Question

One number is 5 less than a second number. Twice the second number is 11 less than 5 times the first. Find the two numbers

Answer 1

The first number = 7

The second number = 12

Explanation:

Let the first number = x

Let the second number = y

x + 5 = y

2*y + 11 = 5x

Solve for x in the first equation

x = y - 5

Substitute this result into the second equation

2y + 11 = 5(y - 5)

Remove the brackets

2y + 11 = 5y - 25

Subtract 2y from both sides

11 = 5y - 2y - 25

11 = 3y - 25

Add 25 to both sides.

11 + 25 = 3y

36 = 3y

Divide by 3

36/3 = 3y/3

y = 12

Rewrite the first equation. Put 12 in for y

x + 5 = y

x + 5 = 12

Subtract 5 from both sides

x = 12 - 5

x = 7