help Find the real numbers $x,y,z$ such that $(x+y)\cdot x= \frac{a^{2}}{2}$ and $(z+y)\cdot z= \frac{b^{2}}{2}$ and $x+y+z= \frac{a+b}{\sqrt{2}}$

Question

Find the real numbers $x,y,z$ such that $(x+y)\cdot x= \frac{a^{2}}{2}$ and $(z+y)\cdot z= \frac{b^{2}}{2}$ and $x+y+z= \frac{a+b}{\sqrt{2}}$ where $a$ and $b$ are positive real numbers .

MY IDEAS

That is all I could think of. I tried reducing the $a$ and $b$ and replacing the numbers every time I had the oportunity to.

Hope one of you can help me! Thank you!

Answer 1

Steps towards a solution. If you're stuck, show what you've tried.

1. Substitute $ x+y = \frac{ a+b}{\sqrt{2}} - z $ and $z+y = \frac{a+b}{\sqrt{2}} - x$ into the given equations. This eliminates $y$.

$(\frac{a+b}{\sqrt{2}} - z ) x = \frac{a^2}{2}, ( \frac{ a+b}{\sqrt{2}} - x) z = \frac{b^2}{2}.$

2. Take their difference and show that $x - z = \frac{ a - b } { \sqrt{2}}$.
3. Substitute back into the equations from step 1 to eliminate $z$ and get $ x = \frac{a}{\sqrt{2}}$.

$ ( \frac{ a+b}{\sqrt{2}} - x) (x - \frac{ a - b } { \sqrt{2}}) = \frac{b^2}{2} \Rightarrow ( x - \frac{a}{\sqrt{2}} ) ^2 = 0 .$

4. Likewise, z = $\frac{b}{\sqrt{2}}$.
5. Hence, $y = 0$.
6. Finally, verify that $ (\frac{a}{\sqrt{2}}, 0, \frac{b}{\sqrt{2}})$ does indeed satisfy the original conditions.