Question

Solve the following systems of equations.
$u+2v=2,|2u-3v|=1$

Answer 1

Given equations,

$u+2v=2$

$|2u-3v|=1$

$2u-3v=\pm 1$

If $2u-3v=-1$,

$(2u-3v)-2(u+2v)=-1-2\left(2\right)$

$-7v=-5$

$v=\frac{5}{7}$

$u=2-2v$

$u=\frac{4}{7}$

If $2u-3v=1$,

$2(u+2v)-(2u-3v)=2\left(2\right)-1$

$7v=3$

$v=\frac{3}{7}$

$u=2-2v$

$u=\frac{8}{7}$

$\therefore$ $(u=\frac{4}{7},v=\frac{5}{7})$ and $(u=\frac{8}{7},v=\frac{3}{7})$ are possible solution of given equations.