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 $x$ :

$x = 31 - 3*7$

$x = 31 - 21$

$x = 10$