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2x - y - 4 = 0 3x + y - 9 = 0 What is the solution set of the given system? (A.) {(6/5, 13/5)} (B.) {(13/5, 6/5)} (C.) {(5, 6)} (D.) {(6, 5)}
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2x - y - 4 = 0 3x + y - 9 = 0 What is the solution set of the given system? (A.) {(6/5, 13/5)} (B.) {(13/5, 6/5)} (C.) {(5, 6)} (D.) {(6, 5)}
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Jan 6, 2018
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2x - y - 4 = 0
3x + y - 9 = 0
What is the solution set of the given system?
(A.) {(6/5, 13/5)}
(B.) {(13/5, 6/5)}
(C.) {(5, 6)}
(D.) {(6, 5)}
Mathematics
high-school
Varatis
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Varatis
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So to start you should get x and y on the same side and get the numbers on one side. Then you should cancel out a variable. Because both y’s are 1 and they are opposite charges they will both cancel out to 0. Then add what is remaining which is the 2x and the 3x and equal that to +9-4 which equals 5. You want x to be all by itself so you would divide both sides by 5 and get that x=1. Then you would just substitue and solve for y. Hope this helped!
Toastie
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Jan 8, 2018
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2x-y=4
3x+y=9
5x=13
x=13/5
3(13/5)+y=9
39/5+y=45/5
y=45/5-39/5
y=6/5
the answer is b
Nvd
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Jan 10, 2018
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Nvd
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