One number is 12 more than twice another number. If the sum of the numbers is 132, find both numbers

Answer 1

1) use the provided information to create inequalities.

x = 2y +12

x + y = 132

2) There's a couple of ways to solve for these inequalities. The simplest way is to get both variables on the same side and either add or subtract the inequalities in order to eliminate one variable to make it solvable.

x - 2y = 12

- (x + y = 132)

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0 - 3y = - 120

3) solve for y

y = (-120)/(-3) = 40

4) plug in y in either of the inequalities to solve for x

x - 2(40) = 12 --> x = 12 + 80 = 92

5) check by plugging x and y into the other inequality

x + y = 132

92 + 40 = 132

So, x is 92 and y is 40.

Answer 2

Set up a system of equations:

a = 2b + 12

a + b = 132

The first equation solves for a in terms of b. Plug that into the second equation and solve for b:

a + b = 132

(2b + 12) + b = 132

3b + 12 = 132

3b = 120

b = 120 / 3 = 40

Plug b = 40 back into one of the equations to solve for a (I used the first equation since it's already set up to solve for a):

a = 2b + 12 = 2(40) + 12 = 92

So one number is 40 and the other number is 92