Question

Perform the indicated operation. Express the answer in simplest form. Show all necessary work.

Answer 1

To answer this question, we can see that we have mixed fractions here. Then we have:

First Case

$\begin{gathered} -2\frac{1}{3}\Rightarrow2\frac{1}{3}=2+\frac{1}{3}=\frac{2\cdot3+1\cdot1}{3}=\frac{6+1}{3}=\frac{7}{3} \\ \Rightarrow-2\frac{1}{3}=-\frac{7}{3} \end{gathered}$$\begin{gathered} -1\frac{3}{4}\Rightarrow1\frac{3}{4}=1+\frac{3}{4}=\frac{1\cdot4+3\cdot1}{4}=\frac{4+3}{4}=\frac{7}{4} \\ -1\frac{3}{4}=-\frac{7}{4} \end{gathered}$

Therefore, we have:

$-2\frac{1}{3}-1\frac{3}{4}=-\frac{7}{3}-\frac{7}{4}$

Now, we need to find the least common multiplier of 3 and 4. In this case, it is 12. Using this, we can obtain two fractions with the same denominator, and it is easier to add both fractions algebraically.

Then, we need to divide 12 by 3 - in the first fraction, and then we need to multiply the result by 7 (the numerator). We can do the same with the other fraction:

$-\frac{28}{12}-\frac{21}{12}=-\frac{49}{12}$

Therefore, in (a) we can express the result as:

$-\frac{49}{12}=-\frac{48}{12}-\frac{1}{12}=-4\frac{1}{12}$

Then the result for the first case can be expressed as:

Improper Fraction:

$-\frac{49}{12}$

Mixed Fraction:

$-4\frac{1}{12}$

Second Case

We can proceed similarly as before:

$-2\frac{1}{3}=-\frac{7}{3}$

We got this result in the previous case.

And we also know that:

$1\frac{3}{4}=\frac{7}{4}$

Then we have:

$-\frac{7}{3}+\frac{7}{4}=\frac{-28+21}{12}=-\frac{7}{12}$

Therefore:

$-2\frac{1}{3}+1\frac{3}{4}=-\frac{7}{12}$

In summary, we can say that:

In case (a) the result of the operation is:

As an improper fraction:

$-\frac{49}{12}$

As a mixed fraction:

$-4\frac{1}{12}$

In case (b) the result of the operation is:

$-\frac{7}{12}$