Solve {l}{5=4a+2b+c}{-1-a/4a=-17/4}{b=2a}

$$\left\{ \begin{array} { l } { 5 = 4 a + 2 b + c } \\ { \frac { - 1 - a } { 4 a } = - \frac { 17 } { 4 } } \\ { b = 2 a } \end{array} \right.$$

Solve for a, b, c

$$a=\frac{1}{16}=0.0625$$

$$b=\frac{1}{8}=0.125$$

$$c = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Solution Steps

Consider the second equation. Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4a, the least common multiple of 4a,4.

$$-1-a=-17a$$

Add 17a to both sides.

$$-1-a+17a=0$$

Combine -a and 17a to get 16a.

$$-1+16a=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$16a=1$$

Divide both sides by 16.

$$a=\frac{1}{16}$$

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

$$b=2\times \frac{1}{16}$$

Multiply 2 and \frac{1}{16} to get \frac{1}{8}.

$$b=\frac{1}{8}$$

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

$$5=4\times \frac{1}{16}+2\times \frac{1}{8}+c$$

Multiply 4 and \frac{1}{16} to get \frac{1}{4}.

$$5=\frac{1}{4}+2\times \frac{1}{8}+c$$

Multiply 2 and \frac{1}{8} to get \frac{1}{4}.

$$5=\frac{1}{4}+\frac{1}{4}+c$$

Add \frac{1}{4} and \frac{1}{4} to get \frac{1}{2}.

$$5=\frac{1}{2}+c$$

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

$$\frac{1}{2}+c=5$$

Subtract \frac{1}{2} from both sides.

$$c=5-\frac{1}{2}$$

Subtract \frac{1}{2} from 5 to get \frac{9}{2}.

$$c=\frac{9}{2}$$

The system is now solved.

$$a=\frac{1}{16} b=\frac{1}{8} c=\frac{9}{2}$$