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Solve in terms of $z$ $$ \begin{cases} 4z&= x + 2y \\ 3z^2&=\frac{1}{2}x^2 + y^2 \\ \end{cases} $$

Solution: $x = 2z/3$ and $y = 5z/3$.

I don't understand how they got to the solution with just those two equations I tried using substitution but was unable to single out a variable. I substituted $x = 4z-2y$ into eq.2 but got a quadratic I was unable to solve The problem itself on Khan Academy

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  • $\begingroup$ I don't know, why you are not able to solve it. $x=4z-2y$, put this value of $x$ in second equation. Then you'll get a quadratic in $y$. Solve for $y$(I myself have solved it) $y$ will turn out to be $z\ or\ {5\over3}z$. $\endgroup$
    – user585765
    Apr 12, 2019 at 11:16
  • $\begingroup$ Even though it is MathSE but I'll give you a hint how to solve such questions quickly, without even bothering for conservation of kinetic energy. You might be knowing that coefficient of restitution $(e)$ is $1$ for elastic collisions. Therefore for elastic collisions, relative velocity of approach=relative velocity of separation. Therefore $2v_0-v_0=v_2-v_1$. Solve this equation with $4mv_0=mv_1+2mv_2$ and you'll quickly get your answer. $\endgroup$
    – user585765
    Apr 12, 2019 at 11:21

1 Answer 1

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From the first equation we get $$z=\frac{x+2y}{4},$$ subsituting that into equation two we get $$3\Big(\frac{x+2y}{4}\Big)^2=\frac{1}{2}x^2+y^2,$$ or equivalently $$\frac{3x^2+12yx+12y^2}{16}=\frac{1}{2}x^2+y^2\Rightarrow 3x^2+12yx+12y^2=8x^2+16y^2,$$ which simplifies to $$5x^2-12xy+4y^2=0.$$ This is a quadratic in $x$ and this factorises to (by inspection or, if you like, using the quadratic formula) $$(5x-2y)(x-2y)=0\Rightarrow x=\frac{2y}{5},\quad x=2y.$$ Now, from equation one, observe that $x=4z-2y$, so $$\frac{2y}{5}=4z-2y\Rightarrow\frac{12y}{5}=4z\Rightarrow y=\frac{5z}{3}.$$ Then $$x=4z-2\Big(\frac{5z}{3}\Big)=4z-\frac{10z}{3}=\frac{2z}{3}.$$

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  • $\begingroup$ Thanks, you helped with my main issue just a quick question how did you go from 1/2x^2 + y^2 to 8x^2 + 8y^2? I get you multiplied by 16 but wouldn't it be 8x^2 and 16y^2? $\endgroup$
    – Josh Shoemaker
    Apr 12, 2019 at 12:14
  • $\begingroup$ No problem, and apologies, that's my mistake. Should indeed be $16y^2$, will amend it. $\endgroup$
    – thesmallprint
    Apr 12, 2019 at 12:44

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