Question

Add the following expressions using both the horizontal method and column method:

$\frac{3}{2}p-\frac{1}{2}q+r$ and $-2p+4q-\frac{3}{2}r$

Answer 1

Step 1: Adding them by the horizontal method

Horizontal method: Taking care of signs, write all expressions in a horizontal line and group like terms together. Now, add these like terms and get the required sum.

We have $\frac{3}{2}p-\frac{1}{2}q+r$ and $-2p+4q-\frac{3}{2}r$

Now,

$$\begin{gathered} (\frac{3}{2}p-\frac{1}{2}q+r)+(-2p+4q-\frac{3}{2}r)=\frac{3}{2}p-\frac{1}{2}q+r-2p+4q-\frac{3}{2}r \\ =\frac{3}{2}p-2p-\frac{1}{2}q+4q+r-\frac{3}{2}r \\ =(\frac{3}{2}-2)p+(-\frac{1}{2}+4)q+(1-\frac{3}{2})r \\ =\left(\frac{3\times 1}{2\times 1}-\frac{2\times 2}{1\times 2}\right)p+\left(\frac{-1\times 1}{2\times 1}+\frac{4\times 2}{1\times 2}\right)q+\left(\frac{1\times 2}{1\times 2}-\frac{3\times 1}{2\times 1}\right)r \\ =\left(\frac{3}{2}-\frac{4}{2}\right)p+\left(\frac{-1}{2}+\frac{8}{2}\right)q+\left(\frac{2}{2}-\frac{3}{2}\right)r \\ =\left(\frac{3-4}{2}\right)p+\left(\frac{-1+8}{2}\right)q+\left(\frac{2-3}{2}\right)r \\ =\frac{-1}{2}p+\frac{7}{2}q-\frac{1}{2}r \end{gathered}$$

Step 2: Adding them by the column method

Column method: Write each expression in separate lines such that their like terms lie in the same vertical column. Taking care of signs, add each column.

$$\begin{gathered} \frac{3}{2}p-\frac{1}{2}q+r \\ +-2p+4q-\frac{3}{2}r \\ \_\_\_\_\_\_\_\_\_\_\_\_\_ \\ -\frac{1}{2}p+\frac{7}{2}q-\frac{1}{2}r \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$

Hence, the required sum is $-\frac{1}{2}p+\frac{7}{2}q-\frac{1}{2}r$