Cookies mean peni* in hungarian lmaoNo cookies.
No help.
For the second one, square the xy=8 to get (xy)^2 = 64. Then times the first equation by x^2 to getMy first questions
Solve simultaneously
7/x-5/y=3 and 2/x+25/2y=12
and
9x^2+y^2=52 and xy=8
For the first one, make 5/y the subject for equation 1. Then sub into equation 2 because equation 2 is equal to 2/x+(5/2 x 5/y)=12. Solve for x then sub back in to equation 1 for y.My first questions
Solve simultaneously
7/x-5/y=3 and 2/x+25/2y=12
and
9x^2+y^2=52 and xy=8
I've reached this part, problem is I don't know how to solve it after this.For the second one, square the xy=8 to get (xy)^2 = 64. Then times the first equation by x^2 to get
Solve the resulting quadratic.
He showed that x^2y^2=64, so replace x^2y^2 with 64 to be left with the quadratic 9x^4-52x^2+64=0. Then solve that quadratic (it's a reducible quadratic, let u=x^2)I've reached this part, problem is I don't know how to solve it after this.
Still not following (sorry if I'm being a pest)He showed that x^2y^2=64, so replace x^2y^2 with 64 to be left with the quadratic 9x^4-52x^2+64=0. Then solve that quadratic (it's a reducible quadratic, let u=x^2)
I'm latexing now...Still not following (sorry if I'm being a pest)
thanksI'm latexing now...
9x^4-52^2+64=0Still not following (sorry if I'm being a pest)