Question

1/3 - 1/2 (7 - 2/15) + 3/10

Answer 1

You are asked to perform:

$\frac{1}{3}-\frac{1}{2}(7-\frac{2}{15})+\frac{3}{10}$

Wee need to start with the parenthesis.

$7-\frac{2}{15}=\frac{7}{1}-\frac{2}{15}$

We the numerator will be 7 times 15 minus 2 (2x1), and the denominator will be 15 (15x1)

$\frac{105-2}{15}=\frac{103}{15}$

Now, the expression to solve is:

$\frac{1}{3}-\frac{1}{2}\cdot\frac{103}{15}+\frac{3}{10}$

We need to perform the multiplication first (the second term)

$\frac{1}{2}\cdot\frac{103}{15}=\frac{103}{2\cdot15}=\frac{103}{30}$

Now, the expression to solve will be:

$\frac{1}{3}-\frac{103}{30}+\frac{3}{10}$

Now we can solve either the substraction or the sum. Let's go first with the substraction:

$\frac{1}{3}-\frac{103}{30}=\frac{1\cdot30-3\cdot103}{3\cdot30}=\frac{30-309}{90}=\frac{-279}{90}$$\begin{gathered} \frac{-279}{90}\text{ can be simplified since both numerator and denominator are divisible by 3} \\ \end{gathered}$$\frac{-279}{90}=\frac{-93}{30}=\frac{-31}{10}$

Note that it could be simplified twice, dividing both numerator and denominator by 3.

Then, the expression to solve is reduced to:

$-\frac{31}{10}+\frac{3}{10}=\frac{3}{10}-\frac{31}{10}$

We can make that substraction easily since they have the same denominator. 3 - 31 = -28

Then, the first expressions equals to:

$\frac{-28}{10}=\frac{-14}{5}$

Then, the full answer will be:

$\frac{1}{3}-\frac{1}{2}\cdot(7-\frac{2}{15})+\frac{3}{10}=-\frac{14}{5}$