Algebra 2 Help!! ASAP!?
1. Show all work to simplify (2/x) - (2/x-1) + (2/x-2). Use complete sentences to explain how to simplify this expression. Remember to list all restrictions.
2. Part 1: Simplify the complex fractions. (4 points)
[(x^2 - x - 20)/(4)]/[(x - 5)/(10)]
and
[(x^2 - x - 20)/(x - 5)]/[(4)/(10)]
Part 2: Are the complex fractions equivalent? Explain why or why not. (4 points)
3. Show all work to solve 1/x + 2/9 = 1/3. Use complete sentences to explain how to solve this equation.
I really really need help with all of these right now!!! Please and thank you so so so much!!
Update:Okay, I got number two. And the steps to number one, but the signs are confusing me!!!
And I still don't have number 3.
Comments
1.
(2/x) - (2/x-1) + (2/x-2)
Take LCM of x, x-1 and x-2,the three denominators(Dr) which is, multiplying Nr(numerator) and Dr of each term by the other two..
it means,
2(x-1)(x-2)/x(x-1)(x-2) - 2x(x-2)/x(x-1)(x-2) + 2x(x-1)/x(x-1)(x-2)
as the Dr of each term is same we can club all the three terms
[2(x-1)(x-2) - 2x(x-2) + 2x(x-1)] / [x(x-1)(x-2)]
Simplifying Nr. we get
2(x-1)(x-2) - 2x(x-2) + 2x(x-1)
= 2x²-6x +4 - (2x²-4x) + (2x²-2x)
= 2x²-6x +4 - 2x²+4x + 2x²-2x
=2x²-4x +4
Dr. we get x(x-1)(x-2) = x(x²-3x+2) = x³-3x²+2x
So simplified expression is Nr/Dr = (2x²-4x +4)/(x³-3x²+2x)
2. Part 1:
[(x² - x - 20)/(4)]/[(x - 5)/(10)]
=[(x² -5x +4x - 20)/(4)]*[(10)/(x-5)]
=[{x(x-5) +4(x-5)} /(4)]*[(10)/(x-5)]
=[(x+4)(x-5) /(4)]*[(10)/(x-5)]
=[(x+4)(x-5)*10 /(4*(x-5))]
= 5(x+4)/2
= (5x/2) + 10
and
[(x^2 - x - 20)/(x - 5)]/[(4)/(10)]
=[(x² -5x +4x - 20)/(x-5)]*[(10)/(4)]
=[{x(x-5) +4(x-5)} /(x-5)]*[(5)/(2)]
=[(x+4)(x-5) /(x-5)]*[(5)/(2)]
=(x+4)*(5/2)
=(5x/2) +10
Part 2: Yes the complex fractions are equivalent as both the expressions have same terms in the numberator and the denominator ultimately.
3.
1/x + 2/9 = 1/3
Taking LCM on left hand side,
9/9x + 2x/9x = 1/3
(9+2x)/9x = 1/3
By cross multiplication
3(9+2x)=1*9x
27 + 6x = 9x
27 = 9x-6x
27 = 3x
Dividing both sides by 3
9 = x
so x = 9
Hope it helped ... ... ...
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sparkling up for the two x or y in the two equation then plug what you get in the different equation for that comparable variable. as an occasion: a million.12x=3.sixty 4+a million.4y->divide the two aspects by a million.12 after which you get x=(3.sixty 4/a million.12)+(a million.4y/a million.12) (Sorry, i've got not got a calculator). Plug that final equation in for x and then slove for y. once you sparkling up for y in the surprising equation plug it in in the different equation and discover x.