# 1.2 Linear Equations – Manipulating and Solving

You are shopping at Old Navy for seven new outfits. The price points are $10 and $30, with the addition of sales tax of 13%. You really like the $30 outfits; however, your total budget can’t exceed $110. How do you spend $110 to acquire all the needed outfits without exceeding your budget while getting as many $30 items as possible?

This is a problem of linear equations, and it illustrates how you can use them to make an optimal decision. Let [latex]L[/latex] represent the quantity of clothing at the low price point of $10$, and [latex]H[/latex] represent the quantity of clothing at the high price point of $30. These conditions result in the following algebraic equations:

*the total number of outfits you need:* [latex]L + H = 7[/latex]

*your total budget:* [latex]10L + 30H +0.13(10L + 30H)= 110[/latex]

By simultaneously solving these equations you can determine how many outfits at each price point you can purchase.

You will encounter many situations like this in your business career, such as in making the best use of a manufacturer’s production capacity. For example, assume your company makes two products on the same production line and sells all of its output. Each product contributes differently to your profitability, and each product takes a different amount of time to manufacture. A natural question arises: what combination of each of these products should you make such that you operate your production line at capacity while also maximizing the profits earned?

We will start that discussion in this section by first exploring how to solve linear equations for unknown variables.

## Understanding Equations

To manipulate algebraic equations and solve for unknown variables, you must first become familiar with some important language. This includes terms such as *linear* versus* nonlinear equations*, and *sides of the equation*.

For example, consider the equation [latex]4x+3=-2x-3[/latex].

Every equation has two sides:

[latex]4x+3[/latex] is the left side

[latex]−2x − 3[/latex] is the right side

[latex]x[/latex] is the variable:

Note that the variable has an exponent of 1. Such equations, where all terms are either constants or monomials with a single variable with exponent of 1 are are called **linear equations**. All other equations are **nonlinear equations**, such as, for example, those that involve a variable with an exponent other than 1**.**

Important: [latex]4x+3=-2x-3[/latex] is the same thing as [latex]-2x-3=4x+3[/latex]. The only thing that changes is how we refer to expressions on each side. In this case, what was the left side before now becomes the right side, and vice versa.

Equations represent a condition on some variables and so typically, when working with equations, the goal is to algebraically manipulate the equation to solve it for a value for the unknown variable that makes the equation true. If you substitute a value of [latex]x = −1[/latex] into the above example, the left-hand side of the equation equals the right-hand side of the equation. The value of [latex]x = −1[/latex] is known as the **root**, or solution, to the linear equation.

## Solving One Linear Equation with One Unknown Variable

In your study of solving linear equations, you need to start by manipulating a single equation to solve for a single unknown variable. Later in this section you will build on this to problems involving solutions to two linear equations with two unknowns.

Remember that “to solve an equation” means to find the value of an unknown variable in the equation using the other quantities in the equation. We can achieve that by applying a sequence of algebraic operations on both sides of the equation simultaneously (so that the original equation still holds) in a way that allows us to isolate the unknown variable on one side of the equation, having it equal to some algebraic expression on the other side.

The fundamental idea underlying this process is that, if two quantities are equal, then the result of applying the same operation on each (such as adding, subtracting, multiplying or dividing by another quantity) will again result in equal quantities.

For example, if [latex]A=B[/latex], then [latex]A+C=B+C[/latex] for any value [latex]C[/latex].

Here are a couple of examples, before we attempt to formulate the general process.

For example, to solve [latex]4x + 3 = −2x − 3[/latex], consider the following:

Analysis | Strategy | Algebraic process | |
---|---|---|---|

Question | Answer | ||

What is the task? | Solve for [latex]x[/latex] | Rearrange equation with [latex]x[/latex] on one side and all other quantities on the other. Continue analyzing. | Starting with:
[latex]4x + 3 = −2x − 3,[/latex] |

Can either side of the equation be simplified? | No. | Continue analyzing. | |

Where is [latex]x[/latex] in the equation? | On both sides, as part of two terms. | Rearrange to have terms with [latex]x[/latex] on one side only. Add [latex]2x[/latex] to both sides to remove [latex]-2x[/latex] from the right side. | [latex]\Rightarrow 4x + 3+2x = − 3[/latex] |

