How to calculate $(-\frac{1}{3} - 3)(-\frac{1}{3} + 5) - (-\frac{1}{3} + 4)(-\frac{1}{3} - 5)$

Question

$(-\frac{1}{3} - 3)(-\frac{1}{3} + 5) - (-\frac{1}{3} + 4)(-\frac{1}{3} - 5)$

1. multiply and calculate left $(-\frac{1}{3} - 3)(-\frac{1}{3} + 5)$ = $-15\frac{5}{9}$

2. Same for right $(-\frac{1}{3} + 4)(-\frac{1}{3} - 5)$ = $-21 \frac{1}{3}$

3. $-15\frac{5}{9} - -21 \frac{1}{3}$

Is it correct vision?

Answer 1

Let's avoid working with mixed numbers. Using them introduces extra steps into our calculations and, with them, extra opportunities to make an error. It is easier to work with improper fractions, then convert the final answer to a mixed number should that prove necessary. \begin{align*} \left(-\frac{1}{3} - 3\right)&\left(-\frac{1}{3} + 5\right) - \left(-\frac{1}{3} + 4\right)\left(-\frac{1}{3} - 5\right)\\ & = \left(-\frac{1}{3} - \frac{9}{3}\right)\left(-\frac{1}{3} + \frac{15}{3}\right) - \left(-\frac{1}{3} + \frac{12}{3}\right)\left(-\frac{1}{3} - \frac{15}{3}\right)\\ & = \left(-\frac{10}{3}\right)\left(\frac{14}{3}\right) - \left(\frac{11}{3}\right)\left(-\frac{16}{3}\right)\\ & = -\frac{140}{9} - \left(-\frac{176}{9}\right)\\ & = -\frac{140}{9} + \frac{176}{9}\\ & = \frac{36}{9}\\ & = 4 \end{align*}

You correctly calculated that $$\left(-\frac{1}{3} - 3\right)\left(-\frac{1}{3} + 5\right) = -15\frac{5}{9}$$ However, your calculation that $$\left(-\frac{1}{3} + 4\right)\left(-\frac{1}{3} - 5\right) = -21\frac{1}{3}$$ is incorrect. You should have obtained $-19\frac{5}{9}$. Note that $$-15\frac{5}{9} - \left(-19\frac{5}{9}\right) = -15 - \frac{5}{9} - \left(-19 - \frac{5}{9}\right) = -15 - \frac{5}{9} + 19 + \frac{5}{9} = 4$$

Answer 2

I also recommend avoiding working with fractions as long as possible

\begin{array}{l} (-\frac{1}{3} - 3)(-\frac{1}{3} + 5) - (-\frac{1}{3} + 4)(-\frac{1}{3} - 5) \\ \qquad = \dfrac 19[3(-\frac{1}{3} - 3)\cdot 3(-\frac{1}{3} + 5) - 3(-\frac{1}{3} + 4) \cdot 3(-\frac{1}{3} - 5)] \\ \qquad = \dfrac 19[(-1-9)(-1+15) -(-1+12)(-1-15)]\\ \qquad =\dfrac 19[(-10)(14) - (11)(-16)] \\ \qquad = \dfrac 19[-140+176] \\ \qquad = \dfrac{36}{9} \\ \qquad = 4 \end{array}

Answer 3

Taking $x=\dfrac{-1}{3}$, we can write the equation as

$$(x-3)(x+5)-(x+4)(x-5)=(x^2+2x-15)-(x^2-x-20)=3x+5$$

Now, back-substitute $x=\dfrac{-1}{3}$, to get the answer as $$3x+5=3 \cdot \dfrac{-1}{3}+5=-1+5=4$$