Question 193397: ``Before, solving the problem, I just want to give you the heads up. It's
Systems of Equation: Application. I can't find the specific title to relate
with my story problems. My chapter in the book is Chapter Eight, Solving Systems by Substiution, Additional. Hope it helps!
*Two numbers are such that twice the first and 3 times the second is 31. When 4 times the second number is subtracted from 3 times the first number the result is 4. Find the two numbers.*
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let the first number be a and the second number, b.
From the problem description, you can write:
2a+3b = 31 "...(the sum of) twice the first number and 3 times the second is 31."
3a-4b = 4 "When 4 times the second number is subtracted from 3 times the first number the result is 4."
So you have a system of equations in two unknowns, a and b.
2a+3b = 31 Multiply this equation by 4.
3a-4b = 4 Multiply this equation by 3.
8a+12b = 124
9a-12b = 12 Add these two equations.
17a = 136 Divide both sides by 17
a = 8
2a+3b = 31
2(8)+3b = 31
16+3b = 31 Subtract 16 from both sides.
3b = 15 Divide both sides by 3.
b = 5.
The two numbers are: 8 and 5
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