Solve x^2-7/5+8x-3/3<4+3x^2/15

$$\frac { x ^ { 2 } - 7 } { 5 } + \frac { 8 x - 3 } { 3 } < \frac { 4 + 3 x ^ { 2 } } { 15 }$$

Solve for x

$$x<1$$

Solution Steps

Multiply both sides of the equation by 15, the least common multiple of 5,3,15. Since 15 is positive, the inequality direction remains the same.

$$3\left(x^{2}-7\right)+5\left(8x-3\right)<4+3x^{2}$$

Use the distributive property to multiply 3 by x^{2}-7.

$$3x^{2}-21+5\left(8x-3\right)<4+3x^{2}$$

Use the distributive property to multiply 5 by 8x-3.

$$3x^{2}-21+40x-15<4+3x^{2}$$

Subtract 15 from -21 to get -36.

$$3x^{2}-36+40x<4+3x^{2}$$

Subtract 3x^{2} from both sides.

$$3x^{2}-36+40x-3x^{2}<4$$

Combine 3x^{2} and -3x^{2} to get 0.

$$-36+40x<4$$

Add 36 to both sides.

$$40x<4+36$$

Add 4 and 36 to get 40.

$$40x<40$$

Divide both sides by 40. Since 40 is positive, the inequality direction remains the same.

$$x<\frac{40}{40}$$

Divide 40 by 40 to get 1.

$$x<1$$