Solve for x
x\geq \frac{1}{3}
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45x\geq 12+3
Add 3 to both sides.
45x\geq 15
Add 12 and 3 to get 15.
x\geq \frac{15}{45}
Divide both sides by 45. Since 45 is positive, the inequality direction remains the same.
x\geq \frac{1}{3}
Reduce the fraction \frac{15}{45} to lowest terms by extracting and canceling out 15.
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