Solve 3x-5/3leq(2x+1)^2-2/2-x*(8x-3)+1/4

$$\frac { 3 x - 5 } { 3 } \leq \frac { ( 2 x + 1 ) ^ { 2 } - 2 } { 2 } - \frac { x \cdot ( 8 x - 3 ) + 1 } { 4 }$$

Solve for x

$$x\geq -\frac{11}{21}$$

Solution Steps

Multiply both sides of the equation by 12, the least common multiple of 3,2,4. Since 12 is positive, the inequality direction remains the same.

$$4\left(3x-5\right)\leq 6\left(\left(2x+1\right)^{2}-2\right)-3\left(x\left(8x-3\right)+1\right)$$

Use the distributive property to multiply 4 by 3x-5.

$$12x-20\leq 6\left(\left(2x+1\right)^{2}-2\right)-3\left(x\left(8x-3\right)+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$12x-20\leq 6\left(4x^{2}+4x+1-2\right)-3\left(x\left(8x-3\right)+1\right)$$

Subtract 2 from 1 to get -1.

$$12x-20\leq 6\left(4x^{2}+4x-1\right)-3\left(x\left(8x-3\right)+1\right)$$

Use the distributive property to multiply 6 by 4x^{2}+4x-1.

$$12x-20\leq 24x^{2}+24x-6-3\left(x\left(8x-3\right)+1\right)$$

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

$$12x-20\leq 24x^{2}+24x-6-3\left(8x^{2}-3x+1\right)$$

Use the distributive property to multiply -3 by 8x^{2}-3x+1.

$$12x-20\leq 24x^{2}+24x-6-24x^{2}+9x-3$$

Combine 24x^{2} and -24x^{2} to get 0.

$$12x-20\leq 24x-6+9x-3$$

Combine 24x and 9x to get 33x.

$$12x-20\leq 33x-6-3$$

Subtract 3 from -6 to get -9.

$$12x-20\leq 33x-9$$

Subtract 33x from both sides.

$$12x-20-33x\leq -9$$

Combine 12x and -33x to get -21x.

$$-21x-20\leq -9$$

Add 20 to both sides.

$$-21x\leq -9+20$$

Add -9 and 20 to get 11.

$$-21x\leq 11$$

Divide both sides by -21. Since -21 is negative, the inequality direction is changed.

$$x\geq -\frac{11}{21}$$