Two numbers have a sum of 17. Four times the smaller number is 8 more than double the larger. Find the numbers.

Answer 1

Hello Janvi

Let the smaller number = x and the larger number = y. Therefore, x + y = 17 and 4x = 2y + 8.

Solve 4x = 2y + 8 for x, i.e., x = (2y + 8)/4 = (1/2)y + 2. Then substitute (1/2)y + 2 for x into x + y = 17, i.e.,

(1/2)y + 2 + y = 17. Now solve for y, i.e., (3/2)y = 15, y = (15)(2/3) = 10. Finally, solve for x, i.e., x + 10 = 17,

x = 7.

Check: 7 + 10 = 17, (4)(7) = (2)(10) + 8, 28 = 28.