Solve {l}{a_{2}=3+d}{a_{2}=2(3+a_{2})+3}

$$\left. \begin{array} { l } { a _ { 2 } = 3 + d } \\ { a _ { 2 } = 2 ( 3 + a _ { 2 } ) + 3 } \end{array} \right.$$

Solve for a_2, d

$$a_{2}=-9$$

$$d=-12$$

Solution Steps

Consider the second equation. Use the distributive property to multiply 2 by 3+a_{2}.

$$a_{2}=6+2a_{2}+3$$

Add 6 and 3 to get 9.

$$a_{2}=9+2a_{2}$$

Subtract 2a_{2} from both sides.

$$a_{2}-2a_{2}=9$$

Combine a_{2} and -2a_{2} to get -a_{2}.

$$-a_{2}=9$$

Divide both sides by -1.

$$a_{2}=-9$$

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

$$-9=3+d$$

Swap sides so that all variable terms are on the left hand side.

$$3+d=-9$$

Subtract 3 from both sides.

$$d=-9-3$$

Subtract 3 from -9 to get -12.

$$d=-12$$

The system is now solved.

$$a_{2}=-9 d=-12$$