Solve {l}{150*2/5+x=2{(150*3/5-x)}}{y=x}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 150 \cdot \frac{2}{5} + x = 2 {(150 \cdot \frac{3}{5} - x)} }\\ { y = x }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b=40$$

Solution Steps

Consider the first equation. Multiply 150 and \frac{2}{5} to get 60.

$$60+x=2\left(150\times \frac{3}{5}-x\right)$$

Multiply 150 and \frac{3}{5} to get 90.

$$60+x=2\left(90-x\right)$$

Use the distributive property to multiply 2 by 90-x.

$$60+x=180-2x$$

Add 2x to both sides.

$$60+x+2x=180$$

Combine x and 2x to get 3x.

$$60+3x=180$$

Subtract 60 from both sides.

$$3x=180-60$$

Subtract 60 from 180 to get 120.

$$3x=120$$

Divide both sides by 3.

$$x=\frac{120}{3}$$

Divide 120 by 3 to get 40.

$$x=40$$

Consider the second equation. Insert the known values of variables into the equation.

$$y=40$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=40$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$a=40$$

Consider the fifth equation. Insert the known values of variables into the equation.

$$b=40$$

The system is now solved.

$$x=40 y=40 z=40 a=40 b=40$$