Solve {l}{2-{(x-0*8)}=-2{(0*9+x)}}{y=1}{z=y}{a=z}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { 2 - {(x - 0 \cdot 8)} = -2 {(0 \cdot 9 + x)} }\\ { y = 1 }\\ { z = y }\\ { a = z }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

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

$$d=1$$

Solution Steps

Consider the first equation. Multiply 0 and 8 to get 0.

$$2-\left(x-0\right)=-2\left(0\times 9+x\right)$$

Multiply 0 and 9 to get 0.

$$2-\left(x-0\right)=-2x$$

Add 2x to both sides.

$$2-\left(x-0\right)+2x=0$$

Reorder the terms.

$$2-x+2x=0$$

Combine -x and 2x to get x.

$$2+x=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$x=-2$$

The system is now solved.

$$x=-2 y=1 z=1 a=1 b=1 c=1 d=1$$