Grade 8 Linear Equations – Real-Life Applications, Problems, and Solutions

Grade 8 questions on applications of linear equations with solutions and explanations included.

Questions 1

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 \(3x + 10\). "Is equal" is \(=\). "Twenty less than six times the number" is \(6x - 20\). Therefore: \[ 3x + 10 = 6x - 20 \] Solving: \[ 3x - 6x = -20 - 10 \] \[ -3x = -30 \] \[ x = 10 \] Check: \(3 \times 10 + 10 = 40\) and \(6 \times 10 - 20 = 40\).

Questions 2

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 \(2(x - 3) + 4\). "The result is" is \(=\). "22 more than four times the number" is \(4x + 22\). Thus: \[ 2(x - 3) + 4 = 4x + 22 \] Solving: \[ 2x - 6 + 4 = 4x + 22 \] \[ 2x - 4x = 22 - 4 + 6 \] \[ -2x = 24 \] \[ x = -12 \]

Questions 3

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 number. The larger number is \(x + 18\). The sum of the two numbers is: \[ x + (x + 18) = 64 \] \[ 2x + 18 = 64 \] \[ 2x = 46 \] \[ x = 23 \] Larger number: \(x + 18 = 41\).

Questions 4

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. Length: \(L = 2W + 10\). Perimeter formula: \[ 62 = 2L + 2W \] Substitute \(L\): \[ 62 = 2(2W + 10) + 2W \] \[ 62 = 4W + 20 + 2W \] \[ 62 = 6W + 20 \] \[ 6W = 42 \] \[ W = 7 \] Length: \(L = 2(7) + 10 = 24\).

Questions 5

The average of 35, 45 and \( x \) is equal to five more than twice \( x \). Find \( x \).

Solution:

Average: \[ \frac{35 + 45 + x}{3} = 2x + 5 \] Multiply both sides by 3: \[ 35 + 45 + x = 6x + 15 \] \[ 80 + x = 6x + 15 \] \[ 65 = 5x \] \[ x = 13 \]

Questions 6

The difference in the measures of two supplementary angles is \( 102^{\circ} \). Find the two angles.

Solution:

Let the smaller angle be \(y\). Then the larger angle is \(y + 102^\circ\). Supplementary angles sum to \(180^\circ\): \[ y + (y + 102) = 180 \] \[ 2y + 102 = 180 \] \[ 2y = 78 \] \[ y = 39 \] Larger angle: \(39 + 102 = 141^\circ\).

Questions 7

Two complementary angles are such that one is \( 14^{\circ} \) more than three times the second angle. What is the measure of the larger angle.

Solution:

Let the smaller angle be \(y\). Larger angle: \(3y + 14^\circ\). Complementary angles sum to \(90^\circ\): \[ 3y + 14 + y = 90 \] \[ 4y = 76 \] \[ y = 19 \] Larger angle: \(3(19) + 14 = 71^\circ\).

Questions 8

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 even integer. The third next even integer is \(x + 6\). Sum: \[ x + (x + 6) = 150 \] \[ 2x + 6 = 150 \] \[ 2x = 144 \] \[ x = 72 \]

Questions 9

The average of three odd successive numbers is equal to 129. What is the largest of the three numbers?

Solution:

Let the numbers be \(x, x+2, x+4\). Average: \[ \frac{x + (x+2) + (x+4)}{3} = 129 \] \[ \frac{3x + 6}{3} = 129 \] \[ 3x + 6 = 387 \] \[ 3x = 381 \] \[ x = 127 \] Largest: \(127 + 4 = 131\).

Questions 10

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:

Let the smaller be \(x\), then the larger is \(x + 42\). Average: \[ \frac{x + (x + 42)}{2} = 40 \] \[ \frac{2x + 42}{2} = 40 \] \[ 2x + 42 = 80 \] \[ 2x = 38 \] \[ x = 19 \] Numbers: \(19\) and \(61\).