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Mathematics 19 Online
OpenStudy (anonymous):

subtracting rational expressions: 7x x-2 --- - ------ 2x 20x+16 please include steps if possible

OpenStudy (anonymous):

\[\frac{7x}{2x} - \frac{x-2}{20x+16}\] ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

just take Lcm 2x(20x+16) and solve in as usual way

OpenStudy (anonymous):

\[\frac{7}{2} - \frac{x-2}{4(5x+4)}\]

OpenStudy (anonymous):

multiply 2(5x+4) both top and bottom of fraction 7/2

OpenStudy (mertsj):

Reduce the first fraction to 7/2. Now the common denominator is 2(10x+8) When you rewrite the first fraction with that denominator, it becomes 7(10x+8)/2(10x+8). since the second fraction already has that denominator, it does not change so now the problem is (70x+56-x+2)/2(10x+8) or (69x+58)/2(10x+8)

OpenStudy (anonymous):

why 2(5x+4) and not 4(5x+4)?why 2(5x+4) and not 4(5x+4)?

OpenStudy (anonymous):

why 2(5x+4) and not 4(5x+4)?

OpenStudy (anonymous):

when i factor 20x+16 I get 4(5x+4)

OpenStudy (anonymous):

why do you multiply times 2(5x+4) anmd not 4(5x+4)?

OpenStudy (mertsj):

Because the first fraction has a denominator of two. So rewrite the denominator of the second fraction as 2(10x+8). Now ask yourself: By what should I multiply the first denominator which is 2 to get the second denominator which is 2(10x+8)? Isn't it clear that the answer is "multiply by (10x+8)"? Then do the same thing to the numerator.

OpenStudy (anonymous):

how di you get 2(10x+8)?

OpenStudy (anonymous):

did*

OpenStudy (mertsj):

I took the denominator of the second fraction which is 20x + 16 and factored out a 2 . Since 20x+16 = 2(10x+8) they are equivalent

OpenStudy (anonymous):

you're just trying to make it equal to the other denomitaor?

OpenStudy (mertsj):

The reason I factored out the two is because the denominator of the first fraction is 2 and I wanted to see what I should multiply the first fraction by to get the same denominator the second fraction has.

OpenStudy (anonymous):

or it always hast to be the lcm?

OpenStudy (anonymous):

or it always hast to be the lcm?

OpenStudy (mertsj):

Yes, love. You cannot subtract two fractions unless they have the same denominator.

OpenStudy (anonymous):

but i mean you divide it to the lcm or just to make it equal to the other fraction?

OpenStudy (mertsj):

It is most efficient if the denominator is the LCM of both denominators but not mandatory. If the denominators are the same then the fractions can be subtracted.

OpenStudy (anonymous):

ok th

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

thanks

OpenStudy (mertsj):

You can see that 20x+16 is the LCM of 2 and 20x+16. I only wrote 20x+16 as 2(10x+8) to make it clear that 20x+16 IS the LCM

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