Question #e4ccd

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))#

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)#