Solve {l}{a+b=3}{20+c=4}{b+c=1}

$$\left\{ \begin{array} { l } { a + b = 3 } \\ { 20 + c = 4 } \\ { b + c = 1 } \end{array} \right.$$

Solve for a, b, c

$$a=-14$$

$$b=17$$

$$c=-16$$

Solution Steps

Consider the second equation. Subtract 20 from both sides.

$$c=4-20$$

Subtract 20 from 4 to get -16.

$$c=-16$$

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

$$b-16=1$$

Add 16 to both sides.

$$b=1+16$$

Add 1 and 16 to get 17.

$$b=17$$

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

$$a+17=3$$

Subtract 17 from both sides.

$$a=3-17$$

Subtract 17 from 3 to get -14.

$$a=-14$$

The system is now solved.

$$a=-14 b=17 c=-16$$