# Solving a system of equations: $\begin{cases}\frac xy-\frac yx=\frac{15}4\\2x-5y=9\end{cases}$

How should I approach this type of a system equation?

$$\begin{cases} \dfrac xy-\dfrac yx=\dfrac{15}4\\ 2x-5y=9 \end{cases}$$

I tried to multiply the first equation by $4xy$ and divide the second one by $2$. After that I got this system: $$\begin{cases} 4x^2 - 4y^2 = 15xy\\ x - 2.5y = 4.5 \Longrightarrow x = 4.5 + 2.5y \end{cases}$$ Then I put $x$ from the second equation in the first one:

$$4(4.5 + 2.5y)^2 - 4y^2 = 15y(4.5 + 2.5y)$$

When I solved it I got these results:

1. $$x_1 = \frac{189}{22},\ y_1 = \frac{18}{11}.$$
2. $$x_2 = -3,\ y_2 = -3.$$

But these results are incorrect.

These are the answers from my book:

1. $$x_1 = 12,\ y_1 = 3.$$
2. $$x_2 = \frac{9}{22},\ y_2 = -\frac{18}{11}.$$

The substitution is correct. Continuing: \begin{align*} 4(4.5 + 2.5y)^2 - 4y^2 &= 15y(4.5 + 2.5y) \\ (25 y^2 + 90 y + 81) - 4y^2 &= 37.5 y^2 + 67.5 y \\ -16.5 y^2 + 22.5 y + 81 &= 0 \\ -33 y^2 + 45 y + 162 &= 0 \\ 11 y^2 - 15 y - 54 &= 0 \\ (11 y + 18) (y - 3) &= 0 \end{align*}

For $y = -18/11$, we obtain $x = 9/22$. For $y = 3$, we obtain $x = 12$.

• Thank you very much!!! I did a mistake when I was solving this equation... – Vitalii Paprotskyi Apr 20 '18 at 8:07

Put $x/y =t$ then $$t-{1\over t} = {15\over 4} \implies 4t^2-15t-4 =0$$

Since $(4t+1)(t-4)=0$ so $t=-1/4$ or $t=4$.

1. case $y=-4x$ put in to $2x+20x=9$ so $x=9/22$ and $y=-18/11$.

2. case $x=4y$ and $y=3$ and $x=12$.

• Please, take a look at "Answers from my book". I just added it to my question. Thanks! – Vitalii Paprotskyi Apr 20 '18 at 8:03
• Nice, very compact. You can even dispense with the factorization of the $t$ polynomial. – Yves Daoust Apr 20 '18 at 8:27

Even shorter:

With $t:=\dfrac xy$, the first equation becomes $t^2-1=\dfrac{15}4t$ which you solve for $t$ (two solutions). Then from $2t-5=\dfrac9y$, you can draw $y$, then $x$.

• wow) That's amazing) Thanks! – Vitalii Paprotskyi Apr 20 '18 at 8:54
• @VitaliiPaprotskyi: yep, but a little "cheating" as this was obtained after other attempts and lengthy reflection. – Yves Daoust Apr 20 '18 at 8:56

Multiply the first equation by $4xy$ and rewrite as

$$4x^2-15xy-4y^2=0,$$ which factors as

$$(4x+y)(x-4y)=0.$$

(you can obtain this be solving the quadratic equation in $x$).

Now you have two easy linear systems

$$\begin{cases}4x+y=0,\\2x-5y=9,\end{cases}$$

$$\begin{cases}x-4y=0,\\2x-5y=9.\end{cases}$$

For the first, $y=-4x$ and $22x=9$. For the second, $x=4y$ and $3y=9$.

• Cool. This solution is so short. Thanks! – Vitalii Paprotskyi Apr 20 '18 at 8:30
• @VitaliiPaprotskyi: thanks for the appraisal. ChristianF worked out the same method, in a more compact way. I wanted to show how the question can be reduced to standard subproblems. – Yves Daoust Apr 20 '18 at 8:37
• @VitaliiPaprotskyi: the "trick" to achieve this is to observe that the first equation is homogeneous and represents a pair of lines through the origin. Which you intersect with the one in the second equation. – Yves Daoust Apr 20 '18 at 8:41
• Thanks. It's a great experience for me! I'm going to investigate all these answers more deeply! – Vitalii Paprotskyi Apr 20 '18 at 8:49

Hint:

$$2x-5y=9\iff2(x-2)=5(y+1),$$

$\implies\dfrac{x-2}5=\dfrac{y+1}2=z$(say)

$$\dfrac{15}4=\dfrac{5z+2}{2z-1}-\dfrac{2z-1}{5z+2}$$

$$\iff15(2z-1)(5z+2)=(5z+2)^2-(2z-1)^2$$

which is a Quadratic equation in $z$ on rearrangement

• This solution is quite interesting. Thanks! – Vitalii Paprotskyi Apr 20 '18 at 8:10