# Algebra 2 help;; solve:?

Asked by rachelmusil (40) May 4th, 2009

Homework wants me to solve 2x-1/5 = x+1/2
(that’s not two-x minus one-fifth, its two-x minus one all over five;; same with the second part)
How would I do this?

Observing members: 0 Composing members: 0

cross multiply or multiply everything by the LCD to get rid of the fractions, either way works

miasmom (3490)

Cross multiplying is indeed what you should do. If you are uncomfortable with this, let me show you how to get there.
You have:
(2x-1)/5 = (x+1)/2

Remember how when you first learned fractions, you looked for a common denominator? We do the same here. The common denominator is 5*2 = 10. Multiplying the numerator and denominator of each fraction by the same number, you get
2*(2x-1)/(2*5) = 5*(x+1)/(5*2) or
2*(2x-1)/10 = 5*(x+1)/10
Since the denominators of the fractions are the same, they will be equal if the numerators are equal, so you get
2*(2x-1) = 5*(x+1)

In the general case, what you end doing is multiplying the numerator of each fraction by the denominator of the other one and setting these two quantities equal to one another. That is cross multiplication.

For future reference, you need parentheses to properly notate this equation. As written, the equation must refer to 2x minus 1/5, etc., because multiplication and division are always performed before addition and subtraction. The equation should be written as (2x-1)/5=(x+1)/2 This shows that the (2x-1) is to be treated as a single term being divided.

Jayne (6751)

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