Solve {l}{6a+3a-3b=8}{3b+3b-3a=6d}{d=a+b}

$$\left\{ \begin{array} { l } { 6 a + 3 a - 3 b = 8 } \\ { 3 b + 3 b - 3 a = 6 d } \\ { d = a + b } \end{array} \right.$$

Solve for a, b, d

$$a=0$$

$$b = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$d = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

Short Steps Using Substitution

Reorder the equations.

$$d=a+b 3b+3b-3a=6d 6a+3a-3b=8$$

Substitute a+b for d in the equation 3b+3b-3a=6d.

$$3b+3b-3a=6\left(a+b\right)$$

Solve the second equation for a and the third equation for b.

$$a=0 b=3a-\frac{8}{3}$$

Substitute 0 for a in the equation b=3a-\frac{8}{3}.

$$b=3\times 0-\frac{8}{3}$$

Calculate b from b=3\times 0-\frac{8}{3}.

$$b=-\frac{8}{3}$$

Substitute 0 for a and -\frac{8}{3} for b in the equation d=a+b.

$$d=0-\frac{8}{3}$$

Calculate d from d=0-\frac{8}{3}.

$$d=-\frac{8}{3}$$

The system is now solved.

$$a=0 b=-\frac{8}{3} d=-\frac{8}{3}$$