10 Linear Equations in One Variable

Solving Linear Equations in One Variable

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form[latex]\,ax+b=0\,[/latex]and are solved using basic algebraic operations.

We begin by classifying linear equations in one variable as one 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.

The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for[latex]\,x\,[/latex]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[latex]\,5x+2=3x-6,[/latex]we have the following:

The solution set consists of one number:[latex]\,\left\{-4\right\}.\,[/latex]It is the only solution and, therefore, we have solved a conditional equation.

An inconsistent equation results in a false statement. For example, if we are to solve[latex]\,5x-15=5\left(x-4\right),[/latex]we have the following:

Indeed,[latex]\,-15\ne \,-20.\,[/latex]There is no solution because this is an inconsistent equation.

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

Solving a Rational Equation

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

Recall that a rational number is the ratio of two numbers, such as[latex]\,\frac{2}{3}\,[/latex]or[latex]\,\frac{7}{2}.\,[/latex]A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

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 by the least 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 do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as[latex]\,\left(x+1\right).\,[/latex]Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are[latex]\,x,[/latex][latex]x-1,[/latex]and[latex]\,3x-3.\,[/latex]First, factor all denominators. We then have[latex]\,x,[/latex][latex]\left(x-1\right),[/latex]and[latex]\,3\left(x-1\right)\,[/latex]as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of[latex]\,\left(x-1\right).\,[/latex]The[latex]\,x\,[/latex]in the first denominator is separate from the[latex]\,x\,[/latex]in the[latex]\,\left(x-1\right)\,[/latex]denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the[latex]\,x,[/latex]one factor of[latex]\,\left(x-1\right),[/latex]and the 3. Thus, the LCD is the following:

So, both sides of the equation would be multiplied by[latex]\,3x\left(x-1\right).\,[/latex]Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as[latex]\,x\,[/latex]and[latex]\,{x}^{2}+2x.\,[/latex]Once the second denominator is factored as[latex]\,{x}^{2}+2x=x\left(x+2\right),[/latex]there is a common factor of x in both denominators and the LCD is[latex]\,x\left(x+2\right).[/latex]

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

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

Multiply[latex]\,a\left(d\right)\,[/latex]and[latex]\,b\left(c\right),[/latex]which results in[latex]\,ad=bc.[/latex]

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

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as[latex]\,y=mx+b,[/latex]where[latex]\,m=\text{slope}\,[/latex]and[latex]\,b=y\text{−intercept}\text{.}\,[/latex]Let us begin with the slope.

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 any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

[latex]m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex]

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in (Figure). The lines indicate the following slopes:[latex]\,m=-3,[/latex][latex]m=2,[/latex]and[latex]\,m=\frac{1}{3}.[/latex]

The Slope of a Line

The slope of a line, m, represents the change in y over the change in x. Given two points,[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]and[latex]\,\left({x}_{2},{y}_{2}\right),[/latex]the following formula determines the slope of a line containing these points:

[latex]m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex]

Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points[latex]\,\left(2,-1\right)\,[/latex]and[latex]\,\left(-5,3\right).[/latex]

[reveal-answer q=”fs-id2715022″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id2715022″]

We substitute the y-values and the x-values into the formula.

[latex]\begin{array}{ccc}\hfill m& =& \frac{3-\left(-1\right)}{-5-2}\hfill \\ & =& \frac{4}{-7}\hfill \\ & =& -\frac{4}{7}\hfill \end{array}[/latex]

The slope is[latex]\,-\frac{4}{7}.[/latex][/hidden-answer]

Analysis

It does not matter which point is called[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]or[latex]\,\left({x}_{2},{y}_{2}\right).\,[/latex]As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.

Try It

Find the slope of the line that passes through the points[latex]\,\left(-2,6\right)\,[/latex]and[latex]\,\left(1,4\right).[/latex]

[reveal-answer q=”fs-id2437070″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id2437070″]

[latex]m=-\frac{2}{3}[/latex]

[/hidden-answer]

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

Identify the slope and y-intercept, given the equation[latex]\,y=-\frac{3}{4}x-4.[/latex]

[reveal-answer q=”fs-id1763671″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1763671″]

As the line is in[latex]\,y=mx+b\,[/latex]form, the given line has a slope of[latex]\,m=-\frac{3}{4}.\,[/latex]The y-intercept is[latex]\,b=-4.[/latex]

[/hidden-answer]

Analysis

The y-intercept is the point at which the line crosses the y-axis. On the y-axis,[latex]\,x=0.\,[/latex]We can always identify the y-intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute[latex]\,x=0\,[/latex]and solve for y.

Vertical and Horizontal Lines

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

[latex]x=c[/latex]

where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c.

