Solve {l}{4b-1*5=0*2-0*6{(3-4b)}}{c=2}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { 4 b - 1 \cdot 5 = 0 \cdot 2 - 0 \cdot 6 {(3 - 4 b)} }\\ { c = 2 }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for b, c, d

$$d=2$$

Solution Steps

Consider the first equation. Multiply 1 and 5 to get 5.

$$4b-5=0\times 2-0\times 6\left(3-4b\right)$$

Multiply 0 and 2 to get 0.

$$4b-5=0-0\times 6\left(3-4b\right)$$

Multiply 0 and 6 to get 0.

$$4b-5=0-0\left(3-4b\right)$$

Anything times zero gives zero.

$$4b-5=0-0$$

Subtracting 0 from itself leaves 0.

$$4b-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$4b=5$$

Divide both sides by 4.

$$b=\frac{5}{4}$$

The system is now solved.

$$b=\frac{5}{4} c=2 d=2$$