Solve for x
x\leq \frac{10}{3}
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12x+0\left(72-2x\right)\leq 40
Multiply 0 and 5 to get 0.
12x+0\leq 40
Anything times zero gives zero.
12x\leq 40
Anything plus zero gives itself.
x\leq \frac{40}{12}
Divide both sides by 12. Since 12 is positive, the inequality direction remains the same.
x\leq \frac{10}{3}
Reduce the fraction \frac{40}{12} to lowest terms by extracting and canceling out 4.
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