Suppose that we want to find the equation of a line containing the following points:[latex]\,\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right),[/latex]and[latex]\,\left(-3,5\right).\,[/latex]First, we will find the slope.

[latex]m=\frac{5-3}{-3-\left(-3\right)}=\frac{2}{0}[/latex]

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through[latex]\,x=-3.\,[/latex]See (Figure).

The equation of a horizontal line is given as

[latex]y=c[/latex]

where c is a constant. The slope of a horizontal line is zero, and for any x-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 following set of points:[latex]\,\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right),[/latex]and[latex]\,\left(5,-2\right).[/latex]We can use the point-slope formula. First, we find the slope using any two points on the line.

[latex]\begin{array}{ccc}\hfill m& =& \frac{-2-\left(-2\right)}{0-\left(-2\right)}\hfill \\ & =& \frac{0}{2}\hfill \\ & =& 0\hfill \end{array}[/latex]

Use any point for[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]in the formula, or use the y-intercept.

[latex]\begin{array}{ccc}\hfill y-\left(-2\right)& =& 0\left(x-3\right)\hfill \\ \hfill y+2& =& 0\hfill \\ \hfill y& =& -2\hfill \end{array}[/latex]

The graph is a horizontal line through[latex]\,y=-2.\,[/latex]Notice that all of the y-coordinates are the same. See (Figure).

Finding the Equation of a Line Passing Through the Given Points

Find the equation of the line passing through the given points:[latex]\,\left(1,-3\right)\,[/latex]and[latex]\,\left(1,4\right).[/latex]

[reveal-answer q=”fs-id2764547″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id2764547″]

The x-coordinate of both points is 1. Therefore, we have a vertical line,[latex]\,x=1.[/latex]

[/hidden-answer]

Try It

Find the equation of the line passing through[latex]\,\left(-5,2\right)\,[/latex]and[latex]\,\left(2,2\right).[/latex]

[reveal-answer q=”fs-id1551241″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1551241″]

Horizontal line:[latex]\,y=2[/latex]

[/hidden-answer]

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines have the same slope and different y-intercepts. Lines that are parallel to each other will never intersect. For example, (Figure) shows the graphs of various lines with the same slope,[latex]\,m=2.[/latex]

All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts.

Lines that are perpendicular intersect to form a[latex]\,90°[/latex]-angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is[latex]\,-1:{m}_{1}\cdot {m}_{2}=-1.\,[/latex]For example, (Figure) shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of[latex]\,-\frac{1}{3}.[/latex]

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:[latex]\,3y=-4x+3\,[/latex]and[latex]\,3x-4y=8.[/latex]

[reveal-answer q=”fs-id2931413″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id2931413″]

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

[latex]\begin{array}{ccc}\hfill 3y& =& -4x+3\hfill \\ \hfill y& =& -\frac{4}{3}x+1\hfill \end{array}[/latex]

Second equation:

[latex]\begin{array}{ccc}\hfill 3x-4y& =& 8\hfill \\ \hfill -4y& =& -3x+8\hfill \\ \hfill y& =& \frac{3}{4}x-2\hfill \end{array}[/latex]

See the graph of both lines in (Figure)

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

[latex]\begin{array}{ccc}\hfill {m}_{1}& =& -\frac{4}{3}\hfill \\ \hfill {m}_{2}& =& \frac{3}{4}\hfill \\ \hfill {m}_{1}\cdot {m}_{2}& =& \left(-\frac{4}{3}\right)\left(\frac{3}{4}\right)=-1\hfill \end{array}[/latex]

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.[/hidden-answer]

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

As we have 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 same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

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

Write the equation of line parallel to a[latex]\,5x+3y=1\,[/latex]and passing through the point[latex]\,\left(3,5\right).[/latex]

[reveal-answer q=”fs-id1531339″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1531339″]

First, we will write the equation in slope-intercept form to find the slope.

[latex]\begin{array}{ccc}\hfill 5x+3y& =& 1\hfill \\ \hfill 3y& =& 5x+1\hfill \\ \hfill y& =& -\frac{5}{3}x+\frac{1}{3}\hfill \end{array}[/latex]

The slope is[latex]\,m=-\frac{5}{3}.\,[/latex]The y-intercept is[latex]\,\frac{1}{3},[/latex]but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

[latex]\begin{array}{ccc}\hfill y-5& =& -\frac{5}{3}\left(x-3\right)\hfill \\ \hfill y-5& =& -\frac{5}{3}x+5\hfill \\ \hfill y& =& -\frac{5}{3}x+10\hfill \end{array}[/latex]

The equation of the line is[latex]\,y=-\frac{5}{3}x+10.\,[/latex]See (Figure).

[/hidden-answer]