Skills Refresher:  Applications of Linear Equations(3.5 – 3.7) –  — Intermediate Algebra

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11 Skills Refresher: Applications of Linear Equations(3.5 – 3.7) – — Intermediate Algebra

3.5 Constructing Linear Equations

Quite often, students are required to find the equation of a line given only a data point and a slope or two data points. The simpler of these problems is to find the equation when given a slope and a data point. To do this, use the equation that defines the slope of a line:

$m=\dfrac{y_2-y_1}{x_2-x_1},\text{ where }y_2\text{ is replaced by }y\text{ and }x_2\text{ is replaced by }x$

This becomes:

$m=\dfrac{y-y_1}{x-x_1},\text{ where }x_1\text{ and }y_1\text{ are replaced by the coordinates and }m \text{ by the given slope}$

To illustrate this method, consider the following example.

Example 3.5.1

Find the equation that has slope $m = 2$ and passes through the point $(2, 5)$ .

First, replace $m$ with 2 and $x_1, y_1$ with $(2, 5)$ .

$m=\dfrac{y_2-y_1}{x_2-x_1} \text{ becomes } 2=\dfrac{y-5}{x-2}$

Next, reduce the resulting equation.

First, multiply both sides by the denominator to eliminate the fraction:

$(x-2)2=\dfrac{y-5}{x-2}(x-2)$

This leaves:

$(x - 2) 2 = y - 5$

Which simplifies to:

$2x - 4 = y - 5$

This can be written in the $y$ -intercept form by isolating the variable $y$ :

$\begin{array}{rrrrrrr} 2x&-&4&=&y&-&5 \\ &+&5&&&+&5 \\ \midrule 2x&+&1&=&y&& \\ \\ &&&\text{or}&&& \\ \\ &&y&=&2x&+&1 \end{array}$

It is also useful to write the equation in the general form of $Ax + By + C = 0$ , where $A$ , $B$ , and $C$ are integers and $A$ is positive.

In general form, $2x - 4 = y - 5$ becomes:

$\begin{array}{rrrrrrrrr} 2x&-&4&&&=&y&-&5 \\ &-&y&+&5&&-y&+&5 \\ \midrule 2x&-&y&+&1&=&0&& \end{array}$

The standard form of a linear equation is written as $Ax + By = C$ .

In standard form, $2x - 4 = y - 5$ becomes:

$\begin{array}{rrrrrrr} 2x&-&4&=&y&-&5 \\ -y&+&4&&-y&+&4 \\ \midrule 2x&-&y&=&-1&& \end{array}$

The three common forms that a linear equation can be written in are:

$\begin{array}{rl} \text{Slope-intercept form:}& y = mx + b \\ \\ \text{General form:}&Ax + By + C = 0 \\ \\ \text{Standard form:}&Ax + By = C \end{array}$

Example 3.5.2

Find the equation having slope $m = -\dfrac{1}{2}$ that passes though the point $(1, -4).$

Write the solutions in slope-intercept form and in both general and standard forms.

First, replace $m$ with $-\dfrac{1}{2}$ and $x_1, y_1$ with $(1, -4).$

$m=\dfrac{y_2-y_1}{x_2-x_1}\text{ becomes }-\dfrac{1}{2}=\dfrac{y-4}{x-1}$

Multiplying both sides by $2(x - 1)$ to eliminate the denominators yields:

$-1 (x - 1) = 2 (y + 4)$

Which simplifies to:

$-x + 1 = 2y + 8$

Writing this solution in all three forms looks like:

$\begin{array}{rl} \text{Slope-intercept form:}&y=-\dfrac{1}{2}x-7 \\ \\ \text{General form:}&x+2y+7=0 \\ \\ \text{Standard form:}&x+2y=-7 \end{array}$

The more difficult variant of this type of problem is that in which the equation of a line that connects two data points must be found. However, this is simpler than it may seem.

The first step is to find the slope of the line that would connect those two points. Use the slope equation, as has been done previously in this textbook. After this is done, use this slope and one of the two data points given at the beginning of the problem. The following example illustrates this.

Example 3.5.3

Find the equation of the line that runs through $(-1, -2)$ and $(3, 8)$ .

