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$$\begin{align*}\sqrt{7x+y}+\sqrt{x+y}=6\\\sqrt{x+y}-y+x=2\end{align*}$$ I have tried various things squaring, summing but nothing really helped, got some weird intermediate results which are probably useless such as: $$(y-x)(y+x+4)+4-x-y=0$$ or $$x_{1,2}=\frac{2y-3\pm\sqrt{8y-7}}{2}$$ Could you please give me some hints?

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Some hints: the second equation leads to $\sqrt{x + y} = 2 + y - x$, which can be substituted into the first equation. Rearrange the first equation and square it. Combine this with $x + y = (2 + y - x)^2$. Two equations in two variables.... –  Zarrax Sep 9 '11 at 13:29

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Hint: If you write the second as $\sqrt{x+y}=2+y-x$ you can substitute that into the first. Now you have only one radical in each equation. If you isolate the radical, you can square to get rid of it, which will give two quadratic equations. Make sure to check your solutions to make sure they are not extraneous, introduced by squaring. You can also introduce $z=x+y, w=y-x$, which simplifies the algebra a bit.

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yeah I already have done this, but the problem was I had $x^2$, $xy$ and $y^2$ and didn't know how to deal with them but now understood when @Zarrax wrote that I should combine this too $x+y=(2+y-x)^2$ –  Templar Sep 9 '11 at 13:49
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That is the beauty of introducing $z$ and $w$-the cross terms go away. It works because $x$ and $y$ only come in as $x-y$ in the terms you are going to square, so now you only square terms in one variable. –  Ross Millikan Sep 9 '11 at 13:52

This is the way I would solve the given system in my school days. It is very similar to the hint provided by Ross Millikan. It turns out that one of the equations becomes linear. I omit some intermediate steps. The system

$$\left\{ \begin{array}{c} \sqrt{7x+y}+\sqrt{x+y}=6 \\ \sqrt{x+y}-y+x=2 \end{array} \right. \tag{1}$$

is equivalent to

$$\left\{ \begin{array}{c} \sqrt{7x+y}=4+x-y \\ \sqrt{x+y}=2-x+y. \end{array} \right. $$

By squaring both sides, we get

$$\left\{ \begin{array}{c} 7x+y=\left( 4+x-y\right) ^{2} \\ x+y=\left( 2-x+y\right) ^{2} \end{array} \right. \tag{2}$$

or

$$\left\{ \begin{array}{c} y=\left( 4+x-y\right) ^{2}-7x \\ \left( 4+x-y\right) ^{2}-7x=\left( 2-x+y\right) ^{2}-x, \end{array} \right. $$

which is equivalent to

$$\left\{ \begin{array}{c} y=\left( 4+x-y\right) ^{2}-7x \\ -12y+6x+12=0 \end{array} \right. $$

and to

$$\left\{ \begin{array}{c} 1+\frac{1}{2}x=9-4x+\frac{1}{4}x^{2} \\ y=1+\frac{1}{2}x. \end{array} \right. \tag{3}$$

This system has two pairs of solutions $\left( x,y\right) =(2,2)$ and $ \left( x,y\right) =(16,9)$, but only $\left( x,y\right) =(2,2)$ is a solution of $(1)$.

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When we see a square root, it is hard to resist the impulse to square. But resist we will, and such a decision is often useful.

The kind of solution we are looking is not specified, but the expressions $\sqrt{7x+y}$ and $\sqrt{x+y}$ suggest we are looking for real solutions. Let $$7x+y=p^2 \qquad\text{and}\qquad x+y=q^2,$$ where we can assume that $p$ and $q$ are non-negative.

Note that $(7x+y)-4(x+y)=3x-3y$, so $3(x-y)=p^2-4q^2$. So our equations can be rewritten as $$p+q=6\qquad \text{and}\qquad q+\frac{1}{3}(p^2-4q^2)=2.$$ Now in principle it's all over! (Substitute $6-q$ for $p$ in the second equation. We get a quadratic in the variable $q$.)


For fun we continue in another way. Observe that the second equation can be written as $p^2-4q^2+3q=6$. Is that a $6$ on the right? Yes, so we have $p^2-4q^2+3q=p+q$, and therefore $$p^2-4q^2=p-2q.$$ Since $p^2-4q^2=(p-2q)(p+2q)$, we conclude that (a) $p-2q=0$ or (b) $p+2q=1$.

Case (a): We have $p+q=6$ and $p=2q$, so $p=4$, $q=2$. That gives $7x+y=16$, $x+y=4$, so $x=y=2$. This works.

Case (b): We have $p+q=6$ and $p+2q=1$, and therefore $p=11$, $q=-5$. This contradicts the fact that $q$ is non-negative.

Comment: It is perhaps worth looking at Case (b) more closely. We get $7x+y=121$, $x+y=25$, which yields $x=16$, $y=9$. If we interpret $\sqrt{121}$ as being $11$, and $\sqrt{25}$ as being $-5$, then $x=16$, $y=9$ is a solution. But that is a quite peculiar interpretation of $\sqrt{25}$ if we are working in the reals.

However, the manipulations that we did apply equally well to complex numbers. So we have found all complex solutions. But if we are working in the complex numbers, interpreting $\sqrt{121}$ as $11$ and $\sqrt{25}$ as $-5$ is not a problem. Thus we have the slightly paradoxical fact that if we are working in the reals, there is one real solution, while if we are working in the complex numbers, there are two real solutions!

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