Solve the system x+y=9, x^2 - y^2 = 36 for x and y

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To solve the system of equations x + y = 8 and x^2 - y^2 = 36, the first equation can be rearranged to express y as y = 8 - x. Substituting this into the second equation leads to x^2 - (8 - x)^2 = 36, simplifying to 16x - 64 = 36. This results in x = 25/4 or 6.25, and subsequently y = 1.75. The discussion emphasizes the importance of having two equations to solve for the two unknowns, confirming that the approach taken is correct.
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Hi, i just sat my maths exam, and i had a very strange question... i don't know how to do it...

x+y=8
x^2-y^2=36

find the values for x and y

Thanks
 
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y = 8-x

x^2 - (8-x)^2 = 36

x^2 - (64-16x+x^2) = 36

16x - 64 = 36
 
is that it?

i did y^2-x^2=36
(x-y)(x+y)=36
(x-y)(8)=36
x-y=36/8

is any of that correct?
 
According to Whozum, 16x- 64= 36 so x= 100/16= 25/4= 6.25 and y= 8- 6.25= 1.75.

x-y= 6.25- 1.75= 4.5 and 36/8= 9/2= 4.5.

Your statement is correct but is not a solution to the "problem" which, I suppose, was to solve the two equations.

In fact, the only thing strange I see about your "question" is that there was no "question"! Are you sure you didn't leave something out- like "solve this pair of equations" or "what are x and y"?
 
aricho said:
is that it?

i did y^2-x^2=36
(x-y)(x+y)=36
(x-y)(8)=36
x-y=36/8

is any of that correct?
Yup. This is correct. Since you have 2 unknowns, you need 2 equations. And you have already had 2 equations. You can then solve:
\left\{ \begin{array}{l}x + y = 8 \\ x - y = \frac{9}{2} \end{array} \right.
for x, and y.
Viet Dao,
 
HallsofIvy said:
Your statement is correct but is not a solution to the "problem" which, I suppose, was to solve the two equations.

Me or him?
 
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