Solve {l}{3a+2b+d=1}{9a+b+2d=1}{a+2b+d=4}

$$\left. \begin{array} { l } { 3 a + 2 b + d = 1 } \\ { 9 a + b + 2 d = 1 } \\ { a + 2 b + d = 4 } \end{array} \right.$$

Solve for a, b, d

$$a = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$b = -\frac{7}{6} = -1\frac{1}{6} \approx -1.166666667$$

$$d = \frac{47}{6} = 7\frac{5}{6} \approx 7.833333333$$

Short Steps Using Substitution

Solve 3a+2b+d=1 for d.

$$d=-3a-2b+1$$

Substitute -3a-2b+1 for d in the second and third equation.

$$9a+b+2\left(-3a-2b+1\right)=1 a+2b-3a-2b+1=4$$

Solve these equations for b and a respectively.

$$b=a+\frac{1}{3} a=-\frac{3}{2}$$

Substitute -\frac{3}{2} for a in the equation b=a+\frac{1}{3}.

$$b=-\frac{3}{2}+\frac{1}{3}$$

Calculate b from b=-\frac{3}{2}+\frac{1}{3}.

$$b=-\frac{7}{6}$$

Substitute -\frac{7}{6} for b and -\frac{3}{2} for a in the equation d=-3a-2b+1.

$$d=-3\left(-\frac{3}{2}\right)-2\left(-\frac{7}{6}\right)+1$$

Calculate d from d=-3\left(-\frac{3}{2}\right)-2\left(-\frac{7}{6}\right)+1.

$$d=\frac{47}{6}$$

The system is now solved.

$$a=-\frac{3}{2} b=-\frac{7}{6} d=\frac{47}{6}$$