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Iwouldhighly appreciate it if someone could solve these forme.I think I typed themout correctly. What I'm looking at,for instan...Asked by mike
Iwouldhighly appreciate it if someone could solve these forme.I think I typed themout correctly. What I'm looking at,for instance,the first one,is displayed 5n over 3 times 4n over 3
1.(5n/3)- (4n/3)
2.(3x/10)+(9x/10)
3.(n-2/n+4)+(8/[n+4])
4.(5m+8/[m-1]) - (m-3/[m-1])
5.(x^2/[x-1]) - (1/[x-1])
6.(4x+y/[2x+3y])-(2x -2y/[2x+3y])
7.(x^2+3x+2/[x^2-16] + (3x++6/[x^2-16])
8.7/(x-5) - (2+x/[x-5])
First, "(5n/3)- (4n/3)" is not "5n over 3 <B>times</B> 4n over 3."
If the fractions have the same denominator, just do the required operations in the numerator.
(5n/3)- (4n/3) = n/3
Use the same principle for the remaining problems.
I hope this helps. Thanks for asking.
1.(5n/3)- (4n/3)
2.(3x/10)+(9x/10)
3.(n-2/n+4)+(8/[n+4])
4.(5m+8/[m-1]) - (m-3/[m-1])
5.(x^2/[x-1]) - (1/[x-1])
6.(4x+y/[2x+3y])-(2x -2y/[2x+3y])
7.(x^2+3x+2/[x^2-16] + (3x++6/[x^2-16])
8.7/(x-5) - (2+x/[x-5])
First, "(5n/3)- (4n/3)" is not "5n over 3 <B>times</B> 4n over 3."
If the fractions have the same denominator, just do the required operations in the numerator.
(5n/3)- (4n/3) = n/3
Use the same principle for the remaining problems.
I hope this helps. Thanks for asking.
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