# Solving in terms of z , three variable two equation system

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

• 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 $z\ or\ {5\over3}z$.
– user585765
Apr 12, 2019 at 11:16
• 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.
– user585765
Apr 12, 2019 at 11:21

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 $$x=4z-2\Big(\frac{5z}{3}\Big)=4z-\frac{10z}{3}=\frac{2z}{3}.$$
• No problem, and apologies, that's my mistake. Should indeed be $16y^2$, will amend it. Apr 12, 2019 at 12:44