Solve (3x+1)(3x-1)/4-9(x+2)^2/4leq1/2[3x(1/3-frac{1}{2})^-1]

$$\frac { ( 3 x + 1 ) ( 3 x - 1 ) } { 4 } - \frac { 9 ( x + 2 ) ^ { 2 } } { 4 } \leq \frac { 1 } { 2 } [ 3 x ( \frac { 1 } { 3 } - \frac { 1 } { 2 } ) ^ { - 1 } ]$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Multiply both sides of the equation by 4, the least common multiple of 4,2. Since 4 is positive, the inequality direction remains the same.

$$\left(3x+1\right)\left(3x-1\right)-9\left(x+2\right)^{2}\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Consider \left(3x+1\right)\left(3x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(3x\right)^{2}-1-9\left(x+2\right)^{2}\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-1-9\left(x+2\right)^{2}\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-1-9\left(x+2\right)^{2}\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$9x^{2}-1-9\left(x^{2}+4x+4\right)\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Use the distributive property to multiply -9 by x^{2}+4x+4.

$$9x^{2}-1-9x^{2}-36x-36\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Combine 9x^{2} and -9x^{2} to get 0.

$$-1-36x-36\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Subtract 36 from -1 to get -37.

$$-37-36x\leq 2\times 3x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Multiply 2 and 3 to get 6.

$$-37-36x\leq 6x\left(\frac{1}{3}-\frac{1}{2}\right)^{-1}$$

Subtract \frac{1}{2} from \frac{1}{3} to get -\frac{1}{6}.

$$-37-36x\leq 6x\left(-\frac{1}{6}\right)^{-1}$$

Calculate -\frac{1}{6} to the power of -1 and get -6.

$$-37-36x\leq 6x\left(-6\right)$$

Multiply 6 and -6 to get -36.

$$-37-36x\leq -36x$$

Add 36x to both sides.

$$-37-36x+36x\leq 0$$

Combine -36x and 36x to get 0.

$$-37\leq 0$$

This is true for any x.

$$x\in \mathrm{R}$$