Solve {l}{2*{(x-1)}+3=5x-29}{y=x+1x-2}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { 2 \cdot {(x - 1)} + 3 = 5 x - 29 }\\ { y = x + 1 x - 2 }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

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

$$c=18$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 2 by x-1.

$$2x-2+3=5x-29$$

Add -2 and 3 to get 1.

$$2x+1=5x-29$$

Subtract 5x from both sides.

$$2x+1-5x=-29$$

Combine 2x and -5x to get -3x.

$$-3x+1=-29$$

Subtract 1 from both sides.

$$-3x=-29-1$$

Subtract 1 from -29 to get -30.

$$-3x=-30$$

Divide both sides by -3.

$$x=\frac{-30}{-3}$$

Divide -30 by -3 to get 10.

$$x=10$$

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

$$y=10+1\times 10-2$$

Multiply 1 and 10 to get 10.

$$y=10+10-2$$

Add 10 and 10 to get 20.

$$y=20-2$$

Subtract 2 from 20 to get 18.

$$y=18$$

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

$$z=18$$

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

$$a=18$$

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

$$b=18$$

Consider the equation (6). Insert the known values of variables into the equation.

$$c=18$$

The system is now solved.

$$x=10 y=18 z=18 a=18 b=18 c=18$$