Solving systems of equations

[RHSLilSweetie07]

Find all solutions to the system of equations.

a)x-y^2=0
y-x^2=0

b)(4/x^2) + (6/y^4) = (7/2)
(1/x^2) - (2/y^4) = 0

For the first one, I set x to equal y^2 and substituted it into the other equation.

y - (y^2)^2 = 0
y - y^4 = 0
Then what should I do, factor? Please help.

Also, please help me get started on the second problem.

 

[Unco]

For the first one, I set x to equal y^2 and substituted it into the other equation.

y - (y^2)^2 = 0
y - y^4 = 0
Then what should I do, factor?

Absolutely. Rememeber that y^3-1 can be factorised. Hint: y-1 is a factor.

Two roots are are non-real.

 

[Unco]

b)(4/x^2) + (6/y^4) = (7/2) [1]
(1/x^2) - (2/y^4) = 0 [2]
Also, please help me get started on the second problem.

Rewrite [2] as (1/x^2) = (2/y^4).

Make y^4 the subject and subsitute into [1] to solve for x.

 

[soroban]

Hello, RHSLilSweetie07!

Find all solutions to the system of equations.

\(\displaystyle b)\;\frac{4}{x^2}\,+\,\frac{6}{y^4}\:=\;\frac{7}{2}\)
. . \(\displaystyle \frac{1}{x^2}\,-\,\frac{2}{y^4}\;=\;0\)

Multiply the second equation by 3:

. . . \(\displaystyle \frac{4}{x^2}\,+\,\frac{6}{y^4}\;=\;\frac{7}{2}\)
. . . \(\displaystyle \frac{3}{x^2}\,-\,\frac{6}{y^4}\;=\;0\)

Add the equations: .\(\displaystyle \frac{7}{x^2}\:=\:\frac{7}{2}\)

Multiply by \(\displaystyle 2x^2:\;\;14\,=\,7x^2\qquad\Rightarrow\qquad x^2\,=\,2\qquad\Rightarrow\qquad x\,=\,\pm\sqrt{2}\)

Substitute into the second equation: . \(\displaystyle \frac{1}{(\pm\sqrt{2})^2}\,-\,\frac{2}{y^4}\:=\:0\qquad\Rightarrow\qquad\frac{1}{2}\,-\,\frac{2}{y^4}\:=\:0\)

Multiply by \(\displaystyle 2y^4:\;\;y^4\,-\,4\:=\:0\qquad\Rightarrow\qquad y^4\,=\,4\qquad\Rightarrow\qquad y\,=\,\pm\sqrt{2}\)

Answers: .\(\displaystyle (\pm\sqrt{2},\,\pm\sqrt{2})\)

 

[Unco]

Multiply by \(\displaystyle 2y^4:\;\;y^4\,-\,4\:=\:0\qquad\Rightarrow\qquad y^4\,=\,4\qquad\Rightarrow\qquad y\,=\,\pm\sqrt{2}\)

Answers: .\(\displaystyle (\pm\sqrt{2},\,\pm\sqrt{2})\)

Are we to ignore non-real roots?