Solve {l}{31=2(4)^2+3*4+c}{-2=8-2B-1-4B}

$$\left. \begin{array} { l } { 31 = 2 ( 4 ) ^ { 2 } + 3 \cdot 4 + c } \\ { - 2 = 8 - 2 B - 1 - 4 B } \end{array} \right.$$

Solve for c, B

$$c=-13$$

$$B = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Consider the first equation. Calculate 4 to the power of 2 and get 16.

$$31=2\times 16+3\times 4+c$$

Multiply 2 and 16 to get 32.

$$31=32+3\times 4+c$$

Multiply 3 and 4 to get 12.

$$31=32+12+c$$

Add 32 and 12 to get 44.

$$31=44+c$$

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

$$44+c=31$$

Subtract 44 from both sides.

$$c=31-44$$

Subtract 44 from 31 to get -13.

$$c=-13$$

Consider the second equation. Subtract 1 from 8 to get 7.

$$-2=7-2B-4B$$

Combine -2B and -4B to get -6B.

$$-2=7-6B$$

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

$$7-6B=-2$$

Subtract 7 from both sides.

$$-6B=-2-7$$

Subtract 7 from -2 to get -9.

$$-6B=-9$$

Divide both sides by -6.

$$B=\frac{-9}{-6}$$

Reduce the fraction \frac{-9}{-6} to lowest terms by extracting and canceling out -3.

$$B=\frac{3}{2}$$

The system is now solved.

$$c=-13 B=\frac{3}{2}$$