- #1

aricho

- 71

- 0

x+y=8

x^2-y^2=36

find the values for x and y

Thanks

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- Thread starter aricho
- Start date

- #1

aricho

- 71

- 0

x+y=8

x^2-y^2=36

find the values for x and y

Thanks

- #2

whozum

- 2,221

- 1

y = 8-x

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

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

16x - 64 = 36

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

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

16x - 64 = 36

- #3

aricho

- 71

- 0

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?

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?

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

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"?

- #5

VietDao29

Homework Helper

- 1,426

- 3

Yup. This is correct. Since you have 2 unknowns, you need 2 equations. And youaricho 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?

[tex]\left\{ \begin{array}{l}x + y = 8 \\ x - y = \frac{9}{2} \end{array} \right.[/tex]

for x, and y.

Viet Dao,

- #6

whozum

- 2,221

- 1

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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