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Linear Equations in One Variable with Applications

Solve linear equations in one variable using properties of equations.

Learning Objectives

After completing this section, you should be able to:

Solve linear equations in one variable using properties of equations.
Construct a linear equation to solve applications.
Determine equations with no solution or infinitely many solutions.
Solve a formula for a given variable.

In this section, we will study linear equations in one variable. There are several real-world scenarios that can be represented by linear equations: taxi rentals with a flat fee and a rate per mile; cell phone bills that charge a monthly fee plus a separate rate per text; gym memberships with a monthly fee plus a rate per class taken; etc. For example, if you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month?

Linear Equations and Applications

Solving any equation is like discovering the answer to a puzzle. The purpose of solving an equation is to find the value or values of the variable that makes the equation a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle! There are many types of equations that we will learn to solve. In this section, we will focus on a linear equation, which is an equation in one variable that can be written as

\[ax+b=0\]

where $a$ and $b$ are real numbers and $a\ne 0$, such that $a$ is the coefficient of $x$ and $b$ is the constant.

To solve a linear equation, it is a good idea to have an overall strategy that can be used to solve any linear equation. In the Example 5.12, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Solving a Linear Equation Using a General Strategy

Try it.

Solve $7(n-3)-8=-15$

Solution

Step 1: Simplify each side of the equation as much as possible.	Use the Distributive Property. Notice that each side of the equation is now simplified as much as possible.	$\begin{array}{lll}7(n-3)-8 & = & -15 \\ 7n-21-8 & = & -15 \\ 7n-29 & = & -15\end{array}$
Step 2: Collect all variable terms on one side of the equation.	Nothing to do; all $n$-terms are on the left side.	$7n-29=-15$
Step 3: Collect constant terms on the other side of the equation.	To get constants only on the right, add 29 to each side. Simplify.	$\begin{array}{lll}7n-29+29 & = & -15+29 \\ 7n & = & 14\end{array}$
Step 4: Make the coefficient of the variable term equal to 1.	Divide each side by 7. Simplify.	$\frac{7n}{7}=\frac{14}{7}$ \[n=2\]
Step 5: Check the solution.	Let $n=2$ Subtract.	Check: $\begin{array}{lll}7(n-3)-8 & = & -15 \\ 7(2-3)-8 & \overset{?}{=} & -15 \\ 7(-1)-8 & \overset{?}{=} & -15 \\ -7-8 & \overset{?}{=} & -15 \\ -15 & = & -15✓\end{array}$

In , we used both the addition and division property of equations. All the properties of equations are summarized in table below. Basically, what you do to one side of the equation, you must do to the other side of the equation to preserve equality.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Linear Equations with No Solutions or Infinitely Many Solutions

Every linear equation we have solved thus far has given us one numerical solution. Now we'll look at linear equations for which there are no solutions or infinitely many solutions.

Solving a Linear Equation with No Solution

Try it.

Solve $3(x+4)=4x+8-x$.

Solution

Step 1: Simplify each side. $3(x+4)=4x+8-x$ \[3x+12-3x=3x+8-3x\]

Step 2: Collect all variables to one side. $3x+12-3x=3x+8-3x$

\[12=8\]

The variable $x$ disappeared! When this happens, you need to examine what remains. In this particular case, we have $12=8$, which is not a true statement. When you have a false statement, then you know the equation has no solution; there does not exist a value for $x$ that can be put into the equation that will make it true.

Solving a Linear Equation with Infinitely Many Solutions

Try it.

Solve $2(x+5)=4(x+3)-2x-2$.

Solution

Step 1: \[\begin{array}{lll}2(x+5) & = & 4(x+3)-2x-2 \\ 2x+10 & = & 4x+12-2x-2 \\ 2x+10 & = & 2x+10\end{array}\]

Step 2: \[\begin{array}{lll}2x+10-2x & = & 2x+10-2x \\ 10 & = & 10\end{array}\]

As with the previous example, the variable disappeared. In this case, however, we have a true statement ($10=10$). When this occurs we say there are infinitely many solutions; any value for $x$ will make this statement true.

Solving a Formula for a Given Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be able to manipulate formulas and solve for specific variables.

To solve a formula for a specific variable means to isolate that variable on one side of the equal sign with a coefficient of 1. All other variables and constants are on the other side of the equal sign. To see how to solve a formula for a specific variable, we will start with the distance, rate, and time formula.

Solving for a Given Variable with Distance, Rate, and Time

Try it.

Solve the formula $d=rt$ for $t$. This is the distance formula where $d$ = distance, $r$ = rate, and $t$ = time.

Solution

Divide both sides by $r$: $d/r=rt/r$

\[d/r=t\]

Solving for a Given Variable in the Area Formula for a Triangle

Try it.

