Solve {l}{a_{n}=42-4{(n-1)}}{n=1}{a=2}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { a_{n} = 42 - 4 {(n - 1)} }\\ { n = 1 }\\ { a = 2 }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for a_n, n, a, b, c, d

$$d=2$$

Solution Steps

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

$$a_{n}=42-4\left(1-1\right)$$

Subtract 1 from 1 to get 0.

$$a_{n}=42-4\times 0$$

Multiply -4 and 0 to get 0.

$$a_{n}=42+0$$

Add 42 and 0 to get 42.

$$a_{n}=42$$

The system is now solved.

$$a_{n}=42 n=1 a=2 b=2 c=2 d=2$$