Question

can me explain the procedure to resolve?

1/a + 1/2 = 2/a

Answer 1

You can multiply both sides by the LCD 2a to get...
\[\Large \frac{1}{a} + \frac{1}{2} = \frac{2}{a}\]
\[\Large 2a\left(\frac{1}{a} + \frac{1}{2}\right) = 2a\left(\frac{2}{a}\right)\]
\[\Large 2a\left(\frac{1}{a}\right) + 2a\left(\frac{1}{2}\right) = 2a\left(\frac{2}{a}\right)\]
\[\Large \frac{2a}{a} + \frac{2a}{2} = \frac{4a}{a}\]
\[\Large \frac{2\cancel{a}}{\cancel{a}} + \frac{\cancel{2}a}{\cancel{2}} = \frac{4\cancel{a}}{\cancel{a}}\]
\[\Large 2 + a = 4\]
Notice how those pesky fractions have been eliminated after multiplying every term by the LCD.

Answer 2

ok thanks

Answer 3

What solution do you get?

Answer 4

A= 2 ?

Answer 5

Check:
\[\Large \frac{1}{a} + \frac{1}{2} = \frac{2}{a}\]
\[\Large \frac{1}{2} + \frac{1}{2} = \frac{2}{2}\]
\[\Large \frac{1+1}{2} = \frac{2}{2}\]
\[\Large \frac{2}{2} = \frac{2}{2} \ \ {\color{green}{\checkmark}}\]
so a = 2 is definitely the answer

Answer 6

it's always a good idea to check the answer back into the ORIGINAL equation because sometimes you'll get potential solutions that don't work at all (they are called extraneous solutions)

Answer 7

ok thanks you

Answer 8

np

Answer 9

just contribute another way to solve