Applications of Linear Equations
Problems with Answers for Grade 8
Solutions and explanations to Grade 8 questions on applications of linear equations .

Three times a number increased by ten is equal to twenty less than six times the number. Find the number.
Solution
Let the number be x. "Three times a number increased by 10" is mathematically translated as
3x + 10
"is equal" is mathematically translated as
=
"Twenty less than six times the number" is mathematically translated as
6x  20
The whole sentence "Three times a number increased by ten is equal to twenty less than six times the number" is translated as
3x + 10 = 6x  20
We now solve the above linear equation to find the number x
3x  6x =  20  10
 3x =  30
x = 10
Check answer
Three times a number increased by ten : 3 × 10 + 10 = 40
Twenty less than six times the number : 6 × 10  20 = 40

If twice the difference of a number and 3 is added to 4, the result is 22 more than four times the number. Find the number.
Solution
Let the number be x. "twice the difference of a number and 3 is added to 4" is mathematically translated as
2(x  3) + 4
"the result is" is mathematically translated as
=
"22 more than four times the number" is mathematically translated as
4x + 22
"twice the difference of a number and 3 is added to 4, the result is 22 more than four times the number" is mathematically translated as
2(x  3) + 4 = 4x + 22
Solve the equation
2x  6 + 4 = 4x + 22
2x  4x = 22  4 + 6
2x = 24
x = 12

The sum of two numbers is 64. The difference of the two numbers is 18. What are the numbers?
Solution
Let x be the smaller of the two numbers. Since the difference of the two numbers is 18, then the larger number is
x + 18
The sum of the two numbers is 64. Hence
smaller number + larger number = 64
or
x + (x + 18) = 64
Solve for x
2x + 18 = 64
2x = 64  18
2x = 46
x = 23 , the smallest of the two numbers.
x + 18 = 23 + 18 = 41 , the largest of the two numbers.

The length of a rectangle is 10 meters more than twice its width. What is the length and width of the rectangle if its perimeter is 62 meters.
Solution
Let W be the width of the rectangle. "the length of a rectangle is 10 meters more than twice its width" is translated as
length = 2 W + 10
The perimeter of the rectangle is given by
Perimeter = 2 length + 2 width
62 = 2 (2 w + 10) + 2 W
Solve the above equation for W
62 = 4 W + 20 + 2 W
62 = 6 W + 20
62  20 = 6 W
42 = 6 W
W = 7
length and width are
width = W = 7 meters , length = 2 W + 10 = 2 (7) + 10 = 24 meters

The average of 35, 45 and x is equal to five more than twice x. Find x.
Solution
The average of 35, 45 and x is given by
(35 + 45 + x) / 3
The average is equal to five more than twice x. Hence
(35 + 45 + x) / 3 = 2x + 5
The above equation can be written as
(35 + 45 + x) / 3 = (2x + 5) / 1
Cross product and solve
1(35 + 45 + x) = 3(2x + 5)
35 + 45 + x = 6x + 15
80 + x = 6x + 15
80  15 = 6x  x
65 = 5x
x = 13

The difference in the measures of two supplementary angles is 102°. Find the two angles.
Solution
If the difference of measures of two angles is 102°, then
larger angle = smaller angle + 102°
The sum of the measures of two supplementary angles is equal to 180°. Hence
Larger angle + smaller angle = 180°
or
smaller angle + 102° + smaller angle = 180°
2 smaller angle = 180  102 = 78°
smaller angle = 78 / 2 = 39°
larger angle = smaller angle + 102 = 141°

Two complementary angles are such that one is 14° more than three times the second angle. What is the measure of the larger angle.
Solution
There are two angles: a larger one and a smaller one. The larger one is such that
larger = 3 × smaller + 14°
The sum of two angles is 90°
. Hence
larger + smaller = 90°
or
3 × smaller + 14° + smaller = 90°
4 × smaller = 90  14
4 × smaller = 76
smaller = 76 / 4 = 19°
larger = 3 × smaller + 14° = 3 × 19 + 14 = 71°

The sum of a positive even integer number and the next third even integer is equal to 150. Find the number.
Solution
Let x be the positive even integer. The next three even integers are
x + 2 , x + 4 , x + 6
The third even integer is x + 6. The sum of x and x + 6 is 150. Hence
x + x + 6 = 150
2x = 150  6
2x = 144
x = 72

The average of three odd successive numbers is equal to 129. What is the largest of the three numbers?
Solution
Three odd successive integers are of the form
x , x + 2 , x + 4
Their average is equal to 129. Hence
(x + x + 2 + x + 4) / 3 = 129
Rewrite above equation as
(x + x + 2 + x + 4) / 3 = 129 / 1
Cross multiply and solve
(3x + 6)1 = 129(3)
3x + 6 = 387
3x = 387  6
3x = 381
x = 127
The largest of the three numbers is
x + 4 = 127 + 4 = 131

Two numbers are such that one number is 42 more that the second number and their average is equal to 40. What are the two numbers?
Solution
If x is the smallest number, then the largest is.
x + 42
The average of x and x + 42 is equal to 40. Hence
(x + x + 40) / 2 = 40
Cross multiply and solve
2x + 40 = 80
2x = 80  40
2x = 40
x = 20
The two numbers are
x = 20 and x + 40 = 60

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