$$\frac{(x-3)}{5}\le3\frac{(2-x)}{4}$$
$x\leq \frac{42}{19}$
$$4\left(x-3\right)\leq 15\left(2-x\right)$$
$$4x-12\leq 15\left(2-x\right)$$
$$4x-12\leq 30-15x$$
$$4x-12+15x\leq 30$$
$$19x-12\leq 30$$
$$19x\leq 30+12$$
$$19x\leq 42$$
$$x\leq \frac{42}{19}$$
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