$$(3x+1)/16+(2x-3)/7=(x+3)/8+(3x-1)/14$$
$x=5$
$$7\left(3x+1\right)+16\left(2x-3\right)=14\left(x+3\right)+8\left(3x-1\right)$$
$$21x+7+16\left(2x-3\right)=14\left(x+3\right)+8\left(3x-1\right)$$
$$21x+7+32x-48=14\left(x+3\right)+8\left(3x-1\right)$$
$$53x+7-48=14\left(x+3\right)+8\left(3x-1\right)$$
$$53x-41=14\left(x+3\right)+8\left(3x-1\right)$$
$$53x-41=14x+42+8\left(3x-1\right)$$
$$53x-41=14x+42+24x-8$$
$$53x-41=38x+42-8$$
$$53x-41=38x+34$$
$$53x-41-38x=34$$
$$15x-41=34$$
$$15x=34+41$$
$$15x=75$$
$$x=\frac{75}{15}$$
$$x=5$$
Show Solution
Hide Solution
