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Solve each one!

 Feb 22, 2019
 #1
avatar+103674 
+1

Why don't you do it yourself?

 Feb 22, 2019
 #2
avatar+8 
-4

I have already. Is that sometimes I used this website to make sure my work is correct.

 

-- 7H3_5H4D0W

GAMEMASTERX40  Feb 22, 2019
 #4
avatar+103674 
+1

ok but you should check your own answers by plugging in the value that you get into both sides of the equation and seeing if the 2 sides are the same.

 

You do not need other people to check questions like this, you can, and should, do it for yourself. 

 

If you want other people to check other types of questions for you  (and you already have your answer) you should state that in your question. 

Melody  Feb 22, 2019
 #5
avatar+8 
-4

Not all math questions are about plugging in numbers in equations and getting a correct answer. There are complicated questions that are new and word problems that are confusing where other people can help to answer them. I don't see a problem of making sure your work is right and you can probably learn a thing or two from it too. It wouldn't hurt to do these kinds of things and I know I can do it myself but this is a Math Forum for people that can freely post their math.

 

-- 7H3_5H4D0W

GAMEMASTERX40  Feb 22, 2019
edited by GAMEMASTERX40  Feb 22, 2019
 #6
avatar+103674 
0

These are plug in questions (as far as checking goes) and you are just being lazy.

Melody  Feb 23, 2019
 #3
avatar+102921 
+3

a)    1/x + 1 / 3x  = 6

        3/3x + 1/3x  = 6

        [ 4] / 3x = 6  multiply both sides by 3x

         4 =  18x       divide both sides by 18

         4/18 = x

         2/9 =   x

 

b)     1/x + 1 /x+1  = 3

         [ x + 1 + x ] / [ x ( x + 1) ] = 3

         [2x + 1 ] =  3 [ x ( x + 1) ]

         2x + 1  = 3 [ x^2 + x]

         2x + 1 =  3x^2 + 3x          rearrange as

         3x^2  + x - 1 = 0

Use the quad formula

 

x = [ -1 ±√ [ 1 - 4(3)(-1) ] / {2 * 3 ]

 

x =   [ -1 ±√13] / 6

 

 

c)   ( x +3 ) / ( x - 1)   -  x / (x + 1)  =    8 / [ (x + 1) ( x - 1) ]

Multiply through by  (x + 1) ( x - 1)

 

(x + 3)(x + 1)  - x(x - 1) =  8       simplify

 

x^2 + 4x  + 3  - x^2 + x = 8

 

4x + 3 + x = 8

 

5x + 3 = 8

 

5x = 5

 

x = 1

 

However....this makes a denominator = 0...so no solution exists

 

 

cool cool cool

 Feb 22, 2019

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