Solve {l}{an+2=a(n+2)+0.8}{text{If}n=-3text{intheequation}}

$$\left. \begin{array} { l } { a n + 2 = a ( n + 2 ) + 0.8 } \\ { \text { If } n = - 3 \text { in the equation } } \end{array} \right.$$

Solve for a, n

$$a=0.6$$

$$n=-3$$

Solution Steps

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

$$a\left(-3\right)+2=a\left(-3+2\right)+0.8$$

Add -3 and 2 to get -1.

$$a\left(-3\right)+2=a\left(-1\right)+0.8$$

Subtract a\left(-1\right) from both sides.

$$a\left(-3\right)+2-a\left(-1\right)=0.8$$

Combine a\left(-3\right) and -a\left(-1\right) to get -2a.

$$-2a+2=0.8$$

Subtract 2 from both sides.

$$-2a=0.8-2$$

Subtract 2 from 0.8 to get -1.2.

$$-2a=-1.2$$

Divide both sides by -2.

$$a=\frac{-1.2}{-2}$$

Expand \frac{-1.2}{-2} by multiplying both numerator and the denominator by 10.

$$a=\frac{-12}{-20}$$

Reduce the fraction \frac{-12}{-20} to lowest terms by extracting and canceling out -4.

$$a=\frac{3}{5}$$

The system is now solved.

$$a=\frac{3}{5} n=-3$$