by Guest » Sat Feb 13, 2016 8:38 am
I'm not willing to give you the entire solutions, because they look like homework questions, but since they are quite difficult I'll give you the outline on how to solve them.
2) Use the substitution [tex]a = x^2+2y+1[/tex], into the top equation.
Rearrange and square to get rid of the root, and get a quadratic in [tex]a[/tex].
Solve to get a value (or values) for [tex]x^2+2y+1[/tex].
Substitute the bottom equation to get rid of the [tex]y[/tex] and get a quadratic for [tex]x[/tex].
3) Very similar to 2, I'll let you work out the details.
4) Simply square the top equation.
Rearrange to get rid of the denominator, you should have a quadratic expression in [tex]x[/tex] and [tex]y[/tex], which factorizes.
Substitute the solution into the bottom equation to get another quadratic in [tex]x[/tex] (or [tex]y[/tex]) and solve.
5) Cube the bottom equation.
The expression you get will have two cube root terms, substitute the original version of the bottom equation into the cubed version of the bottom equation to get rid of the cube roots.
Use the substitution [tex]z=y+1[/tex] and rearrange to get rid of the denominators.
You should have a quadratic expression in [tex]z[/tex] and [tex]x[/tex] which factorizes.
Substitute the solutions into the top equation.
Rearrange and square (twice) to get rid of the square roots, to get a quadratic equation you can solve.
6) Use the substitution [tex]a=x+y[/tex], and [tex]b=xy[/tex] to write both the top and bottom equations in terms of [tex]a[/tex] and [tex]b[/tex] (and constants) only.
Eliminate [tex]b[/tex] to get a quadratic in [tex]a[/tex].
Once the solutions for [tex]a[/tex] and [tex]b[/tex] have been determined, you have a simpler pair of simultaneous equations to solve namely [tex]x+y=[/tex]constant, [tex]xy=[/tex]constant, which should be easy to solve by substitution.
7) Very similar to 6, except use [tex]a=x-y[/tex].
8 ) Very similar to 6, except rearrange and square to get rid of the square root.
9) Very similar to 6, except use [tex]a=x^2+y^2[/tex].
Once you have [tex]a[/tex] and [tex]b[/tex], use the fact that [tex]a+2b=(x+y)^2[/tex] to get a value for [tex]x+y[/tex].
10) Very similar to 6, except multiply the top equation by [tex]\sqrt{xy}[/tex] to get rid of the denominators, and use the substitution [tex]b=\sqrt{xy}[/tex].
Hope this helped,
R. Baber.