SOLUTION: Here's the question: Sorry for it being so long
Solve each system by the addition method
x/3 - y/2 = -5/6
x/5 - y/3 = -3/5
This is what I did:
6*x/3 - 6*y/2 = 6*-5/6
Algebra ->
Linear-equations
-> SOLUTION: Here's the question: Sorry for it being so long
Solve each system by the addition method
x/3 - y/2 = -5/6
x/5 - y/3 = -3/5
This is what I did:
6*x/3 - 6*y/2 = 6*-5/6
Log On
Question 5680: Here's the question: Sorry for it being so long
Solve each system by the addition method
x/3 - y/2 = -5/6
x/5 - y/3 = -3/5
This is what I did:
6*x/3 - 6*y/2 = 6*-5/6
2x - 3y = -5
15*x/3 - 15*y/3 = 15*-3/5
3x - 5y = -9
-3(2x - 3y) = -3*-5
-6x + 9y = 15
2(3x - 5y) = 2*-9
6x - 10y = -18
-6x + 9y = 15
6x - 10y = -18
-1y = -3
y = 3
for the first eq I substituted 3 for y to find x
x/3 - 3/2 = -56
6*x/3 - 6*3/2 = 6*-5/6
2x - 9 + 9 = -5 + 9
2x = 4
x = 2
TO TEST:
2/3 - 3/2 = -5/6
6*2/3 - 6*3/2 = 6*-5/6
4/3 - 9/2 = -5/6 (do you just subtract 9 from 4 and multiply the two denominators to get 6?)
(this is where I get stuck. Or is this how to test?)
2/5 - 1 = -3/5
(2/5 - 1/1 = -3/5)
(I think this tests right. I hope)
I can't figure out how to test this. I think I got the right answers for x and y, but I can't figure out how to test to see if they are correct. I tried, but it's not working out. I must be missing a step somewhere. Please help! Thank you. Maybe I got it right and am looking for something that isn't there. Algebra. ARGHHHHHHHHHH!