\mathrm{-2}\right)$ and $\left(\mathrm{iii},\mathrm{iv}\right)$

44 .

$\left(3,\mathrm{-2}\right)$ and $\left(\mathrm{iii},\mathrm{-2}\right)$

For the following exercises, discover the gradient of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.

45 .

$\begin{array}{l}\left(\mathrm{-ane},3\right)\text{ and}\left(\mathrm{v},\mathrm{ane}\right)\\ \left(\mathrm{-2},3\right)\text{ and}\left(0,9\right)\end{array}$

46 .

$\begin{array}{l}\left(\mathrm{two},\mathrm{five}\right)\text{ and}\left(5,9\right)\\ \left(\mathrm{-i},\mathrm{-1}\right)\text{ and}\left(2,3\right)\end{array}$

Technology

For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths identify). Enter this into a graphing calculator as Y1, and so adjust the ymin and ymax values for your window to include where the y-intercept occurs. State your ymin and ymax values.

47 .

$0.537x-\mathrm{2.nineteen}y=100$

48 .

$\mathrm{4,500}\mathrm{10}-200y=\mathrm{nine,528}$

49 .

$\frac{200-30y}{x}=70$

Extensions

50 .

Starting with the point-gradient formula $y-y_1=\mathrm{yard}(\mathrm{ten}-{\mathrm{ten}}_1),$ solve this expression for $\mathrm{ten}$ in terms of $x_1,y,y_1,$ and $m$ .

51 .

Starting with the standard form of an equation $Ax+By=C$ solve this expression for $y$ in terms of $A,B,C$ and $x$ . Then put the expression in slope-intercept form.

52 .

Apply the above derived formula to put the following standard equation in slope intercept course: $\mathrm{seven}x-5y=25.$

53 .

Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle past showing the slopes of the sides that meet are perpendicular.

$(–1,1),(2,0),(\mathrm{iii},3)$ and $(0,4)$

54 .

Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

Real-World Applications

55 .

The gradient for a wheelchair ramp for a abode has to exist $\frac{1}{12}.$ If the vertical distance from the ground to the door lesser is 2.5 ft, notice the distance the ramp has to extend from the home in order to comply with the needed slope.

56 .

If the turn a profit equation for a pocket-size business organisation selling $x$ number of item ane and $y$ number of particular two is $p=3x+\mathrm{four}y,$ discover the $y$ value when $p=\text{\$}453\text{and}\mathrm{10}=75.$

For the post-obit exercises, apply this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the toll would exist $y=45+.25x,$ where $\mathrm{10}$ is the number of miles traveled.

57 .

What is your cost if you lot travel 50 mi?

58 .

If your cost were $\text{\$}63.75,$ how many miles were you charged for traveling?

59 .

Suppose yous have a maximum of $100 to spend for the auto rental. What would be the maximum number of miles yous could travel?

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Source: https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable

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