Question

Find the sum.

$\left(-\frac{1}{8}c+16\right)-\left(\frac{3}{8}+3c\right)$

Answer 1

Addition of algebraic expressions

$\begin{array}{rcl}\left(-\frac{1}{8}\mathrm{c}+16\right)-\left(\frac{3}{8}+3\mathrm{c}\right)& =& \left(-\frac{1}{8}\mathrm{c}+16\right)+\left(-\frac{3}{8}-3\mathrm{c}\right)\{\mathrm{Add}\mathrm{the}\mathrm{opposites}]\\ & =& \left(-\frac{1}{8}\mathrm{c}\right)+\left(-3\mathrm{c}\right)+16+\left(-\frac{3}{8}\right)\{\mathrm{Commutative}\mathrm{property}\}\\ & =& \left[\left(-\frac{1}{8}\mathrm{c}\right)+(-3\mathrm{c}\right]+\left[16+\left(-\frac{3}{8}\right)\right]\{\mathrm{Group}\mathrm{like}\mathrm{terms}\}\\ & =& \left(-\frac{25}{8}\mathrm{c}\right)+\left(\frac{125}{8}\right)\{\mathrm{Combine}\mathrm{like}\mathrm{terms}\}\end{array}$

Hence, the required sum of algebraic expressions is $-\frac{25}{8}c+\frac{125}{8}$