Solve (1)^6/6+16(1)^12/3+c=-6

$$\frac { ( 1 ) ^ { 6 } } { 6 } + \frac { 16 ( 1 ) ^ { 12 } } { 3 } + c = - 6$$

Solve for c

$$c = -\frac{23}{2} = -11\frac{1}{2} = -11.5$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 6, the least common multiple of 6,3.

$$1^{6}+2\times 16\times 1^{12}+6c=-36$$

Calculate 1 to the power of 6 and get 1.

$$1+2\times 16\times 1^{12}+6c=-36$$

Multiply 2 and 16 to get 32.

$$1+32\times 1^{12}+6c=-36$$

Calculate 1 to the power of 12 and get 1.

$$1+32\times 1+6c=-36$$

Multiply 32 and 1 to get 32.

$$1+32+6c=-36$$

Add 1 and 32 to get 33.

$$33+6c=-36$$

Subtract 33 from both sides.

$$6c=-36-33$$

Subtract 33 from -36 to get -69.

$$6c=-69$$

Divide both sides by 6.

$$c=\frac{-69}{6}$$

Reduce the fraction \frac{-69}{6} to lowest terms by extracting and canceling out 3.

$$c=-\frac{23}{2}$$