Hi, there--

THE PROBLEM:
Find two numbers such that the sum of the first and three times the second is 5 and the sum of the second and two times the first is 8.

A SOLUTION:
Define the variables:
Let x be the first number.
Let y be the second number.

Write two equations modeling the relationships in the problem.
The sum of the first and three times the second is 5.
"The first" is x, and "three times the second" is 3y.

Their sum is 5, so one equation is x + 3y = 5.

For the other equation,
The sum of the second and two times the first is 8.
"The second" is y, and "two times the first" is 2x.

The sum is 8, so another equation is y + 2x = 8.

Solve the System of Equations:

x + 3y = 5
y + 2x = 8

Rewrite the first equation in "x=" form.
x + 3y = 5
x = -3y + 5

Substitute -3y+5 for x in the second equation equation. Simplify and solve for y.
y + 2x = 8 = 8
y + 2(-3y + 5) = 8
y - 6y + 10 = 8
-5y + 10 = 8
-5y = -2
y = 2/5


The second number is 2/5.


Solve for x using the first equation.
x + 3y = 5
x + 3(2/5) = 5
x + 6/5 = 5
x = 5 - 6/5
x = 25/5 - 6/5
x = 19/5

The first number is 19/5.

NOW CHECK your work with the original sentences:

The sum of the first and three times the second is 5.
(19/5) + 3*(2/5) = 5
19/5 + 6/5 = 5
25/5 = 5
5 = 5 CHECK!

The sum of the second and two times the first is 8.
(2/5) + 2*(19/5) = 8
2/5 + 38/5 = 8
40/5 = 8
8 = 8 CHECK!

The two numbers are 2/5 and 19/5.

Hope this helps! Feel free to email if you have any questions about the solution.

Good luck with your math,

Mrs. F
math.in.the.vortex@gmail.com