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