Can either side of the equation be simplified? | Yes. Left side is adding two like terms. | Add the two like terms on the left side. | [latex]\Rightarrow 6x + 3 = − 3[/latex] |

Where is [latex]x[/latex] in the equation? | On left side, as part of a single term. | Continue analyzing. | |

What are the operations that are applied to [latex]x[/latex] and in which order? | [latex]x[/latex] is first multiplied by 6, then 3 is added to the result. | Determine the last operation and undo it by applying the opposite operation to both sides. Here, the last operation is add 3, so should subtract 3 from both sides. | [latex]\Rightarrow 6x = − 3-3[/latex] |

Can either side of the equation be simplified? | Yes. Right side is subtracting two like terms. | Subtract two like terms on right side. | [latex]\Rightarrow 6x = − 6[/latex] |

Where is [latex]x[/latex] in the equation? | On left side only, as part of a single term. | Continue analyzing. | |

What are the operations that are applied to [latex]x[/latex] and in which order? | [latex]x[/latex] is multiplied by 6. | Determine the last operation and undo it by applying the opposite operation to both sides. Here, the last operation is multiply by 6, so should divide both sides by 6. | [latex]\Rightarrow x = − 1[/latex] |

Can either side of the equation be simplified? | No. | Continue analyzing. | |

Where is [latex]x[/latex] in the equation? | On left side only, on its own. | Done. |

Now consider a more general type of example: Solve [latex]px + b = −tx − b[/latex] for the variable [latex]x[/latex]:

Analysis | Strategy | Algebraic process | |
---|---|---|---|

Question | Answer | ||

What is the task? | Solve for [latex]x[/latex] | Rearrange equation with [latex]x[/latex] on one side and all other quantities on the other. Continue analyzing. | Starting with:
[latex]px + b = −tx − b,[/latex] |

Can either side of the equation be simplified? | No. | Continue analyzing. | |

Where is [latex]x[/latex] in the equation? | On both sides, as part of two terms. | Rearrange to have terms with [latex]x[/latex] on one side only. Add [latex]tx[/latex] to both sides to remove [latex]-tx[/latex] from the right side. | [latex]\Rightarrow px + b +tx= − b[/latex] |

Can either side of the equation be simplified? | Yes. Left side is adding two like terms. | Add the two like terms on the left side. | [latex]\Rightarrow (p+t)x + b = − b[/latex] |

Where is [latex]x[/latex] in the equation? | On left side, as part of a single term. | Continue analyzing. | |

What are the operations that are applied to [latex]x[/latex] and in which order? | [latex]x[/latex] is first multiplied by [latex]p+t[/latex], then b is added to the result. | Determine the last operation and undo it by applying the opposite operation to both sides. Here, the last operation is add b, so should subtract b from both sides. | [latex]\Rightarrow (p+t)x = − b-b[/latex] |

Can either side of the equation be simplified? | Yes. Right side is subtracting two like terms. | Subtract two like terms on right side. | [latex]\Rightarrow (p+t)x = − 2b[/latex] |

Where is [latex]x[/latex] in the equation? | On left side only, as part of a single term. | Continue analyzing. | |

What are the operations that are applied to [latex]x[/latex] and in which order? | [latex]x[/latex] is multiplied by [latex]p+t[/latex]. | Determine the last operation and undo it by applying the opposite operation to both sides. Here, the last operation is multiply by [latex]p+t[/latex], so should divide both sides by [latex]p+t[/latex]. | [latex]\Rightarrow x = − \frac{2b}{p+t}[/latex] |

Can either side of the equation be simplified? | No. | Continue analyzing. | |

Where is [latex]x[/latex] in the equation? | On left side only, on its own. | Done. |

As you can see, even though the algebraic process did not include many or complicated steps, the analysis process required is quite involved. Since there infinitely many equation possibilities, there is no one single process that will work in solving every equation. However, some general principles apply, applied in a series of cycles governed by analysis-strategy-deployment cycle.

**General principles in solving an equation for an unknown variable**

Always start by identifying the task, i.e., the deliverable. Which variable are you solving for and from which equation?

Cycle through Analysis-Strategy-Deployment until done: the unknown variable is isolated, on its own, on one side of the equation.

