Solve {l}{x=4}{y=7}{z=20+3{(3y-4x)}}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x = 4 }\\ { y = 7 }\\ { z = 20 + 3 {(3 y - 4 x)} }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z, a

$$a=35$$

Solution Steps

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

$$z=20+3\left(3\times 7-4\times 4\right)$$

Do the multiplications.

$$z=20+3\left(21-16\right)$$

Subtract 16 from 21 to get 5.

$$z=20+3\times 5$$

Multiply 3 and 5 to get 15.

$$z=20+15$$

Add 20 and 15 to get 35.

$$z=35$$

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

$$a=35$$

The system is now solved.

$$x=4 y=7 z=35 a=35$$