How do I solve this fraction question?

Question

If $a = -1/5$, how do I calculate:

$$3 a + 2 a^2$$

I did $3\times(-1/5) + (-1/5) \times (-1/5) \times 2$, but can't figure out what the right way to solve this is.

Answer 1

A helpful tip: You can only add fractions with the same denominator (the numbers at the bottom). If the denominators are not equal, you have to multiply them by a constant number $k$ to add them.

Starting from $$\frac{3 \times -1}{5} + \frac{-1 \times -1 \times 2}{5\times5},$$ since $+a \times -b = -ab$, and $-a \times -b = ab$ for arbitrary numbers $a,b$, we have:

$$-\frac{3}{5} + \frac{2}{25}$$

Since $\frac{1}{5}$ is $5$ times that of $\frac{1}{25}$, $k=5$, and we have to multiply both numerator and denominator by $5$ to simplify:

$$-\frac{15}{25} + \frac{2}{25}$$

Now you can simplify the expression by cancelling $\frac{10}{25}$ and adding the $2$ fractions together:

$$-\frac{15}{25} + \frac{2}{25} = \frac{-15+2}{5}$$

I'll leave the last step for you to figure out.

Answer 2

That's the correct way.

\begin{align} \left(3 \times \frac{-1}{5}\right)+\left(\frac{-1}{5} \times 2 \times \frac{-1}{5}\right) &= \frac{-3}{5}+\frac{2}{25} \\ &=\frac{-15}{25}+\frac{2}{25} \\ &= \frac{-13}{25}\\ \end{align}

Answer 3

It might help you to add a denominator of $1$ to integers: $$\frac{3}{1} \times \frac{-1}{5} = \frac{-3}{5}.$$

So far so good, right? Next, $$\frac{2}{1} \times \left(\frac{-1}{5}\right)^2 = \frac{2}{1} \times \frac{1}{25} = \frac{2}{25}.$$

And then $$\frac{-3}{5} + \frac{2}{25} = \frac{-15}{25} + \frac{2}{25} = ?$$