Question

Find all values of $a$ for each of which the system of equations
$ax+(a-1)y=2+4a,3|x|+2y=a-5$ has a unique solution. Find that solution.

Answer 1

Given equations,

$ax+(a-1)y=2+4a\to \left(1\right)$

$3x+2y=a-5\to \left(2\right)$

$-3x+2y=a-5\to \left(3\right)$

$3\left|x\right|+2y=a-5\to \left(4\right)$

Consider given equations as line equations

$\therefore$ To have unique all the $3$lines should meat at same point,

Other than the intersection of $\left(2\right)$ and $\left(3\right)$, $\left(1\right)$ meet $\left(4\right)$ twice (or) zero times .for all $a$

$\therefore$ x-co-ordinate $=0$, For intersection of $\left(2\right)$ and $\left(3\right)$.

$\therefore x=0$,

$\left(1\right)\Rightarrow y=\frac{4a+2}{a-1}$

$\left(4\right)\Rightarrow y=\frac{a-5}{2}$

Both $y$'s must be equal.

$\therefore 4(2a+1)=(a-5)(a-1)$

$8a+4=a^2-6a+5$

$a^2-14a+1=0$

$a=\frac{14\pm \sqrt{196-4}}{2}=7\pm 4\sqrt{3}$

If $a=7+4\sqrt{3}$

$y=\frac{2+4\sqrt{3}}{2}=1+2\sqrt{3}(\because y=\frac{a-5}{2})$

$a=7-4\sqrt{3}$,

$y=\frac{a-5}{2}=\frac{2-4\sqrt{3}}{2}=1-2\sqrt{3}$

$\therefore$ At $a=7\pm 4\sqrt{3}$, Given set of equations has unique solution.

If $a=7+4\sqrt{3}$, $x=0$ and $y=1+2\sqrt{3}$ is solution.

If $a=7-4\sqrt{3}$, $x=0$ and $y=1-2\sqrt{3}$ is solution.