Question

Question #e4ccd

Answer 1

`((x,y),(-sqrt2, -1/sqrt2),(sqrt2, 1/sqrt2),(-3 sqrt3/4, -sqrt3/4),(3 sqrt3/4, sqrt[3]/4))`

Explanation:

Making `y = lambda x` and substituting

`{(x^2(1-lambda^2)=3/2),((1-lambda)/(1+lambda)=1/(6 lambda)):}`

From `(1-lambda)/(1+lambda)=1/(6 lambda)` we obtain

`lambda = {(1/3),(1/2):}` and now substituting those results into

`x^2=3/2 1/(1-lambda^2)->{(lambda=1/3->x=pm 3 sqrt3/4),(lambda=1/2->x=pmsqrt2):}`

And now as `y = lambda x` we have:

`((x,y),(-sqrt2, -1/sqrt2),(sqrt2, 1/sqrt2),(-3 sqrt3/4, -sqrt3/4),(3 sqrt3/4, sqrt[3]/4))`

Answer 2

`(+-sqrt(2),+-sqrt(2)/2)` and `(+-(3sqrt(3))/4,+-sqrt(3)/4)`

Explanation:

Step 1. Use cross multiplication to eliminate fractions from the second equation.

`(x-y)/(x+y)=x/(6y)`

`(x-y)6y=(x+y)x`

`6xy-6y^2=x^2+xy`

`6y^2-5xy+x^2=0`

Think of this as a quadratic equation: `ay^2+by+c=0`.

Factoring gives

`(2y-x)(3y-x)=0`

or

`y=1/2 x` and `y=1/3x`

Step 2. Plug these linear equations for `y` back into the other equation. First, use `y=x-:2`

`x^2-y^2=3/2`

`x^2-(1/2 x)^2=3/2`

`x^2-1/4 x^2=3/2`

`3/4x^2=3/2`

`x=+-sqrt(2)`

Then use `y=x-:3`

`x^2-(1/3x)^2=3/2`

`x^2-1/9x^2=3/2`

`8/9x^2=3/2`

`x=+-(3sqrt(3))/4`

Step 3. Plug values of `x` back into `y` equations.

`y=1/2x=1/2(+-sqrt(2))=+-sqrt(2)/2`

`y=1/3x=1/3(+-(3sqrt(3))/4)=+-sqrt(3)/4`

ANSWER: `(+-sqrt(2),+-sqrt(2)/2)` and `(+-(3sqrt(3))/4,+-sqrt(3)/4)`