Solve 3x+4^3<17^2-(0^2-7^2)+66

$$3 x + 4 ^ { 3 } < 17 ^ { 2 } - ( 0 ^ { 2 } - 7 ^ { 2 } ) + 66$$

Solve for x

$$x<\frac{340}{3}$$

Solution Steps

Calculate 4 to the power of 3 and get 64.

$$3x+64<17^{2}-\left(0^{2}-7^{2}\right)+66$$

Calculate 17 to the power of 2 and get 289.

$$3x+64<289-\left(0^{2}-7^{2}\right)+66$$

Calculate 0 to the power of 2 and get 0.

$$3x+64<289-\left(0-7^{2}\right)+66$$

Calculate 7 to the power of 2 and get 49.

$$3x+64<289-\left(0-49\right)+66$$

Subtract 49 from 0 to get -49.

$$3x+64<289-\left(-49\right)+66$$

The opposite of -49 is 49.

$$3x+64<289+49+66$$

Add 289 and 49 to get 338.

$$3x+64<338+66$$

Add 338 and 66 to get 404.

$$3x+64<404$$

Subtract 64 from both sides.

$$3x<404-64$$

Subtract 64 from 404 to get 340.

$$3x<340$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x<\frac{340}{3}$$