SOLUTION: Twice one integer plus 3 times a second integer equals 9. Five times the first iteger plus 4 times the second integer equals 5. What are the numbers?

Question 1080935: Twice one integer plus 3 times a second integer equals 9. Five times the first iteger plus 4 times the second integer equals 5. What are the numbers?

Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Twice one integer plus 3 times a second integer equals 9.
2a + 3b = 9
Five times the first iteger plus 4 times the second integer equals 5.
5a + 4b = 5
5a = -4b + 5
divide by 5
a = b + 1
In the first equation, replace a with b + 1
2(b + 1) + 3b = 9
b + 2 + 3b = 9
b + 3b = 9 - 2
b + 3b = 7
multiply by 5
-8b + 15b = 35
7b = 35
b = 35/7
b = 5
Find a
a = (5) + 1
a = -4 + 1
a = -3
:
;
See if that checks out in the first equation
2(-3) + 3(5) =
-6 + 15 = 9
:
What are the numbers? -3 and 5
Answer by MathTherapy(10553) (Show Source): You can put this solution on YOUR website!

Twice one integer plus 3 times a second integer equals 9. Five times the first iteger plus 4 times the second integer equals 5. What are the numbers?

Let the first integer be F, and the second, S
Then we get: 2S + 3S = 9 ------ eq (i)
Also, 5F + 4S = 5 ------- eq (ii)
1) Multiply eq (i) by - 5 to get eq (iii)
2) Multiply eq (ii) by 2 to get eq (iv)
3) Add eqs (iv) & (iii) to ELIMINATE F, and to find the value of S, the second integer
4) Substitute value for S in any of the 2 ORIGINAL equations (i, or ii) to get the value of F, the first integer.
That's it!! Nothing COMPLEX and/or CONFUSING!
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