# solution to equations having 4 variables

I have the following set of equations:

$$\frac{x^2}{2^2-1} + \frac{y^2}{2^2 - 3^2} + \frac{z^2}{2^2-5^2} + \frac{w^2}{2^2-7^2} = 1$$

$$\frac{x^2}{4^2-1} + \frac{y^2}{4^2 - 3^2} + \frac{z^2}{4^2-5^2} + \frac{w^2}{4^2-7^2} = 1$$ $$\frac{x^2}{6^2-1} + \frac{y^2}{6^2 - 3^2} + \frac{z^2}{6^2-5^2} + \frac{w^2}{6^2-7^2} = 1$$ $$\frac{x^2}{8^2-1} + \frac{y^2}{8^2 - 3^2} + \frac{z^2}{8^2-5^2} + \frac{w^2}{8^2-7^2} = 1$$

Well the nature of numbers in the denominator itself freaked me out and now I am unable to approach the question. I tried to simplify the expressions which turned out to be of no use. Any help would be appreciated.

• Just write it in matrix form $Au=b$, where $u = (x^2, y^2,z^2, w^2) ^T$and invert the matrix, that will give you the values of the squares. – Eddy Jul 10 '18 at 14:03
• From where does it come? – Dr. Sonnhard Graubner Jul 10 '18 at 14:13
• ${{x}^{2}}=\frac{11025}{1024},{{y}^{2}}=\frac{10395}{1024},{{z}^{2}}=\frac{9009}{1024}and\ {{w}^{2}}=\frac{6435}{1024}$ – user547564 Jul 10 '18 at 15:04

there are many methods but i prefer to use matrix

Then i try to solve the system $AX=T$, where

$$A=\left( \begin{array}{cccc} \frac{1}{2^2-1} & \frac{1}{2^2-3^2} & \frac{1}{2^2-5^2} & \frac{1}{2^2-7^2} \\ \frac{1}{4^2-1} & \frac{1}{4^2-3^2} & \frac{1}{4^2-5^2} & \frac{1}{4^2-7^2} \\ \frac{1}{6^2-1} & \frac{1}{6^2-3^2} & \frac{1}{6^2-5^2} & \frac{1}{6^2-7^2} \\ \frac{1}{8^2-1} & \frac{1}{8^2-3^2} & \frac{1}{8^2-5^2} & \frac{1}{8^2-7^2} \\ \end{array} \right)$$

and

$$X=\left( \begin{array}{c} x^2 \\ y^2 \\ z^2 \\ w^2 \\ \end{array} \right)$$

and

$$T=\left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ \end{array} \right)$$

to find $X$ we must calculate the inverse matrix of $A$ and then multuply by the $T$ we can get

$$X=\frac{1}{1024} \left( \begin{array}{c} 11025 \\ 10395 \\ 9009 \\ 6435 \\ \end{array} \right)$$

finally

$$\left( \begin{array}{c} x \\ y \\ z \\ w \\ \end{array} \right)=\pm\sqrt{X}=\pm\frac{1}{32}\left( \begin{array}{c} 105 \\ 3 \sqrt{1155} \\ 3 \sqrt{1001} \\ 3 \sqrt{715} \\ \end{array} \right)$$

i used a Math Software

• Don't forget to sprinkle liberally with $\pm$s, to get the full $16$ possible solutions. :-) – Theo Bendit Jul 10 '18 at 15:13
• verification : $\frac{10395}{2^2-3^2}+\frac{9009}{2^2-5^2}+\frac{6435}{2^2-7^2}+\frac{11025}{2^2-1}$ you get 1024 – El Mouden Jul 10 '18 at 15:13
• thanks for the solution but is there a way to solve it by not making use of the huge numbers – saisanjeev Jul 18 '18 at 5:41
• maybe there are a pretty method but sorry i have not time for think in it . – El Mouden Jul 20 '18 at 11:57