Question

Find all values of $a$ for each of which the system of equations
$2x+2(a-1)y=a-4,2|x+1|+ay=2$ has a unique solution. Find that solution.

Answer 1

Given equations,

$2x+2(a-1)=a-4$

$2|x+1|+ay=2$

$x\ge -1$,

$2x+(2a-2)y=a-4$

$2x+ay=0$

$2x=-ay$

$(a-2)y=a-4$

$y=\frac{a-4}{a-2}$

$x=\frac{-a(a-4)}{2(a-2)}$

$x\le 1$, $2x+(2a-2)y=a-4\to \left(1\right)$

$-2x-2+ay=2\to \left(2\right)$

$\left(1\right)+\left(2\right)\Rightarrow (3a-2)y=a$

$y=\frac{a}{3a-2}$

$x=\frac{a^2-12a+8}{2(3a-2)}$

For Unique solution,

Both $y$ values must be equal

$\frac{a-4}{a-2}=\frac{a}{3a-2}$

$3a^2-12a-2a+8=a^2-2a$

$2a^2-12a-2a+8=0$

$a^2-6a+4=0$

$a=3\pm \sqrt{5}$

For $a=3\pm \sqrt{5}$, $x=\frac{-(3+\sqrt{5})(\sqrt{5}-1)}{2(\sqrt{5}+1)}$

$=\frac{-(\sqrt{5}+1)(\sqrt{5}-1)}{4}=-1$

(or)

$x=\frac{-(3-\sqrt{5})(-1-\sqrt{5})}{2(1-\sqrt{5})}$

$=\frac{(1-\sqrt{5})(1+\sqrt{5})}{4}=-1$

If $a=3+\sqrt{5}$ Then $x=-1,y=\frac{3-\sqrt{5}}{2}$ is solution.

If $a=3-\sqrt{5}$ Then $x=-1,y=\frac{3+\sqrt{5}}{2}$ is solution.