As part of each cycle, ask yourself a question, then develop a strategy based on the answer and put the strategy into action. Some analysis-strategy-deployment paths:

- Can the expressions on each side of the equation be simplified through algebraic operations? If yes, simplify. If no, continue analyzing.

Example: [latex]p=?[/latex]

[latex]3p-\frac{4}{9}+1.2p=p-7+5\ \ \ \Rightarrow\ \ \ 4.2p-\frac{4}{9}=p-2[/latex]

- Is the unknown variable in the denominator of a fraction? If yes, multiply both sides of the equation by the term(s) in the denominator(s) that contain the unknown variable and make note that the value of the variable cannot be 0 (otherwise you are multiplying everything by 0 and getting [latex]0=0[/latex]). If no, continue analyzing.

Example: [latex]r=?[/latex]

[latex]7m=\frac{3(t+1)}{mr}+5\ \ \ \Rightarrow\ \ \ 7m^2r=3(t+1)+5mr[/latex]

- Is the unknown variable in a term inside brackets? If yes, get it out of the bracket by multiplying each term in the bracket with the term that multiplies the bracket. If no, continue analyzing.

Example: [latex]k=?[/latex]

[latex]5x+p(r+6)=2m(3-2k)\ \ \ \Rightarrow\ \ \ 5x+p(r+6)=6m-4mk[/latex]

- Is the unknown variable on both sides of the equation, but not in a bracket or in a denominator? If yes, add or subtract the terms with the unknown variable on one side appropriately so that they are removed from that side and their opposites are added to the other side of the equation. If no, continue analyzing.

Example: [latex]p=?[/latex]

[latex]2pk-3r=5p+8k\ \ \ \Rightarrow\ \ \ 2pk-3r-5p=8k[/latex]

- Is the unknown variable only on one side of the equation? If yes, identify which operations are applied to the variable and in what order (follow the order of operations rules); then undo the last operation by applying the opposite of the operation to both sides.

Example: [latex]t=?[/latex]

[latex]\frac{7t-4}{2p}=5-6p\ \ \ \Rightarrow\ \ \ 7t-4=2p(5-6p)[/latex]

- Repeat until done, with the unknown variable isolated to one side of the equation, on its own.

#### Concept in Action – Demo by Video

#### Important Notes

When you are unsure whether your calculated solution is accurate, an easy way to verify your answer is to take the original equation and substitute your solution in place of the variable. If you have the correct solution, the left-hand side of the equation equals the right-hand side of the equation. If you have an incorrect solution, the two sides will be unequal. The inequality typically results from one of the three most common errors in algebraic manipulation:

- The rules of order of operations (also sometimes referred to as BEDMAS) have been broken.
- The rules of algebra have been violated.
- What was done to one side of the equation was not done to the other side of the equation.

#### Things To Watch Out For

When you move a term from one side of the equation to another using multiplication or division, remember that this affects each and every term on both sides of the equation. To remove the [latex]x[/latex] from the denominator in the following equation, multiply both sides of the equation by [latex]x[/latex]:

[latex]\frac{5}{x} + \frac{1}{x} = \frac{2}{x} + 2[/latex] becomes [latex]x\left( \frac{5}{x} + \frac{1}{x} \right) = \left( \frac{2}{x} + 2\right)x[/latex] which then becomes [latex]5 + 1 = 2 + 2x[/latex]

Multiplying every term on both sides by [latex]x[/latex] maintains the equality as long as the value of [latex]x[/latex] is not allowed to be 0.

#### Paths To Success

Negative numbers can cause some people a lot of grief. In moving terms from a particular side of the equation, many people prefer to avoid negative numerical coefficients in front of literal coefficients. Revisiting [latex]4x + 3 = −2x − 3[/latex], you could move the [latex]4x[/latex] from the left side to the right side by subtracting [latex]4x[/latex] from both sides. However, on the right side this results in [latex]−6x[/latex]. The negative is easily overlooked or accidentally dropped in future steps. Instead, move the variable to the left side of the equation, yielding a positive coefficient of [latex]6x[/latex].

