# Solve, $x^2-y^2=176$; $6x+5y=3\sqrt{400+x^2}=3\sqrt{576+y^2}$ [closed]

$x^2-y^2=176$ -- (|)

$26x^2+26y^2+60xy=3424$ -- (||)

$6x+5y=3\sqrt{400+x^2}=3\sqrt{576+y^2}$ -- (|||)

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## closed as not constructive by Alexander Gruber♦, ncmathsadist, Micah, amWhy, Austin MohrJan 5 '13 at 3:58

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Why do you need (|||)? You already had two equations in two unknowns. Or did you derive it from the first two? –  Ross Millikan Jan 4 '13 at 4:44
equation (||) was derived from equation (|||) –  Rajesh K Singh Jan 4 '13 at 4:53
equation (|) is derived from equation (|||) too –  miracle173 Jan 4 '13 at 5:32
you have different equations in the title and the body of your post. $5x+6y=$ in the title and $6x+5y=$ in the body. –  miracle173 Jan 4 '13 at 7:59
It is not a good idey to change title and content of the post permanently.It is not a good idea too to put the same information in the title and the body. –  miracle173 Jan 4 '13 at 8:12

Hint: You might define $z=x+y, w=x-y$. Then the first is $zw=176$. If you think the solutions are integers, you could just factor $176$ and try them, but this fails this time. Otherwise, subtract the first from the second to get $26x^2+26y^2+60xy=3424$ (the point being to get the same coeficient on $x^2, y^2$). Then note that $az^2+bw^2=(a+b)(x^2+y^2)+2(a-b)xy$ to change that into $28z^2-2w^2=3424$. Now you can substitute $w=\frac{176}z$ in and get a quadratic in $z^2$. I get pure imaginary solutions.

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I use equations (|||).

$$\begin{eqnarray} 5x+6y&=&3\sqrt{400+x^2} \tag{1} \\ 5x+6y&=&3\sqrt{576+y^2} \tag{2}\\ \end{eqnarray}$$

squaring them gives

$$\begin{eqnarray} 25y^2+60xy+27x^2-3600&=&0 \tag{3} \\ 16y^2+60xy+36x^2-5184&=&0 \tag{4}\\ \end{eqnarray}$$

$(3)-(4)$ gives $(5)$, an equation in $x^2$ and $y^2$.
$4\cdot(3)-3\cdot(4)$ gives $(6)$, an equation in $y^2$ and $xy$. $$\begin{eqnarray} x^2&=&y^2+176 \tag{5} \\ 52y^2+1152&=&-60xy \tag{6} \\ \end{eqnarray}$$

squaring $(6)$ gives $(7)$

$$\begin{eqnarray} x^2&=&y^2+176 \tag{5} \\ 2704y^4+11980y^2+1327104&=&3600x^2y^2 \tag{7} \\ \end{eqnarray}$$

we substitute $(5)$ in $(7)$ to get $(8)$

$$\begin{eqnarray} x^2&=&y^2+176 \tag{5} \\ y^4+\frac{4014}{7}y^2-\frac{10368}{7}&=&0 \tag{8} \\ \end{eqnarray}$$

$(8)$ can be split in two equations $(9)$ and $(10)$

$$\begin{eqnarray} z^2+\frac{4014}{7}z-\frac{10368}{7}&=&0 \tag{9} \\ y^2&=&z \tag{10} \\ x^2&=&z+176 \tag{11} \\ \end{eqnarray}$$

This gives the solutions $\frac{18}{7}$ and $-576$ for $z$ and therefore the solution pairs $x^2=\frac{1250}{7}, y^2=\frac{18}{7}$ and $x^2=-400, y^2=-576$ and

therefore $x=\pm 25\sqrt{\frac{2}{7}}, y=\pm 3\sqrt{\frac{2}{7}}$ and $x=\pm 20i, y=\pm 24i$ These are 8 pairs but one has to check each of them. One sees that the

signs of the solutions have to be opposite to each other. So the following solutions remain

$$\begin{eqnarray} x&=& 25\sqrt{\frac{2}{7}}, &y& &=& -3\sqrt{\frac{2}{7}} \\ x&=& -25\sqrt{\frac{2}{7}}, &y& &=& 3\sqrt{\frac{2}{7}} \\ x&=& 20i, &y& &=& -24i \\ x&=& -20i, &y& &=& 24i \\ \end{eqnarray}$$

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There is a difference between the equations in the title and in the body. Also the equation changed after several edits. I solved the equation $6x+5y=3\sqrt{400+x^2}=3\sqrt{576+y^2}$ -- (|||) from the body of version 5 –  miracle173 Jan 4 '13 at 8:07