SOLUTION: Find two numbers such that twice the first exceeds three times the second by 1, and three times the first exceeds twice the second by 14.

Question 271438: Find two numbers such that twice the first exceeds three times the second by 1, and three times the first exceeds twice the second by 14.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
let x = the first number
let y = the second number
:
Write an equation for each statement:
:
Find two numbers such that twice the first exceeds three times the second by 1,
2x = 3y + 1
:
and three times the first exceeds twice the second by 14.
3x = 2y + 14
:
Multiply the first equation by 2, and the 2nd equation by 3
4x = 6y + 2
9x = 6y + 42
---------------subtraction eliminates y, find x
-5x = -40
x = +8
:
Find y using the 1st equation
2(8) = 3y + 1
16 - 1 = 3y
3y = 15
y = 5
:
Check solution in the 2nd equation
3(8) = 2(5) + 14
24 = 10 + 14
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Response to the comment
2x = 3y + 1 is exactly the same thing as 2x - 3y = 1
and
3x = 2y + 14 is exactly the same thing as 3x - 2y = 14
:
You can check the solutions in either equation, if you doubt it
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