Question

Solve: $3x+4y\ge 12,x\ge 0,y\ge 1$ and $4x+7y\le 28$

Answer 1

Intersection of $3x+4y=12$ and $4x+7y=28$,

$3x+4y=12\times 4\Rightarrow 12x+16y=48$

$4x+7y=28\times 3\Rightarrow 12x+21y=84$

$\Rightarrow 5y=36\Rightarrow y=\frac{36}{5}$

$x=\frac{12}{3}-\frac{4}{3}\times \frac{36}{5}$

$\Rightarrow 4-\frac{48}{5}=-\frac{28}{5}$.