Solve 27+3+10+6^5x-6x+27geq10

$$27 + 3 + 10 + 6 ^ { 5 } x - 6 x + 27 \geq 10$$

Solve for x

$$x\geq -\frac{19}{2590}$$

Solution Steps

Add 27 and 3 to get 30.

$$30+10+6^{5}x-6x+27\geq 10$$

Add 30 and 10 to get 40.

$$40+6^{5}x-6x+27\geq 10$$

Calculate 6 to the power of 5 and get 7776.

$$40+7776x-6x+27\geq 10$$

Combine 7776x and -6x to get 7770x.

$$40+7770x+27\geq 10$$

Add 40 and 27 to get 67.

$$67+7770x\geq 10$$

Subtract 67 from both sides.

$$7770x\geq 10-67$$

Subtract 67 from 10 to get -57.

$$7770x\geq -57$$

Divide both sides by 7770. Since 7770 is positive, the inequality direction remains the same.

$$x\geq \frac{-57}{7770}$$

Reduce the fraction \frac{-57}{7770} to lowest terms by extracting and canceling out 3.

$$x\geq -\frac{19}{2590}$$