#### Concept in Action – Demo by Video

#### Example 1.2 A: Solving a Linear Equation with One Unknown Variable

Solve the following equation for [latex]m[/latex]:

[latex]\frac{3m}{4} + 2m = 4m − 15[/latex]

**Answer:**

Task: [latex]m=?[/latex]

Condition: [latex]\frac{3m}{4} + 2m = 4m − 15[/latex]

Therefore we have:

[latex]\begin{align*} \frac{3m}{4} + 2m = 4m − 15&\Rightarrow \frac{3}{4}m + 2m -4m= − 15\\ &\Rightarrow -\frac{5}{4}m=-15\\ &\Rightarrow m=-15\cdot\left(-\frac{4}{5}\right)\\ &\Rightarrow m=12 \end{align*}[/latex]

#### Example 1.2 B: Solving a Linear Equation with One Unknown Variable Containing Fractions

Solve the following equation for [latex]b[/latex], rounding the final answer to four decimals:

[latex]\frac{5}{8}b+ \frac{2}{5} = \frac{17}{20} −\frac{b}{4}[/latex]

**Answer:**

Task: [latex]b=?[/latex]

Condition: [latex]\frac{5}{8}b+ \frac{2}{5} = \frac{17}{20} −\frac{b}{4}[/latex]

We don’t have to, but it would be easier to deal with the equation if we could eliminate the fractions. To do that, we find a common denominator and multiply both sides by it.

[latex]\begin{align*} \frac{5}{8}b+ \frac{2}{5} = \frac{17}{20} −\frac{b}{4}&\Rightarrow 40\left(\frac{5}{8}b+ \frac{2}{5}\right) = 40\left(\frac{17}{20} −\frac{b}{4}\right)\\ &\Rightarrow 25b+16=34-10b\\ &\Rightarrow 25b+16+10b=34\\ &\Rightarrow 35b=34-16\\ &\Rightarrow 35b=18\\ &\Rightarrow b=\frac{18}{35}\approx 0.5143 \end{align*}[/latex]

## Solving Two Linear Equations with Two Unknown Variables

The manipulation process you have just practiced works well for solving one linear equation with one variable. But what happens if you need to solve two linear equations with two variables simultaneously? Remember when you were at Old Navy purchasing seven outfits earlier in this chapter and you needed to stay within a pricing budget? Each equation had two unknown variables, one representing the number of lower-priced and the other the number of higher-priced outfits.

When dealing with multiple equations with multiple variables. the goal is to manipulate the equations into a single equation with only one unknown. Once this transformation is complete, you then follow the process of solving an equation for the unknown variable by applying the analysis-strategy-deployment process for solving a linear equation, as just discussed.

When you work with multiple linear equations with multiple unknowns, the rules of algebra permit the following two manipulations:

- Substitution: Solve for one of the variables using one of the equations, then substitute the expression for that variable into the other equation. This will reduce the number of variables in the other equation.
- Elimination: Decide which variable you would like to eliminate. Then multiply each of the equations with the coefficient of that variable from the other equation. This will result in two equations that have the same coefficient for the chosen variable. Then subtract one equation from the other, or their corresponding sides, to eliminate the chosen variable and get a single equation with a single variable.

Let’s consider the following example and solve it using both approaches: Solve for [latex]x[/latex] and [latex]y[/latex] if

[latex]\begin{align*} 4.9x + 1.5y &= 38.3\\ 2.7x − 8.6y &= 17.8 \end{align*}[/latex]

**Solution by substitution:**

Start by stating the system of the equations. | [latex]\begin{align*} 4.9x + 1.5y &= 38.3\\ 2.7x − 8.6y &= 17.8 \end{align*}[/latex] |

Choose one equation and solve for one of the variables. Say we pick the first equation and we solve for [latex]x[/latex]. | [latex]\begin{align*} 4.9x + 1.5y = 38.3&\Rightarrow 4.9x = 38.3- 1.5y\\ &\Rightarrow x = \frac{38.3}{4.9}- \frac{1.5}{4.9}y \end{align*}[/latex] |

Substitute the new expression for [latex]x[/latex] into the other equation and solve for [latex]y[/latex]. | [latex]\begin{align*} 2.7x − 8.6y = 17.8&\Rightarrow 2.7\left(\frac{38.3}{4.9}- \frac{1.5}{4.9}y\right) − 8.6y = 17.8\\ &\Rightarrow \frac{103.41}{4.9}-\frac{4.05}{4.9}y − 8.6y = 17.8\\ &\Rightarrow -\frac{46.19}{4.9}y = 17.8-\frac{103.41}{4.9}\\ &\Rightarrow -\frac{46.19}{4.9}y = -\frac{16.19}{4.9}\\ &\Rightarrow y = -\frac{16.19}{4.9}\cdot\left(-\frac{4.9}{46.19}\right)\\ &\Rightarrow y = \frac{16.19}{46.19} \end{align*}[/latex] |

