Applications of Linear Equations

Site: Saylor Academy
Course: GKT101: General Knowledge for Teachers – Math
Book: Applications of Linear Equations
Printed by: Guest user
Date: Sunday, May 19, 2024, 6:25 PM

Description

This section this textbook explains how to translate the situations described in word problems to equations and provides a variety of examples. Read the chapter and work through the problems. Some examples involved the geometric facts you have learned in Unit 2.

Linear Equations - Number and Geometry

Objective: Solve number and geometry problems by creating and solving a linear equation.

Word problems can be tricky. Often it takes a bit of practice to convert the English sentence into a mathematical sentence. This is what we will focus on here with some basic number problems, geometry problems, and parts problems.

A few important phrases are described below that can give us clues for how to set up a problem.

  • A number (or unknown, a value, etc) often becomes our variable
  • Is (or other forms of is: was, will be, are, etc) often represents equals (=) x is 5 becomes x =5
  • More than often represents addition and is usually built backwards, writing the second part plus the first
    Three more than a number becomes x + 3
  • Less than often represents subtraction and is usually built backwards as well, writing the second part minus the first
    Four less than a number becomes x − 4

Using these key phrases we can take a number problem and set up and equation and solve.


Source: Tyler Wallace, http://www.wallace.ccfaculty.org/book/Beginning_and_Intermediate_Algebra.pdf
Creative Commons License This work is licensed under a Creative Commons Attribution 3.0 License.

Example 102.

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

5 x-28 Subtraction is built backwards, multiply the unknown by 5
5 x-28=232 Is translates to equals
underline +28 \quad +28 Add 28 to both sides
\underline {5x=260} The variable is multiplied by 5
 5 \quad \quad 5 Divide both sides by 5
x=52 The number is 52.


This same idea can be extended to a more involved problem as shown in the next example.

Example 103.

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

3x + 15 First, addition is built backwards
6x − 10 Then, subtraction is also built backwards
3x + 15 = 6x − 10 Is between the parts tells us they must be equal
\underline {− 3x − 3x} Subtract 3x so variable is all on one side
15 = 3x − 10 Now we have a two − step equation
\underline {+10 \quad+10} Add 10 to both sides
 \underline {25 =3x} The variable is multiplied by 3
3 \quad \quad 3 Divide both sides by  3
 \frac {25}{3} = x Our number is \frac {25}{3}


Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation. This is shown in the following example.

Example 104.

The sum of three consecutive integers is 93. What are the integers?

First x Make the first number x.
Second x +1 To get the next number we go up one or +1
Third x +2 Add another 1(2 total) to get the third
F + S + T = 93 First (F) plus Second (S) plus Third (T) equals 93
(x)+(x + 1)+ (x +2) = 93 Replace F with x, S with x +1, and T with x +2
x + x +1+ x +2 = 93 Here the parenthesis aren't needed.
3x +3 = 93 Combine like terms x + x + x and 2 + 1
\underline {− 3 \quad − 3} Add 3 to both sides
\underline {3x = 90} The variable is multiplied by 3
3 \quad \quad 3 Divide both sides by 3
x = 30 Our solution for x
First 30 Replace x in our original list with 30

Second  (30) +1 = 31
Third  (30) +2 = 32

The numbers are 30, 31, and 32


Sometimes we will work consecutive even or odd integers, rather than just consecutive integers. When we had consecutive integers, we only had to add 1 to get to the next number so we had x, x + 1, and x + 2 for our first, second, and third number respectively. With even or odd numbers they are spaced apart by two. So if we want three consecutive even numbers, if the first is x, the next number would be x + 2, then finally add two more to get the third, x + 4. The same is true for consecutive odd numbers, if the first is x, the next will be x + 2, and the third would be x + 4. It is important to note that we are still adding 2 and 4 even when the numbers are odd. This is because the phrase "odd" is referring to our x, not to what is added to the numbers. Consider the next two examples.

Example 105.

The sum of three consecutive even integers is 246. What are the numbers?

First x Make the first x
Second x +2 Even numbers, so we add 2 to get the next
Third x +4 Add 2 more (4 total)to get the third
F + S + T = 246 Sum means add First (F) plus Second (S) plus Third (T)
(x)+ (x +2) +(x + 4)= 246 Replace each F, S, and T with what we labeled them
x + x +2 + x + 4= 246 Here the parenthesis are not needed
3x + 6= 246 Combine like terms x + x + x and 2+ 4
\underline {− 6 \quad − 6} Subtract 6 from both sides
 \underline {3x = 240} The variable is multiplied by 3
3 \quad \quad 3 Divide both sides by 3
x = 80 Our solution for x
First 80 Replace x in the original list with 80.

Second (80)+ 2= 82
Third (80)+ 4= 84

The numbers are 80, 82, and 84

Example 106.