Solve the formula $A=½$ $\text{bh}$ for $h$. This is the area formula of a triangle where $A$ = area, $b$ = base, and $h$ = height.

Solution

Step 1: Multiply both sides by 2. \[\begin{array}{l}2A=2(½\ bh) \\ 2A=bh\end{array}\]

Step 2: Divide both sides by $b$. \[\begin{array}{lll}\frac{2A}{b} & = & \frac{bh}{b} \\ \frac{2A}{b} & = & h \\ h & = & \frac{2A}{b}\end{array}\]

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

Solving linear equations means discovering what the value of the variable in a linear equation represents in the given conditions.
When solving a linear equation, most often you will have one solution; however, a linear equation may have no solutions or infinitely many solutions.

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve $7(n-3)-8=-15$

Jawaby görkez

Step 1: Simplify each side of the equation as much as possible.	Use the Distributive Property. Notice that each side of the equation is now simplified as much as possible.	$\begin{array}{lll}7(n-3)-8 & = & -15 \\ 7n-21-8 & = & -15 \\ 7n-29 & = & -15\end{array}$
Step 2: Collect all variable terms on one side of the equation.	Nothing to do; all $n$-terms are on the left side.	$7n-29=-15$
Step 3: Collect constant terms on the other side of the equation.	To get constants only on the right, add 29 to each side. Simplify.	$\begin{array}{lll}7n-29+29 & = & -15+29 \\ 7n & = & 14\end{array}$
Step 4: Make the coefficient of the variable term equal to 1.	Divide each side by 7. Simplify.	$\frac{7n}{7}=\frac{14}{7}$ \[n=2\]
Step 5: Check the solution.	Let $n=2$ Subtract.	Check: $\begin{array}{lll}7(n-3)-8 & = & -15 \\ 7(2-3)-8 & \overset{?}{=} & -15 \\ 7(-1)-8 & \overset{?}{=} & -15 \\ -7-8 & \overset{?}{=} & -15 \\ -15 & = & -15✓\end{array}$

Solve $9(y-2)-y=16+7y$.

Jawaby görkez

Step 1: Simplify each side. \[\begin{array}{lll}9(y-2)-y & = & 16+7y \\ 9y-18-y & = & 16+7y \\ 8y-18 & = & 16+7y\end{array}\]
Step 2: Collect all variables on one side. \[\begin{array}{lll}8y-18-7y & = & 16+7y-7y \\ y-18 & = & 16\end{array}\]

Step 3: Collect constant terms on one side. \[\begin{array}{lll}y-18+18 & = & 16+18 \\ y & = & 34\end{array}\]
Step 4: Make the coefficient of the variable 1. Already done!
Step 5: Check. \[\begin{array}{lll}9(34)-18-(34) & \overset{?}{=} & 16+7(34) \\ 306-18-34 & \overset{?}{=} & 16+238 \\ 288-34 & \overset{?}{=} & 254 \\ 254 & = & 254✓\end{array}\]
The Beaudrie family has two cats, Basil and Max. Together, they weigh 23 pounds. Basil weighs 16 pounds. How much does Max weigh?

Jawaby görkez

Let $b$ = Basil’s weight and $m$ = Max’s weight.
\[b+m=23\]
We also know that Basil weighs 16 pounds so:
Steps 1 and 2: $16+m=23$
Since both sides are simplified, the variable is on one side of the equation, we start in Step 3 and collect the constants on one side:
Step 3: \[\begin{array}{lll}16+m-16 & = & 23-16 \\ m & = & 7\end{array}\]
Step 4: is already done so we go to Step 5:
Step 5: \[\begin{array}{lll}16+7 & \overset{?}{=} & 23 \\ 23 & = & 23✓\end{array}\]
Basil weighs 16 pounds and Max weighs 7 pounds.
If you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month?

Jawaby görkez

If we let $x$ = number of classes, the expression $5x+10$ would represent what you pay per month if each class is $5 and there’s a $10 monthly fee per class. $10 is your constant. If you want to know how many classes you can take if you have a $75 monthly gym budget, set the equation equal to 75. Then solve the equation $5x+10=75$ for $x$.
Steps 1 and 2: \[5x+10=75\]
Step 3: \[\begin{array}{lll}5x+10-10 & = & 75-10 \\ 5x & = & 65\end{array}\]
Step 4: \[\begin{array}{lll}\frac{5x}{5} & = & \frac{65}{5} \\ x & = & 13\end{array}\]
Step 5: \[\begin{array}{lll}5(13)+10 & \overset{?}{=} & 75 \\ 65+10 & \overset{?}{=} & 75 \\ 75 & = & 75✓\end{array}\]
The solution is 13 classes. You can take 13 classes on a $75 monthly gym budget.
Write an application that can be solved using the equation $50x+35=185$. Then solve your application.

