# How do I eliminate $x$ and $y$ from the system $x^2 y= a$, $x(x+y)= b$, $2x+y=c$ to get a single equation in $a$, $b$, $c$?

Alright, a homework problem.

I'm stuck at this question,

Eliminate $x$ and $y$ from the given equations to get a single equation in terms of $a$ , $b$ and $c$

\begin{align} x^2 y &= a \\ x(x+y) &= b \\ 2x+y &=c \end{align}

Let me tell you what I tried, I tried to get $y$ from one equation and substitute in the other two. Turns out that I'm not able to fully get rid of both $x$ and $y$. Help please.

• can you show what you have obtained currently? – Siong Thye Goh Aug 26 '18 at 9:18
• @SiongThyeGoh yea sure, it's not anything impressive though. I was just trying things out to see where it goes. – William Aug 26 '18 at 9:20
• Why not totally isolate $y$? $y=\frac a{x^2} = \frac bx-x=c-2x$ You can use this to get rid of $x$. – Mohammad Zuhair Khan Aug 26 '18 at 9:21
• @MohammadZuhairKhan yes that's exactly what I tried. But that's the thing, I'm not able to get rid of both. Either $x$ stays in or $y$. – William Aug 26 '18 at 9:24
• Solve 2nd eqn for $y$, sub into 3rd eqn, solve that for $x$, etc. – Gerry Myerson Aug 26 '18 at 9:27

I'm confused why you say in comments that "still $y$ does not get eliminated" -- at a point in the conversation where $y$ should have disappeared long ago.

You can solve your third equation $2x+y=c$ to get $y=c-2x$. When you plug that into the two other equations you get $$x^2(c-2x)=a \\ x(x+c-2x)= b$$ No matter what you do subsequently, there won't be any $y$s left to deal with!

What I would do at this point is put off dividing for as long as possible, so rewrite the second equation to $$x^2 = cx - b$$ This form lets you reduce any polynomial in $x$ to a first-degree polynomial, by repeatedly using it to eliminate the highest-degree term. We will use it to simplify the third-degree first equation. First substitute the leading $x^2$ to get $$(cx-b)(c-2x) = a$$ Multiply out: $$c^2x + 2bx - 2cx^2 - bc = a$$ and then insert $x^2=cx-b$ once again: $$c^2x + 2bx - 2c(cx-b) - bc = a$$ This is finally a linear equation in $x$. You can solve it without running into any square roots, and insert this value for $x$ into $x^2=cx-b$. What is left then is a rational equation in only $a$, $b$, and $c$, as desired.

Unless the coefficient of $x$ in the linear equation was $0$ (which is a case you'll need to handle separately), each step along the way was reversible, so you will know that if you have $a,b,c$ that satisfy the final equation, they will also have a solution for $x$ and $y$.

Combine the first and second equations to eliminate $y$:

$x(x+\frac{a}{x^2})=b$

which tidies to

$x^2+\frac{a}{x}=b$ ....... (1)

combine the second and third equations to eliminate $y$:

$x(x+c-2x)=b$

which tidies to

$x^2-cx+b=0$

$x=\frac{c \pm \sqrt{c^2-4b}}{2}$

which you can substitute into equation (1):

$\left(\frac{c \pm \sqrt{c^2-4b}}{2}\right)^2+\frac{a}{\left(\frac{c \pm \sqrt{c^2-4b}}{2}\right)}=b$

• It might be messy but it is, at least, an answer – Mandelbrot Aug 26 '18 at 10:07

You can notice that the product resp. the sum of $x$ and $x+y$ are $c$ resp. $b$, you solve that and get a simple system in $x$ and $x+y$ and you substitute in the first equation. Pretty same ideas.

I will work with 2nd and 3rd equations to solve for $x$ and $y$.

From (3) we get, $$y=c-2x$$ Substituting the value in (2) we get, $$x(x+(c-2x))=b$$ $$\implies x^2-cx+b=0$$ $$x=\dfrac{1}{2}(c+\sqrt{c^2-4b}) \text{ or, }\dfrac{1}{2}(c+\sqrt{c^2-4b})$$ If $x=\dfrac{1}{2}(c+\sqrt{c^2-4b})$,then, $$y=-\sqrt{c^2-4b}$$

Now,we plug the values of x and y in equation in (1), We get,

$$-\dfrac{1}{4}(c^2+2c\sqrt{c^2-4b}+c^2-4b)\sqrt{c^2-4b}=a$$ And this is the required equation.

[If you use another value of $x$ then,we will get another equation like it.]