Find three consecutive odd integers so that the sum of twice the first, the second and three times the third is 152.

First x Make the first x
Second x +2 Odd numbers so we add 2 (same as even!)
Third x +4 Add 2 more (4 total)to get the third
2F + S + 3T = 152
Twice the first gives 2F and three times the third gives 3T
2(x)+ (x +2) +3(x +4) = 152 Replace F, S, and T with what we labeled them
2x + x +2 +3x + 12 = 152 Distribute through parenthesis
6x + 14 = 152 Combine like terms 2x + x +3x and 2+ 14
 \underline {− 14 − 14} Subtract 14 from both sides
\underline {6x = 138} Variable is multiplied by 6
6 \quad \quad6 Divide both sides by 6
x = 23 Our solution for x
First 23 Replace x with 23 in the original list
Second  (23)+2 = 25
Third  (23)+4 = 27
The numbers are 23, 25, and 27


When we started with our first, second, and third numbers for both even and odd we had x, x + 2, and x + 4. The numbers added do not change with odd or even, it is our answer for x that will be odd or even.

Another example of translating English sentences to mathematical sentences comes from geometry. A well known property of triangles is that all three angles will always add to 180. For example, the first angle may be 50 degrees, the second 30 degrees, and the third 100 degrees. If you add these together, 50 + 30 + 100 = 180. We can use this property to find angles of triangles.

World View Note: German mathematician Bernhart Thibaut in 1809 tried to prove that the angles of a triangle add to 180 without using Euclid’s parallel postulate (a point of much debate in math history). He created a proof, but it was later shown to have an error in the proof.

Example 107.

The second angle of a triangle is double the first. The third angle is 40 less than the first. Find the three angles.

First x With nothing given about the first we make that x
Second 2x The second is double the first,
Third x − 40 The third is 40 less than the first
F + S + T = 180 All three angles add to 180
(x) +(2x)+ (x − 40)= 180 Replace F , S, and T with the labeled values
x +2x + x − 40 = 180 Here the parenthesis are not needed.
4x − 40 = 180 Combine like terms, x +2x + x
\underline {+ 40 + 40} Add 40 to both sides
\underline {4x = 220} The variable is multiplied by 4
4 \quad \quad 4 Divide both sides by 4
x = 55 Our solution for x
First 55 Replace x with 55 in the original list of angles
Second 2(55)= 110
Third (55) − 40 = 15
Our angles are 55, 110, and 15


Another geometry problem involves perimeter or the distance around an object. For example, consider a rectangle has a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides) so we add 8 + 8 + 3 + 3 = 22. As there are two lengths and two widths in a rectangle an alternative to find the perimeter of a rectangle is to use the formula P = 2L + 2W. So for the rectangle of length 8 and width 3 the formula would give, P = 2(8) + 2(3) = 16 + 6 = 22. With problems that we will consider here the formula P = 2L + 2W will be used.

Example 108.

The perimeter of a rectangle is 44. The length is 5 less than double the width. Find the dimensions.

Length x We will make the length x
Width 2x − 5 Width is five less than two times the length
P =2L + 2W The formula for perimeter of a rectangle
(44) =2(x)+ 2(2x − 5) Replace P, L, and W with labeled values
44 = 2x +4x − 10 Distribute through parenthesis
 44 =6x − 10 Combine like terms 2x +4x
\underline {+ 10 + 10} Add 10 to both sides
 \underline {54 =6x} The variable is multiplied by 6
6 \quad \quad 6 Divide both sides by 6
9= x Our solution for x
Length 9 Replace x with 9 in the original list of sides
Width 2(9) − 5 = 13 The dimensions of the rectangle are 9 by 13.


We have seen that it is important to start by clearly labeling the variables in a short list before we begin to solve the problem. This is important in all word problems involving variables, not just consecutive numbers or geometry problems. This is shown in the following example.

Example 109.

A sofa and a love seat together costs $444. The sofa costs double the love seat. How much do they each cost?

Love Seat x With no information about the love seat,this is our x
Sofa 2x Sofa is double the love seat, so we multiply by 2
S + L = 444 Together they cost 444, so we add
(x) +(2x)= 444 Replace S and L with labeled values
\underline {3x = 444} Parenthesis are not needed, combine like terms  x +2x
3 \quad \quad 3 Divide both sides by 3
x = 148 Our solution for x
Love Seat 148 Replace x with 148 in the original list
Sofa 2(148)= 296 The love seat costs $148 and the sofa costs $296.


Be careful on problems such as these. Many students see the phrase "double" and believe that means we only have to divide the 444 by 2 and get $222 for one or both of the prices. As you can see this will not work. By clearly labeling the variables in the original list we know exactly how to set up and solve these problems.