Question

Can someone please help me use substitution to solve for two values of y?

x^2+y^2-16y+39=0
y^2-x^2-9=0

Answer 1

I solved the second equation for x^2 and got y^2-9+y^2-16y+39=0, but I dont know what to do from there

Answer 2

adding the two equations

Answer 3

Ok, good. Now you can use the quadratic formula, or (if you are able to see it) that\[2y^2 - 16y + 39 - 9 = 0 => y^2 - 8y + 15 = (y-3)(y-5)\], with roots y = 3 and y = 5.

Answer 4

The thing to do next is to solve the second equation (because it is the easiest) first for y = 3. You find x = some number, and you put that into the first equation. If the left side of the equation is not equal to the right side, then y = 5 has to be correct, and you can solve equation number two again and find the correct x value. (You should also check this x value by calculating the left and right side of the first equation. Then it has to be correct :))

Answer 5

I'm confused on the finding the x's part :(

Answer 6

OBS! Important, as surhithayer said, you have to add the two equations in order to get rid of the x^2s

Answer 7

*surjithayer

Answer 8

Do you want me to show you how I add the equations?

Answer 9

\[x^2+y^2-16y+39+y^2-x^2-9 = 0\], so we get \[2y^2-16y+30 = 0\]. We can divide by 2, which gives us \[y^2-8y+15 = 0\] with roots y = 3 and y = 5. Now, we have to solutions for y, but only one of them can be true. In order to determine which one is the correct value, we have to use the equations again, this time the second (easier to calculate with). With y = 3, we get by using the second equation:
\[(3)^2-x^2-9 = 0 => x^2 = 0\], so x = 0. Checking with the first equation:
\[0^2+3^2-16*3+39 = 0\], which is equal to the right side of the same equation, so y has to be y = 3, which implies x = 0.
Hope this will clear things up for you :)

Answer 10

@SfunnSneffanS im so sorry, i had to pick my brother up from school! let me just look this all over really quickly!

Answer 11

so the coordinates are (0,3) (0,5)?

Answer 12

No problem :)
I am so sorry too, for forgetting that there were two valid values for y!
No the coordinates are (4,5) AND (0,3). You get x = 4 by solving the second equation for y = 5.

Answer 13

oh thats right! i forgot to do that part! thank you soo much :)

Answer 14

And then by using the first equation with x = 4 and y = 5, you will get:\[4^2 + 5^2 - 16*5 + 39 = 0, which is equal to the right side of the same equation.
No problem, and sorry for not testing for y = 5; thought there only was supposed be one set of coordinates.