3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ can you show what you have obtained currently? $\endgroup$ – Siong Thye Goh Aug 26 '18 at 9:18
  • $\begingroup$ @SiongThyeGoh yea sure, it's not anything impressive though. I was just trying things out to see where it goes. $\endgroup$ – William Aug 26 '18 at 9:20
  • $\begingroup$ Why not totally isolate $y$? $y=\frac a{x^2} = \frac bx-x=c-2x$ You can use this to get rid of $x$. $\endgroup$ – Mohammad Zuhair Khan Aug 26 '18 at 9:21
  • $\begingroup$ @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 $y$. $\endgroup$ – William Aug 26 '18 at 9:24
  • $\begingroup$ Solve 2nd eqn for $y$, sub into 3rd eqn, solve that for $x$, etc. $\endgroup$ – Gerry Myerson Aug 26 '18 at 9:27
5
$\begingroup$

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 $y$.

$\endgroup$
1
$\begingroup$

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

$x^2-cx+b=0$

solving this quadratic gives

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

which you can substitute into equation (1):

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

$\endgroup$
  • 1
    $\begingroup$ It might be messy but it is, at least, an answer $\endgroup$ – Mandelbrot Aug 26 '18 at 10:07
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.