How do you find two numbers such that the first number added to three times the second is 31, while three times the first less twice the second is 16?

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Answer 1 · 2016-11-25T12:00:38

The first number is 10 and the second number is 7

Let's define the two numbers we are looking for as #x# and #y#.

The first number added to three times the second number is 31 can be written as:

#x + 3y = 31#

Three time the first less twice the second is 16 can be written as:

#3x - 2y = 16#

Solve the first equation for #x#L

#x + 3y - 3y = 31 - 3y#

#x = 31 - 3y#

Substitute #31 - 3y# for #x# in the second equation and solve for #y#:

#3(31 - 3y) - 2y = 16#

#93 - 9y - 2y = 16#

#93 - 11y = 16#

#93 - 11y + 11y - 16 = 16 + 11y - 16#

#93 - 16 = 11y#

#77 = 11y#

#77/11 = (11y)/11#

#y = 7#

Now substitute #7# for #y# in the solution to the first equation to calculate #x#:

#x = 31 - 3*7#

#x = 31 - 21#

#x = 10#