Question 252412
Even though these are non-linear equations, we can still use substitution to solve them. I'll do the first one to get you started. If that doesn't help either repost or ask me.



# 1


{{{xy^2=1}}} Start with the second equation.



{{{x=1/(y^2)}}} Divide both sides by {{y^2}}} to isolate 'x'.



{{{7x-8y=24}}} Move onto the first equation.



{{{7(1/(y^2))-8y=24}}} Plug in {{{x=1/(y^2)}}}



{{{7-8y^3=24y^2}}} Multiply EVERY term by the LCD {{{y^2}}} to clear out the fractions.



{{{-8y^3-24y^2+7=0}}} Get every term to the left side.



{{{8y^3+24y^2-7=0}}} Multiply every term by -1.



Now use the rational root theorem to find that {{{y=1/2}}} is a root to the polynomial equation above. In other words, if you plug in {{{y=1/2}}} into {{{8y^3+24y^2-7}}}, you will get 0. Because of this fact, this means that {{{2y-1}}} is a factor of {{{8y^3+24y^2-7}}}



Now use polynomial long division to find that {{{(8y^3+24y^2-7)/(2y-1)=4y^2+14y+7}}}. So {{{8y^3+24y^2-7=(2y-1)(4y^2+14y+7)}}}



This tells us that {{{(2y-1)(4y^2+14y+7)=0}}}. Since we know that {{{2y-1=0}}} gives a root of {{{1/2}}}, we can ignore this equation. So the next step is to solve {{{4y^2+14y+7=0}}} for 'y'. Use the quadratic equation to find the next two solutions of {{{y=(-7+sqrt(21))/4}}} or {{{y=(-7-sqrt(21))/4}}}



So the three solutions in terms of 'y' are {{{y=1/2}}}, {{{y=(-7+sqrt(21))/4}}} or {{{y=(-7-sqrt(21))/4}}}



From here, plug each solution (in terms of 'y') into {{{x=1/(y^2)}}} to find the corresponding solution in terms of 'x'.



I skipped a bit of steps (since they're a bit long and I'm out of time for now), so feel free to ask about any step.