Solve {l}{(a-d)+a+(a+d)=120}{4(a-d)+5=a+d}

$$\left\{ \begin{array} { l } { ( a - d ) + a + ( a + d ) = 120 } \\ { 4 ( a - d ) + 5 = a + d } \end{array} \right.$$

Solve for a, d

$$a=40$$

$$d=25$$

Solution Steps

Consider the first equation. Combine a and a to get 2a.

$$2a-d+a+d=120$$

Combine 2a and a to get 3a.

$$3a-d+d=120$$

Combine -d and d to get 0.

$$3a=120$$

Divide both sides by 3.

$$a=\frac{120}{3}$$

Divide 120 by 3 to get 40.

$$a=40$$

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

$$4\left(40-d\right)+5=40+d$$

Use the distributive property to multiply 4 by 40-d.

$$160-4d+5=40+d$$

Add 160 and 5 to get 165.

$$165-4d=40+d$$

Subtract d from both sides.

$$165-4d-d=40$$

Combine -4d and -d to get -5d.

$$165-5d=40$$

Subtract 165 from both sides.

$$-5d=40-165$$

Subtract 165 from 40 to get -125.

$$-5d=-125$$

Divide both sides by -5.

$$d=\frac{-125}{-5}$$

Divide -125 by -5 to get 25.

$$d=25$$

The system is now solved.

$$a=40 d=25$$