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neverforgetvivistee
subtracting rational expressions: 7x x-2 --- - ------ 2x 20x+16 please include steps if possible
\[\frac{7x}{2x} - \frac{x-2}{20x+16}\] ?
just take Lcm 2x(20x+16) and solve in as usual way
\[\frac{7}{2} - \frac{x-2}{4(5x+4)}\]
multiply 2(5x+4) both top and bottom of fraction 7/2
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)
why 2(5x+4) and not 4(5x+4)?why 2(5x+4) and not 4(5x+4)?
why 2(5x+4) and not 4(5x+4)?
when i factor 20x+16 I get 4(5x+4)
why do you multiply times 2(5x+4) anmd not 4(5x+4)?
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.
how di you get 2(10x+8)?
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
you're just trying to make it equal to the other denomitaor?
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.
or it always hast to be the lcm?
or it always hast to be the lcm?
Yes, love. You cannot subtract two fractions unless they have the same denominator.
but i mean you divide it to the lcm or just to make it equal to the other fraction?
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.
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