Use the solution for [latex]x[/latex] in terms of [latex]y[/latex] to find the value of [latex]x[/latex]. | [latex]\begin{align*} x = \frac{38.3}{4.9}- \frac{1.5}{4.9}y&= \frac{38.3}{4.9}- \frac{1.5}{4.9}\cdot\frac{16.19}{46.19}\\ &=\frac{356.08}{46.19} \end{align*}[/latex] |

State the solution. | [latex]\begin{align*} x &=\frac{356.08}{46.19}\approx 7.709\\ y &= \frac{16.19}{46.19}\approx 0.351 \end{align*}[/latex] |

**Solution by elimination:**

Start by stating the system of the equations. | [latex]\begin{align*} 4.9x + 1.5y &= 38.3\\ 2.7x − 8.6y &= 17.8 \end{align*}[/latex] |

Choose a variable to eliminate and multiply each equation by the coefficient of that variable in the other equation. Say we pick to eliminate [latex]x[/latex]. Then we have to multiply the first equation by 2.7 and the second equation by 4.9 | [latex]\begin{align*} 2.7(4.9x + 1.5y) &= 2.7(38.3)\\ 4.9(2.7x − 8.6y) &= 4.9(17.8)\\ \Rightarrow&\\ 13.23x + 4.05y &= 103.41\\ 13.23x − 42.14y &= 87.22\\ \end{align*}[/latex] |

Subtract left side of second equation from the left side of first equation and same for the right side, then solve for [latex]y[/latex]. | [latex]\begin{align*} \Rightarrow 46.19y &= 16.19\Rightarrow y=\frac{16.19}{46.19} \end{align*}[/latex] |

Use the solution for [latex]y[/latex] in one of the equations to find the value of [latex]x[/latex]. | [latex]\begin{align*} 4.9x + 1.5y = 38.3&\Rightarrow 4.9x + 1.5\cdot\frac{16.19}{46.19} = 38.3\\ &\Rightarrow 4.9x + \frac{24.285}{46.19} = 38.3\\ &\Rightarrow 4.9x = 38.3-\frac{24.285}{46.19}\\ &\Rightarrow 4.9x = \frac{1744.792}{46.19}\\ &\Rightarrow x = \frac{1744.792}{46.19}\cdot\frac{1}{4.9}\\ &\Rightarrow x = \frac{356.08}{46.19} \end{align*}[/latex] |

State the solution. | [latex]\begin{align*} x &=\frac{356.08}{46.19}\approx 7.709\\ y &= \frac{16.19}{46.19}\approx 0.351 \end{align*}[/latex] |

Note that we got the same solution either way. When solving a system of multiple equations with multiple variables, you can use either of the methods. In some cases one of the methods may be simpler to execute than another. Choosing the most efficient method is a matter of practice but one guideline may be useful: If there is a variable in one of the equations that has a coefficient 1 or -1, then it is generally best to use substitution by solving for that variable using that equation. If none of the variables has a coefficient 1 or -1, then elimination is generally the better option.

Remember that what you do to one side of the equation must be done to the other side of the equation to maintain the equality. Therefore, you can multiply or divide any equation by any number without changing the solution of the equation. For example, if you multiply all terms of [latex]x + y = 2[/latex] by [latex]2[/latex] on both sides, resulting in [latex]2x + 2y = 4[/latex], the equality of the equation remains unchanged and the same roots exist.

We can also add the left side of one equation to the left side of the other equation, as long as we are doing the same with the right sides of the two equations. Same holds for subtracting the left side on one equation from the left side of the other equation, as long as the same is done with the right sides of the two equations also to maintain equality.

**Concept Check**

Check your understanding of the solution methods when multiple equation with multiple variables are involved by doing the exercises below.

MathMatize: Solving Systems of Equations

#### Example 1.2D: Buying Those Outfits

You are shopping at Old Navy for seven new outfits. The price points are $10 and $30 for the outfits, with the addition of sales tax of 13%. You really like the $30 outfits; however, your total budget can’t exceed $110. How do you spend $110 to acquire all the needed outfits without exceeding your budget while getting as many $30 items as possible?

