Solve 3*(x+20)/4-5/8-1/2*(2x+0,5x)=x+0,75/2

$$\frac { 3 \cdot ( x + 20 ) } { 4 } - \frac { 5 } { 8 } - \frac { 1 } { 2 } \cdot ( 2 x + 0,5 x ) = \frac { x + 0,75 } { 2 }$$

Solve for x

$$x=14$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 8, the least common multiple of 4;8;2.

$$8\left(\frac{3\left(x+20\right)}{4}-\frac{5}{8}\right)-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

Use the distributive property to multiply 3 by x+20.

$$8\left(\frac{3x+60}{4}-\frac{5}{8}\right)-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 8 is 8. Multiply \frac{3x+60}{4} times \frac{2}{2}.

$$8\left(\frac{2\left(3x+60\right)}{8}-\frac{5}{8}\right)-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

Since \frac{2\left(3x+60\right)}{8} and \frac{5}{8} have the same denominator, subtract them by subtracting their numerators.

$$8\times \frac{2\left(3x+60\right)-5}{8}-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

Do the multiplications in 2\left(3x+60\right)-5.

$$8\times \frac{6x+120-5}{8}-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

Combine like terms in 6x+120-5.

$$8\times \frac{6x+115}{8}-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

Express 8\times \frac{6x+115}{8} as a single fraction.

$$\frac{8\left(6x+115\right)}{8}-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

Cancel out 8 and 8.

$$6x+115-4\left(2x+0,5x\right)=4\left(x+0,75\right)$$

Combine 2x and 0,5x to get 2,5x.

$$6x+115-4\times 2,5x=4\left(x+0,75\right)$$

Multiply 4 and 2,5 to get 10.

$$6x+115-10x=4\left(x+0,75\right)$$

Combine 6x and -10x to get -4x.

$$-4x+115=4\left(x+0,75\right)$$

Use the distributive property to multiply 4 by x+0,75.

$$-4x+115=4x+3$$

Subtract 4x from both sides.

$$-4x+115-4x=3$$

Combine -4x and -4x to get -8x.

$$-8x+115=3$$

Subtract 115 from both sides.

$$-8x=3-115$$

Subtract 115 from 3 to get -112.

$$-8x=-112$$

Divide both sides by -8.

$$x=\frac{-112}{-8}$$

Divide -112 by -8 to get 14.

$$x=14$$