Directions say to prove the algebraic identity by starting with the Left hand side expression and supplying a sequence of equivalent expressions that ends with the right hand expressions

(1/x) - (1/2) = (2-x)/(2x)

Your goal is to prove that this algebraic identity holds true. To do so, simply solve for the left hand side of this equation. Since the left hand side of the equation requires you to subtract two fractions, you need to find the smallest common denominator between these two fractions. Given that the first fraction has a denominator of x and the second fraction has a denominator of 2, it can be concluded that the smallest common denominator is 2x. With that, multiply the first term (1/x) by 2/2 to yield a fraction whose denominator is 2x. Then, multiply the second term (1/2) by x/x to yield a fraction whose denominator is also 2x. Notice that you are multiplying the numerator and the denominator by the same number (or term), which means that you are essentially multiplying it by 1 so you are not changing the value of the fraction.

So we first change the two fractions on the left hand of the equation to obtain two fractions that share a common denominator:

(1/x) * (2/2) = (1 * 2) / (x * 2) = (2) / (2x) = **2/(2x)**

(1/2) * (x/x) = (1 * x) / (2 * x) = (x) / (2x) = **x/(2x)**

Replace the first fraction in the original equation by 2/(2x) and the second fraction by x/(2x):

(1/x) - (1/2) ==> (2/(2x)) - (x/(2x))

Next, we can combine these two fractions since they now have a common denominator:

(2/(2x)) - (x/(2x)) = (2 - x)/(2x) = (2-x)/(2x)

Thus, we have proved that (1/x) - (1/2) = (2-x)/(2x).

## Comments

Thank you so much, this was so helpful! :)

Nice answer.

Also note that a common denominator (not necessarily the smallest, tho) can be found by cross-multiplying the fraction:

1/x - 1/2 ==> (1*2 - 1*x) / (2 * x) = (2 - x)/2x

To get the smallest, then you would just need to simplify the expression. In thise case, there is no simplification possible as it is already the smallest common denominator.