# Thread: Basic Math Problem

1. ## 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

2. Plug x-7 in for y in the first equation so $\displaystyle x^2+(x-7)^2=25$

3. 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
$\displaystyle x^2+y^2 = 25$

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

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

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

4. 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

5. 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

6. 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:

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

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

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

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

7. 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