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mathssydney
They are 5 and minus 2

You have to solve the system of equations
2(a-b)=14
a+b=3
SJ2006
Let, x+y = 3
2(x-y) = 14

Substitute value from eq (1) in eq (2),
2(3-y-y) = 14
2(3-2y) = 14
6-4y = 14
-4y = 8
y = -2

Put it into eq (1),
x-2 = 3
x = 5

So, x = 5 & y = -2

Hope this helps!

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