Solve {l}{4*35={(x*5)}+{(x*30)}}{y=x}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 4 \cdot 35 = {(x \cdot 5)} + {(x \cdot 30)} }\\ { y = x }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b=4$$

Solution Steps

Consider the first equation. Multiply 4 and 35 to get 140.

$$140=x\times 5+x\times 30$$

Combine x\times 5 and x\times 30 to get 35x.

$$140=35x$$

Swap sides so that all variable terms are on the left hand side.

$$35x=140$$

Divide both sides by 35.

$$x=\frac{140}{35}$$

Divide 140 by 35 to get 4.

$$x=4$$

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

$$y=4$$

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

$$z=4$$

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

$$a=4$$

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

$$b=4$$

The system is now solved.

$$x=4 y=4 z=4 a=4 b=4$$