can someone show me the steps to solve x/6=x/7+5

- anonymous

can someone show me the steps to solve x/6=x/7+5

- jamiebookeater

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- anonymous

multiply with 6 and 7 than bring x to one side. voila

- anonymous

\[\frac{ x }{ 6 }=\frac{ x }{ 7 }+5\] to better understand here is the actual problem in the correct equation

- anonymous

Move the x's to one side. Make the denominators equal. Minus the two terms, and move the denominator over to the over side and multiply 5 by the denominator (42) and then divide by the number next to x on the side where 5(42) is. and voila

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## More answers

- anonymous

7x/42 - 6x/42 = 5
x/42=5
x=5(42)

- calculusfunctions

\[\frac{ x }{ 6 }=\frac{ x }{ 7 }+5\]
Multiply both sides by 42 (least common multiple of 6 and 7), to obtain
7x = 6x + 210
∴ x = 210
Do you understand?

- anonymous

so step 2 would look like x/7-x/6=5

- anonymous

yes

- anonymous

Then you make the denominators the same

- anonymous

which would be 42,

- anonymous

yes since 6 X 7 = 42

- anonymous

this is the part i get confused at

- anonymous

would it look like x/42?

- anonymous

but in order to make the denominators equation, you need to times the numerator by the amount that the denominator needs to make 42. eg. 6 X ? = 42 (7)
so times the top x by 7 which becomes 7x.
and the other becomes 6x since you need to times 7 by 6 to make 42.
so it would be 7x/42 - 6x/42
7x-6x = x
so it will eventually become x/42

- anonymous

ahhh ok

- anonymous

so then is x/42=5

- anonymous

Yup

- anonymous

and then simplify from there

- anonymous

yup :)

- anonymous

how do i get x by itself?

- anonymous

move the 42 over to the other side by timesing both sides by 42.

- anonymous

|dw:1349349298325:dw|

- anonymous

42(x/42) = 5(42)
x =210

- anonymous

thank you so much.. fractions is one thing i have an issue with..

- anonymous

that's fine :)

- anonymous

i have even harder ones... like \[\frac{ x}{ 2x+2 }=\frac{ -2x }{ 4x+4 }+\frac{ 2x-3 }{ x+1 }\]

- anonymous

You use the same principle:
so to add two fractions, the denominators must be the same.
make the denominator under -2x simplified:
4x + 4
4(x+1)
and use {-2x over 4(x+1)} + {4(2x-3) over 4(x+1)}
as a first step

- anonymous

\[{-2x \over 4(x+1)} + {4(2x-3) \over 4(x+1)}\]

- anonymous

and add them
\[-2x + 8x -12 \over 4(x+1)\]

- anonymous

do the same with the other side and use 2(x+1) as denominator

- anonymous

\[{x \over 2(x+1)} = {6x -12 \over 4(x+1)}\]

- anonymous

and cross multiply for simplification

- anonymous

2x(x+1) = (6x-12) (x+1)

- anonymous

cancel the x+1
and you have 2x = 6x-12

- anonymous

and that is simplified as
-4x = -12
which means x = 3

- anonymous

understand?

- anonymous

i believe so

- anonymous

great :)

- anonymous

i have one more fraction if anyone is interested in helping

- anonymous

\[\frac{ 5 }{ y-4 }-\frac{ 8 }{ y-4 }=\frac{ 2 }{ y^2-16 }\]

- anonymous

it throws me off to add sqrt

- anonymous

um... you could do it like this:
\[{-3 \over y-4} = {2 \over y^2 -16}\]
\[{-3 \over y -4} = {2 \over (y+4)(y-4)}\]

- anonymous

Then knock off the y-4 so you're left with
\[-3 = {2 \over y+4}\]

- anonymous

and then use -3 (y+4) = 2
and simplify
-3y - 12 = 2
-3y = 2 +12
-3y = 14
y = 14/-3

- anonymous

y = -(14/3)

- anonymous

wow that was easier than i thought

- anonymous

You were only stumped by the y^2 - 16
which is simplified as (y-4)(y+4)

- anonymous

thank you so much for helping me with these!

- anonymous

No probelma :)

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