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

$$a = \frac{32}{27} = 1\frac{5}{27} \approx 1.185185185$$

$$b=\frac{8}{9}\approx 0.888888889$$

$$d=\frac{8}{27}\approx 0.296296296$$

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 b and the third equation for a.

$$b=\frac{3}{4}a a=\frac{1}{3}b+\frac{8}{9}$$

Substitute \frac{3}{4}a for b in the equation a=\frac{1}{3}b+\frac{8}{9}.

$$a=\frac{1}{3}\times \frac{3}{4}a+\frac{8}{9}$$

Solve a=\frac{1}{3}\times \frac{3}{4}a+\frac{8}{9} for a.

$$a=\frac{32}{27}$$

Substitute \frac{32}{27} for a in the equation b=\frac{3}{4}a.

$$b=\frac{3}{4}\times \frac{32}{27}$$

Calculate b from b=\frac{3}{4}\times \frac{32}{27}.

$$b=\frac{8}{9}$$

Substitute \frac{8}{9} for b and \frac{32}{27} for a in the equation d=a-b.

$$d=\frac{32}{27}-\frac{8}{9}$$

Calculate d from d=\frac{32}{27}-\frac{8}{9}.

$$d=\frac{8}{27}$$

The system is now solved.

$$a=\frac{32}{27} b=\frac{8}{9} d=\frac{8}{27}$$