Solve 3-x^2/2+7(1/2+x)<1/2(8-x^2)

$$\frac { 3 - x ^ { 2 } } { 2 } + 7 ( \frac { 1 } { 2 } + x ) < \frac { 1 } { 2 } ( 8 - x ^ { 2 } )$$

Solve for x

$$x<-\frac{1}{7}$$

Solution Steps

Multiply both sides of the equation by 2. Since 2 is positive, the inequality direction remains the same.

$$3-x^{2}+14\left(\frac{1}{2}+x\right)<1\left(8-x^{2}\right)$$

Use the distributive property to multiply 14 by \frac{1}{2}+x.

$$3-x^{2}+7+14x<1\left(8-x^{2}\right)$$

Add 3 and 7 to get 10.

$$10-x^{2}+14x<1\left(8-x^{2}\right)$$

Use the distributive property to multiply 1 by 8-x^{2}.

$$10-x^{2}+14x<8-x^{2}$$

Add x^{2} to both sides.

$$10-x^{2}+14x+x^{2}<8$$

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

$$10+14x<8$$

Subtract 10 from both sides.

$$14x<8-10$$

Subtract 10 from 8 to get -2.

$$14x<-2$$

Divide both sides by 14. Since 14 is positive, the inequality direction remains the same.

$$x<\frac{-2}{14}$$

Reduce the fraction \frac{-2}{14} to lowest terms by extracting and canceling out 2.

$$x<-\frac{1}{7}$$