Question

How do you solve this simultaneous equation?
2x + 3y = 13
2xy + 5y^2 - 4x^2 = 41

Answer 1

Use substitution: y = (13 - 2x)/3 and put it into your second equation. Hopefully you will get a quadratic in x which you can solve (hopefully b^2 - 4ac >= 0). There will be two values for x, each corresponding to a different value of y, unless you have a repeated x, in which case the y will be distinct.

Answer 2

yeah. you could use the first equation for value of y and x, it's optional.

then substitute value of y or x to the second equation to find x or y

Answer 3

I've tried doing that but i can't seem to get the answer which is x=2 or -8/1/2. y=3 or 10. ;/

Answer 4

I'll do it in a few mins after I've had my breakfast if it isn't done it by then...

Answer 5

Okay, thank youu! (:

Answer 6

Right...this is going to take a while to type just to give you the very gory last details...

Answer 7

haha okay :)

Answer 8

Rearrange 2x+3y=13 to give: y=(13-2x)/3. Plug this into your second equation to get:
\[\frac{2x(13-2x)}{3} + \frac{5(13-2x)^2}{9} - 4x^2 - 41 = 0\]
now multiply the whole thing by 3x9 = 27 to remove the fractions:
\[18x(13-2x) + 15(169 - 52x + 4x^2) - 108x^2 - 1107 = 0\]
next isolate the coefficients so we can write our familiar quadratic equation:
\[x^2(60-36-108) + x(234-780) + (2535 - 1107) = 0\]
now just calculate:
\[-84x^2 - 546x + 1428 = 0\]
simplify this by dividing by -6:
\[ 14x^2 + 91x - 238 = 0\]
Now use the quadratic formula because it's a lovely thing :p
\[x = \frac{-91 \pm \sqrt{91^2 + 4(14)(238)}}{28} = \frac{-91 \pm \sqrt{21609}}{28} = \frac{-91 \pm 147}{28}\]
so the positive one is x = 2 and the negative one is x=-8.5. Now we use the equation from before:
\[y = \frac{13 - 2x}{3}\]
in order to find two y-values, one associated with x=2, the other associated with x=-8.5.
Substituting x=2 into y=(13-2x)/3 gives y = 3.
Substituting x=-8.5 into y=(13-2x)/3 gives y = 10.
So (x,y) solutions are: (2,3) and (-8.5, 10).

Answer 9

Hope this is clear!

Answer 10

you can pay me in medals :D

Answer 11

thank you so much :) i understand now