Solve {l}{2x+y=20}{y=20}{z=x+3y+2x}{a=2}{text{Solvefor}b,ctext{where}}{b=z}{c=a}

$$\left. \begin{array} { l } { 2 x + y = 20 }\\ { y = 20 }\\ { z = x + 3 y + 2 x }\\ { a = 2 }\\ { \text{Solve for } b,c \text{ where} } \\ { b = z }\\ { c = a } \end{array} \right.$$

Solve for x, y, z, a, b, c

$$b=60$$

$$c=2$$

Solution Steps

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

$$2x+20=20$$

Subtract 20 from both sides.

$$2x=20-20$$

Subtract 20 from 20 to get 0.

$$2x=0$$

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$x=0$$

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

$$z=0+3\times 20+2\times 0$$

Do the multiplications.

$$z=0+60+0$$

Add 0 and 60 to get 60.

$$z=60+0$$

Add 60 and 0 to get 60.

$$z=60$$

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

$$b=60$$

The system is now solved.

$$x=0 y=20 z=60 a=2 b=60 c=2$$