First, find the slope:

$\begin{array}{rrl} m&=&\dfrac{y_2-y_1}{x_2-x_1} \\ \\ m&=&\dfrac{8-2}{3-1} \text{ or } \dfrac{8+2}{3+1} \\ \\ m&=&\dfrac{10}{4} \text{ or } \dfrac{5}{2} \end{array}$

Now, treat this as a problem of finding a line with a given slope $m$ running through a point $(x, y)$ .

The slope is $\dfrac{5}{2}$ , and there are two points to choose from: $(-1, -2)$ and $(3, 8)$ . Choose the simplest point to work with. For this problem, either point works. For this example, choose $(3, 8)$ .

$\begin{array}{rrl} m&=&\dfrac{y-y_1}{x-x_1} \\ \\ \dfrac{5}{2}&=&\dfrac{y-8}{x-3} \end{array}$

Eliminate the fraction by multiplying both sides by $2(x - 3)$ .

This leaves $5(x-3)=2(y-8),$ which must be simplified:

$\begin{array}{rrrrrrr} 5(x&-&3)&=&2(y&-&8) \\ 5x&-&15&=&2y&-&16 \\ -2y&+&15&&-2y&+&15 \\ \midrule 5x&-&2y&=&-1&& \end{array}$

This answer is in standard form, but it can easily be converted to the $y$ -intercept form or general form if desired.

3.6 Perpendicular and Parallel Lines — Intermediate Algebra

Perpendicular, parallel, horizontal, and vertical lines are special lines that have properties unique to each type. Parallel lines, for instance, have the same slope, whereas perpendicular lines are the opposite and have negative reciprocal slopes. Vertical lines have a constant $x$ -value, and horizontal lines have a constant $y$ -value.

Two equations govern perpendicular and parallel lines:

For parallel lines, the slope of the first line is the same as the slope for the second line. If the slopes of these two lines are called $m_1$ and $m_2$ , then $m_1 = m_2$ .

$\text{The rule for parallel lines is } m_1 = m_2$

Perpendicular lines are slightly more difficult to understand. If one line is rising, then the other must be falling, so both lines have slopes going in opposite directions. Thus, the slopes will always be negative to one another. The other feature is that the slope at which one is rising or falling will be exactly flipped for the other one. This means that the slopes will always be negative reciprocals to each other. If the slopes of these two lines are called $m_1$ and $m_2$ , then $m_1 = \dfrac{-1}{m_2}$ .

$\text{The rule for perpendicular lines is } m_1=\dfrac{-1}{m_2}$

Example 3.6.1

Find the slopes of the lines that are parallel and perpendicular to $y = 3x + 5.$

The parallel line has the identical slope, so its slope is also 3.

The perpendicular line has the negative reciprocal to the other slope, so it is $-\dfrac{1}{3}.$

Example 3.6.2

Find the slopes of the lines that are parallel and perpendicular to $y = -\dfrac{2}{3}x -4.$

The parallel line has the identical slope, so its slope is also $-\dfrac{2}{3}.$

The perpendicular line has the negative reciprocal to the other slope, so it is $\dfrac{3}{2}.$

Typically, questions that are asked of students in this topic are written in the form of “Find the equation of a line passing through point $(x, y)$ that is perpendicular/parallel to $y = mx + b$ .” The first step is to identify the slope that is to be used to solve this equation, and the second is to use the described methods to arrive at the solution like previously done. For instance:

Example 3.6.3

Find the equation of the line passing through the point $(2,4)$ that is parallel to the line $y=2x-3.$

The first step is to identify the slope, which here is the same as in the given equation, $m=2$ .

Now, simply use the methods from before:

$\begin{array}{rrl} m&=&\dfrac{y-y_1}{x-x_1} \\ \\ 2&=&\dfrac{y-4}{x-2} \end{array}$

Clearing the fraction by multiplying both sides by $(x-2)$ leaves:

$2(x-2)=y-4 \text{ or } 2x-4=y-4$

Now put this equation in one of the three forms. For this example, use the standard form:

$\begin{array}{rrrrrrr} 2x&-&4&=&y&-&4 \\ -y&+&4&&-y&+&4 \\ \midrule 2x&-&y&=&0&& \end{array}$

Example 3.6.4

Find the equation of the line passing through the point $(1, 3)$ that is perpendicular to the line $y = \dfrac{3}{2}x + 4.$

