Solve (5)4-3x-1/4leq5(x+5)/8+1

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

Solve for x

$$x\geq \frac{129}{11}$$

Solution Steps

Multiply both sides of the equation by 8, the least common multiple of 4,8. Since 8 is positive, the inequality direction remains the same.

$$40\times 4-2\left(3x-1\right)\leq 5\left(x+5\right)+8$$

Multiply 40 and 4 to get 160.

$$160-2\left(3x-1\right)\leq 5\left(x+5\right)+8$$

Use the distributive property to multiply -2 by 3x-1.

$$160-6x+2\leq 5\left(x+5\right)+8$$

Add 160 and 2 to get 162.

$$162-6x\leq 5\left(x+5\right)+8$$

Use the distributive property to multiply 5 by x+5.

$$162-6x\leq 5x+25+8$$

Add 25 and 8 to get 33.

$$162-6x\leq 5x+33$$

Subtract 5x from both sides.

$$162-6x-5x\leq 33$$

Combine -6x and -5x to get -11x.

$$162-11x\leq 33$$

Subtract 162 from both sides.

$$-11x\leq 33-162$$

Subtract 162 from 33 to get -129.

$$-11x\leq -129$$

Divide both sides by -11. Since -11 is negative, the inequality direction is changed.

$$x\geq \frac{-129}{-11}$$

Fraction \frac{-129}{-11} can be simplified to \frac{129}{11} by removing the negative sign from both the numerator and the denominator.

$$x\geq \frac{129}{11}$$