Solve {l}{2*{(frac{3*{(2)}+1}{2})}={(frac{1*{(2)}+1}{2})}x-1}{y=x+2}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 2 \cdot {(\frac{3 \cdot {(2)} + 1}{2})} = {(\frac{1 \cdot {(2)} + 1}{2})} x - 1 }\\ { y = x + 2 }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b = \frac{22}{3} = 7\frac{1}{3} \approx 7.333333333$$

Solution Steps

Consider the first equation. Multiply both sides of the equation by 2.

$$2\left(3\times 2+1\right)=\left(1\times 2+1\right)x-2$$

Multiply 3 and 2 to get 6.

$$2\left(6+1\right)=\left(1\times 2+1\right)x-2$$

Add 6 and 1 to get 7.

$$2\times 7=\left(1\times 2+1\right)x-2$$

Multiply 2 and 7 to get 14.

$$14=\left(1\times 2+1\right)x-2$$

Multiply 1 and 2 to get 2.

$$14=\left(2+1\right)x-2$$

Add 2 and 1 to get 3.

$$14=3x-2$$

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

$$3x-2=14$$

Add 2 to both sides.

$$3x=14+2$$

Add 14 and 2 to get 16.

$$3x=16$$

Divide both sides by 3.

$$x=\frac{16}{3}$$

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

$$y=\frac{16}{3}+2$$

Add \frac{16}{3} and 2 to get \frac{22}{3}.

$$y=\frac{22}{3}$$

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

$$z=\frac{22}{3}$$

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

$$a=\frac{22}{3}$$

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

$$b=\frac{22}{3}$$

The system is now solved.

$$x=\frac{16}{3} y=\frac{22}{3} z=\frac{22}{3} a=\frac{22}{3} b=\frac{22}{3}$$