Jawaby görkez

Answers will vary. Let’s say you want to rent a snowblower for a huge winter storm coming up. If $x$ = the number of days you rent a snowblower, then the expression $50x+35$ represents what you pay if, for each day, it costs $50 to rent the snowblower and there is a $35 flat rental fee. $35 is the constant. To find out how many days you can rent a snowblower for $185, set the expression equal to 185. Then solve the equation $50x+35=185$ for $x$.
Steps 1 and 2: \[50x+35=185\]
Step 3: \[\begin{array}{lll}50x+35-35 & = & 185-35 \\ 50x & = & 150\end{array}\]
Step 4: \[\begin{array}{lll}\frac{50x}{50} & = & \frac{150}{50} \\ x & = & 3\end{array}\]
Step 5: \[\begin{array}{lll}50(3)+35 & \overset{?}{=} & 185 \\ 150+35 & \overset{?}{=} & 185 \\ 185 & = & 185✓\end{array}\]
The equation is $50x+35=185$ and the solution is 3 days. You can rent a snowblower for 3 days on a $185 budget.
Solve $3(x+4)=4x+8-x$.

Jawaby görkez

Step 1: Simplify each side. $3(x+4)=4x+8-x$ \[3x+12-3x=3x+8-3x\]
Step 2: Collect all variables to one side. $3x+12-3x=3x+8-3x$
\[12=8\]
The variable $x$ disappeared! When this happens, you need to examine what remains. In this particular case, we have $12=8$, which is not a true statement. When you have a false statement, then you know the equation has no solution; there does not exist a value for $x$ that can be put into the equation that will make it true.
Solve $2(x+5)=4(x+3)-2x-2$.

Jawaby görkez

Step 1: \[\begin{array}{lll}2(x+5) & = & 4(x+3)-2x-2 \\ 2x+10 & = & 4x+12-2x-2 \\ 2x+10 & = & 2x+10\end{array}\]
Step 2: \[\begin{array}{lll}2x+10-2x & = & 2x+10-2x \\ 10 & = & 10\end{array}\]
As with the previous example, the variable disappeared. In this case, however, we have a true statement ($10=10$). When this occurs we say there are infinitely many solutions; any value for $x$ will make this statement true.
Solve the formula $d=rt$ for $t$. This is the distance formula where $d$ = distance, $r$ = rate, and $t$ = time.

Jawaby görkez

Divide both sides by $r$: $d/r=rt/r$
\[d/r=t\]
Solve the formula $A=½$ $\text{bh}$ for $h$. This is the area formula of a triangle where $A$ = area, $b$ = base, and $h$ = height.

Jawaby görkez

Step 1: Multiply both sides by 2. \[\begin{array}{l}2A=2(½\ bh) \\ 2A=bh\end{array}\]
Step 2: Divide both sides by $b$. \[\begin{array}{lll}\frac{2A}{b} & = & \frac{bh}{b} \\ \frac{2A}{b} & = & h \\ h & = & \frac{2A}{b}\end{array}\]

Symbols used here

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Linear Equations in One Variable with Applications

Solve linear equations in one variable using properties of equations.
Construct a linear equation to solve applications.
Determine equations with no solution or infinitely many solutions.
Solve a formula for a given variable.
Did the trick work? How do you know?
Why did this occur? In other words, how does this trick work?
linear equation
Solving linear equations means discovering what the value of the variable in a linear equation represents in the given conditions.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Özüňi synla

Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

_Ýaşa Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Step 1: Simplify each side of the equation as much as possible.	Use the Distributive Property. Notice that each side of the equation is now simplified as much as possible.	\(\begin{array}{lll}7(n-3)-8 & = & -15 \\ 7n-21-8 & = & -15 \\ 7n-29 & = & -15\end{array}\)
Step 2: Collect all variable terms on one side of the equation.	Nothing to do; all \(n\)-terms are on the left side.	\(7n-29=-15\)
Step 3: Collect constant terms on the other side of the equation.	To get constants only on the right, add 29 to each side. Simplify.	\(\begin{array}{lll}7n-29+29 & = & -15+29 \\ 7n & = & 14\end{array}\)
Step 4: Make the coefficient of the variable term equal to 1.	Divide each side by 7. Simplify.	\(\frac{7n}{7}=\frac{14}{7}\) \[n=2\]
Step 5: Check the solution.	Let \(n=2\) Subtract.	Check: \(\begin{array}{lll}7(n-3)-8 & = & -15 \\ 7(2-3)-8 & \overset{?}{=} & -15 \\ 7(-1)-8 & \overset{?}{=} & -15 \\ -7-8 & \overset{?}{=} & -15 \\ -15 & = & -15✓\end{array}\)