# Basic Math Problem

• May 16th 2009, 11:57 AM
Aslam Anjum
Basic Math Problem
I need help on a question the questions says

by eliminating Y, find the solution to the Simulataneous Eqautions.

X(squared)+Y(squared)=25
Y=X-7

Aslam
• May 16th 2009, 11:59 AM
artvandalay11
Plug x-7 in for y in the first equation so $x^2+(x-7)^2=25$
• May 16th 2009, 12:00 PM
e^(i*pi)
Quote:

Originally Posted by Aslam Anjum
I need help on a question the questions says

by eliminating Y, find the solution to the Simulataneous Eqautions.

X(squared)+Y(squared)=25
Y=X-7

Aslam

$x^2+y^2 = 25$

As you've stated y=x-7 so we can sub that in for where we see 'y' in the first equation.

$x^2 + (x-7)^2 = 25 \: , \: x^2 + x^2 - 14x + 49 = 25$

$2x^2 - 14x + 24 = 0$
• May 16th 2009, 12:08 PM
Aslam Anjum
Thanks for that but I think this is a quadratic as there is space to put two sets of values for X and Y.

can you please work it out and tell what the answers are so i can check to see if I'm gettting the right answers

Thanks again,

Aslam
• May 16th 2009, 12:16 PM
artvandalay11
You can tell us what you get for answers and we'll tell you if you are right. This forum is to help people not to do it for you
• May 16th 2009, 12:22 PM
e^(i*pi)
Quote:

Originally Posted by Aslam Anjum
Thanks for that but I think this is a quadratic as there is space to put two sets of values for X and Y.

can you please work it out and tell what the answers are so i can check to see if I'm gettting the right answers

Thanks again,

Aslam

Well you can solve that equation to find x and from there find y:

$2x^2 - 14x + 24 = 0$

Divide by 2: $x^2 - 7x+12 = 0$

I can see that this factorises to $(x-4)(x-3) = 0$

therefore either x-4=0 or x-3=0 and therefore x = 4 or x=3.

We then put these back into y = x-7

y = 4-7 = -3 or y = 3-7 = -3.

Our solutions are therefore x=4, y=-3 or x=3, y=-4. There are two pairs of answers for this
• May 16th 2009, 12:48 PM
Aslam Anjum
thanks for that i didnt get anything close to that I started using the other eqaution i cant tell which equation to use when i see the questions.

thanks for the help,

Aslam