hi,

im not too sure how to solve this question:

can someone please tell me how to do this one!

im okay at doing them with Y= at the start but haven't done one with x^2 and y^2 before.

thanks!

2. Originally Posted by andyboy179
hi,

im not too sure how to solve this question:

can someone please tell me how to do this one!

im okay at doing them with Y= at the start but haven't done one with x^2 and y^2 before.

thanks!
Do you know how to substitute? What do your class notes/text reveal?

3. no i don't know how to do it for this question because it has x^2 and y^2 in it!

4. Originally Posted by andyboy179
no i don't know how to do it for this question because it has x^2 and y^2 in it!
Solve for x (which is equal to 5 - y), substitute into the second equation (which is quadratic - I hope you know how to solve a quadratic equation). Once you figure out y, then you can get x from the first equation.

5. simultaneous equations ...

$x+y = 5$ ... this is a linear equation

$x^2+y^2 = 13$ ... this is the equation of a circle

using the linear equation, note that $y = 5-x$

substitute $(5-x)$ for $y$ in the circle equation ...

$x^2 + (5-x)^2 = 13
$

solve the quadratic equation for x , then you can determine y.

6. so i would do this to find X:

is this correct so far?

7. and to get Y would i do:

5-1.2= 3.8
1.2^2= 1.44
3.8^2= 14.44
14.44-1.44=13

Y= 3.8??

8. Originally Posted by andyboy179
so i would do this to find X:

is this correct so far?
Unfortunately no.

1. According to the binomial formula

$(5-x)^2 = (5-x)(5-x) = 25-10x+x^2$

$2x^2-10x+12 = 0~\implies~x^2-5x+6=0~\implies~(x-2)(x-3)=0$

A product equals zero if one of the factors equals zero.

9. Originally Posted by andyboy179
hi,

im not too sure how to solve this question:

can someone please tell me how to do this one!

im okay at doing them with Y= at the start but haven't done one with x^2 and y^2 before.

thanks!
$x+y=5$

$x^2+y^2=13$

$(x+y)^2=(x+y)(x+y)=x(x+y)+y(x+y)=x^2+xy+xy+y^2$

$(x+y)^2=x^2+y^2+2xy=5^2=25$

$(x+y)^2-\left(x^2+y^2\right)=2xy=25-13=12\Rightarrow\ xy=6$

So you could also solve

$x+y=5$

$xy=6$

10. oh i see where i went wrong! i got -x^2 and it should be 2x^2. i can take it from here thanks!