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Alja [10]
3 years ago
11

On the same day at school, 1/6 of the students were wearing skirts and 5/8 of the students were wearing pants. The rest of the s

tudents were wearing shorts. What fraction of the students were wearing shorts?
Mathematics
2 answers:
anastassius [24]3 years ago
4 0
What you do is you find the common denominator between the two fractions given.  the easiest is to just multiply them 

6*8=48

and then multiply each numerator by the number it's denominator is multiplied by

1*8=8  and 5*6=30

and now you have your new fractions, add them together

8/48+30/48=38/48

this leaves 10/48 as the people who were wearing shorts,  just simplify this and it's done

10/2    =5/24
48/2
Jlenok [28]3 years ago
4 0
The answer to this problem is 5/24. First you find the LCM that is 48. Then you multiply 1 by 8 and get 8 so that fraction is 8/48. Then you multiply 5 by 4 and  get 20 put that into a fraction and subtract then you reduce the fraction.

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3 years ago
Two problems here I need solved! I need every step, so please have that with your answers!!
Free_Kalibri [48]
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

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We multiply through by LCM of
(x-4)(x - 2)

(x - 2) + x(x-4) = 2

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
x = 4
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It is an extraneous solution.

\therefore \: x = - 1
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QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

We square both sides,

x + 11 = (x - 1)^{2}

We expand to get,

x + 11 = {x}^{2} - 2x + 1

This implies,

{x}^{2} - 3x - 10 = 0

We solve this quadratic equation by factorization,

{x}^{2} - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x + 2 = 0 \: or \: x - 5 = 0

x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
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