**Answer:**

Task: # of $10-outfits=? [latex]\leftarrow L[/latex]; # of $30-outfits=? [latex]\leftarrow H[/latex]

Conditions:

total number of outfits: [latex]L+H=7[/latex]

total budget, including 13% tax: [latex]10L+0.13(10L)+30H+0.13(30H)=110[/latex]

This gives us:

[latex]\begin{align*} L+H&=7\\ 11.3L+33.9H&=110 \end{align*}[/latex]

Since [latex]L[/latex] in the first equation has the coefficient 1, best approach is through substitution:

[latex]\begin{align*} L+H=7&\Rightarrow L=7-H\\ 11.3L+33.9H=110&\Rightarrow 11.3(7-H)+33.9H=110\\ &\Rightarrow 79.1-11.3H+33.9H=110\\ &\Rightarrow 22.6H=110-79.1\\ &\Rightarrow 22.6H=30.9\\ &\Rightarrow H=\frac{30.9}{22.6}\approx 1.37\\ \end{align*}[/latex]

Since the number of suits must be a whole number, the number of $30-suits can’t be larger than 1. Therefore, they can buy six suits at $10 per suit and one suit at $30 per suit if they wish to stay within the budget and buy as many of $30-suits as possible.

#### Paths To Success

One of the most difficult areas of mathematics involves translating words into mathematical symbols and operations. To assist in this translation, the table below lists some common language and the mathematical symbol that is typically associated with the word or phrase.

Language | Math Symbol | ||
---|---|---|---|

Sum Addition |
In addition to In excess |
Increased by Plus |
[latex]+[/latex] |

Subtract Decreased by Diminished by |
Less Minus |
Difference Reduced by |
[latex]-[/latex] |

Divide Division |
Divisible Quotient |
Per | [latex]/[/latex] |

Multiplied by Times |
Percentage of | Product of Of |
[latex]\cdot[/latex] |

Becomes Is/Was/Were |
Will be | Results in Totals |
[latex]=[/latex] |

Less than | Lower than | Fewer than | [latex]<[/latex] |

Less/lower/fewer than or equal to | At most | Up to No more than |
[latex]\le[/latex] |

More than | Higher than | [latex]>[/latex] | |

More/higher than or equal to | At least | No less than | [latex]\ge[/latex] |

#### Example 1.2E: Solving Two Linear Equations with Two Unknowns

Tinkertown Family Fun Park charges $15 for a child wrist band and $10.50 for an adult wrist band. On a warm summer day, the amusement park had total wrist band revenue of $15,783 from sales of 1,279 wrist bands. How many adult and child wrist bands did the park sell that day?

**Answer:**

Task: # of child wrist bands=? [latex]\leftarrow c[/latex]; # of adult wrist bands=? [latex]\leftarrow a[/latex]

Conditions:

total revenue: [latex]15c+10.50a=15783[/latex]

total number of wrist bands: [latex]c+a=1279[/latex]

This gives us:

[latex]\begin{align*} 15c+10.50a&=15783\\ c+a&=1279 \end{align*}[/latex]

Since [latex]c[/latex] in the second equation has the coefficient 1, best approach is through substitution by solving for [latex]c[/latex]:

[latex]\begin{align*} c+a=1279&\Rightarrow c=1279-a\\ 15c+10.50a=15783&\Rightarrow 15(1279-a)+10.50a=15783\\ &\Rightarrow 19185-15a+10.50a=15783\\ &\Rightarrow -4.5a=15783-19185\\ &\Rightarrow -4.5a=-3402\\ &\Rightarrow a=\frac{-3402}{-4.5}=756\\ \end{align*}[/latex]

We can use the value for [latex]a[/latex] to solve for [latex]c[/latex]:

[latex]c=1279-a=c=1279-756=523[/latex]

Therefore they sold 523 child wrist bands and 756 adult wrist bands.

## Section Exercises

Work on section 1.2 exercises in Fundamentals of Business Math Exercises. Discuss your solutions with your peers and/or course instructor.

You may consult answers to select exercises: Fundamentals of Business Math Exercises – Select Answers