Applications of Linear Equations
Problems with Answers for Grade 8

Solutions and explanations to Grade 8 questions on applications of linear equations .

  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 \(6 \times 10 - 20 = 40\).
  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: \[ 2x - 6 + 4 = 4x + 22 \] \[ 2x - 4x = 22 - 4 + 6 \] \[ -2x = 24 \] \[ x = -12 \]
  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: \(x + 18 = 41\).
  4. The length of a rectangle is 10 meters more than twice its width. What is the length and width 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: \(L = 2(7) + 10 = 24\).
  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: \[ 35 + 45 + x = 6x + 15 \] \[ 80 + x = 6x + 15 \] \[ 65 = 5x \] \[ x = 13 \]
  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: \(39 + 102 = 141^\circ\).
  7. Two complementary angles are such that one is \(14^\circ\) more than three times the other. What is 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: \(3(19) + 14 = 71^\circ\).
  8. The sum of a positive even integer and the third next even integer is 150. Find the number.

    Solution

    Let \(x\) be the even integer. The third next even integer is \(x + 6\). Sum: \[ x + (x + 6) = 150 \] \[ 2x + 6 = 150 \] \[ 2x = 144 \] \[ x = 72 \]
  9. The average of three odd successive numbers is 129. What is the largest number?

    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: \(127 + 4 = 131\).
  10. Two numbers are such that one is 42 more than the other and their average is 40. Find the 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 \(61\).

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