The first step is to identify the slope, which here is the negative reciprocal to the one in the given equation, so $m = -\dfrac{2}{3}.$

Now, simply use the methods from before:

$\begin{array}{rrl} m&=&\dfrac{y-y_1}{x-x_1} \\ \\ -\dfrac{2}{3}&=&\dfrac{y-3}{x-1} \end{array}$

First, clear the fraction by multiplying both sides by $3(x - 1)$ . This leaves:

$-2(x - 1) = 3(y - 3)$

which reduces to:

$-2x + 2 = 3y - 9$

Now put this equation in one of the three forms. For this example, choose the general form:

$\begin{array}{rrrrrrrrr} -2x&&&+&2&=&3y&-&9 \\ &&-3y&+&9&&-3y&+&9 \\ \midrule -2x&-&3y&+&11&=&0&& \end{array}$

For the general form, the coefficient in front of the $x$ must be positive. So for this equation, multiply the entire equation by −1 to make $-2x$ positive.

$(-2x -3y + 11 = 0)(-1)$

$2x + 3y - 11 = 0$

Questions that are looking for the vertical or horizontal line through a given point are the easiest to do and the most commonly confused.

Vertical lines always have a single $x$ -value, yielding an equation like $x = \text{constant.}$

Horizontal lines always have a single $y$ -value, yielding an equation like $y = \text{constant.}$

Example 3.6.5

Find the equation of the vertical and horizontal lines through the point $(-2, 4).$

The vertical line has the same $x$ -value, so the equation is $x = -2$ .

The horizontal line has the same $y$ -value, so the equation is $y = 4$ .

3.7 Numeric Word Problems — Intermediate Algebra

Number-based word problems can be very confusing, and it takes practice to convert a word-based sentence into a mathematical equation. The best strategy to solve these problems is to identify keywords that can be pulled out of a sentence and use them to set up an algebraic equation.

Variables that are to be solved for are often written as “a number,” “an unknown,” or “a value.”

“Equal” is generally represented by the words “is,” “was,” “will be,” or “are.”

Addition is often stated as “more than,” “the sum of,” “added to,” “increased by,” “plus,” “all,” or “total.” Addition statements are quite often written backwards. An example of this is “three more than an unknown number,” which is written as $x + 3.$

Subtraction is often written as “less than,” “minus,” “decreased by,” “reduced by,” “subtracted from,” or “the difference of.” Subtraction statements are quite often written backwards. An example of this is “three less than an unknown number,” which is written as $x - 3.$

Multiplication can be seen in written problems with the words “times,” “the product of,” or “multiplied by.”

Division is generally found by a statement such as “divided by,” “the quotient of,” or “per.”

Example 3.7.1

28 less than five times a certain number is 232. What is the number?

28 less means that it is subtracted from the unknown number (write this as −28)
five times an unknown number is written as $5x$
is 232 means it equals 232 (write this as = 232)

Putting these pieces together and solving gives:

$\begin{array}{rrrrrr} 5x&-&28&=&232& \\ &+&28&&+28& \\ \midrule &&5x&=&260& \\ \\ &&x&=&\dfrac{260}{5}&\text{or }52 \end{array}$

Example 3.7.2

Fifteen more than three times a number is the same as nine less than six times the number. What is the number?

Fifteen more than three times a number is $3x + 15$ or $15 + 3x$
is means =
nine less than six times the number is $6x-9$

Putting these parts together gives:

$\begin{array}{rrrrrrr} 3x&+&15&=&6x&-&9 \\ -6x&-&15&=&-6x&-&15 \\ \midrule &&-3x&=&-24&& \\ \\ &&x&=&\dfrac{-24}{-3}&\text{or }8& \\ \end{array}$

Another type of number problem involves consecutive integers, consecutive odd integers, or consecutive even integers. Consecutive integers are numbers that come one after the other, such as 3, 4, 5, 6, 7. The equation that relates consecutive integers is:

$x, x + 1, x + 2, x + 3, x + 4$

Consecutive odd integers and consecutive even integers both share the same equation, since every second number must be skipped to remain either odd (such as 3, 5, 7, 9) or even (2, 4, 6, 8). The equation that is used to represent consecutive odd or even integers is:

$x, x + 2, x + 4, x + 6, x + 8$