Solve 27+3+10+6*-6x+27geq10

$$27 + 3 + 10 + 6 \times - 6 x + 27 \geq 10$$

Solve for x

$$x\leq \frac{19}{12}$$

Solution Steps

Add 27 and 3 to get 30.

$$30+10+6\left(-6\right)x+27\geq 10$$

Add 30 and 10 to get 40.

$$40+6\left(-6\right)x+27\geq 10$$

Multiply 6 and -6 to get -36.

$$40-36x+27\geq 10$$

Add 40 and 27 to get 67.

$$67-36x\geq 10$$

Subtract 67 from both sides.

$$-36x\geq 10-67$$

Subtract 67 from 10 to get -57.

$$-36x\geq -57$$

Divide both sides by -36. Since -36 is negative, the inequality direction is changed.

$$x\leq \frac{-57}{-36}$$

Reduce the fraction \frac{-57}{-36} to lowest terms by extracting and canceling out -3.

$$x\leq \frac{19}{12}$$