boomerang285
can someone show me the steps to solve x/6=x/7+5
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carl_gauss
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multiply with 6 and 7 than bring x to one side. voila
boomerang285
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\[\frac{ x }{ 6 }=\frac{ x }{ 7 }+5\] to better understand here is the actual problem in the correct equation
order
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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
order
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7x/42 - 6x/42 = 5
x/42=5
x=5(42)
calculusfunctions
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\[\frac{ x }{ 6 }=\frac{ x }{ 7 }+5\]
Multiply both sides by 42 (least common multiple of 6 and 7), to obtain
7x = 6x + 210
∴ x = 210
Do you understand?
boomerang285
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so step 2 would look like x/7-x/6=5
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yes
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Then you make the denominators the same
boomerang285
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which would be 42,
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yes since 6 X 7 = 42
boomerang285
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this is the part i get confused at
boomerang285
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would it look like x/42?
order
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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
boomerang285
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ahhh ok
boomerang285
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so then is x/42=5
order
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Yup
boomerang285
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and then simplify from there
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yup :)
boomerang285
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how do i get x by itself?
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move the 42 over to the other side by timesing both sides by 42.
magepker728
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|dw:1349349298325:dw|
order
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42(x/42) = 5(42)
x =210
boomerang285
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thank you so much.. fractions is one thing i have an issue with..
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that's fine :)
boomerang285
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i have even harder ones... like \[\frac{ x}{ 2x+2 }=\frac{ -2x }{ 4x+4 }+\frac{ 2x-3 }{ x+1 }\]
order
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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
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\[{-2x \over 4(x+1)} + {4(2x-3) \over 4(x+1)}\]
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and add them
\[-2x + 8x -12 \over 4(x+1)\]
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do the same with the other side and use 2(x+1) as denominator
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\[{x \over 2(x+1)} = {6x -12 \over 4(x+1)}\]
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and cross multiply for simplification
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2x(x+1) = (6x-12) (x+1)
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cancel the x+1
and you have 2x = 6x-12
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and that is simplified as
-4x = -12
which means x = 3
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understand?
boomerang285
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i believe so
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great :)
boomerang285
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i have one more fraction if anyone is interested in helping
boomerang285
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\[\frac{ 5 }{ y-4 }-\frac{ 8 }{ y-4 }=\frac{ 2 }{ y^2-16 }\]
boomerang285
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it throws me off to add sqrt
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um... you could do it like this:
\[{-3 \over y-4} = {2 \over y^2 -16}\]
\[{-3 \over y -4} = {2 \over (y+4)(y-4)}\]
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Then knock off the y-4 so you're left with
\[-3 = {2 \over y+4}\]
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and then use -3 (y+4) = 2
and simplify
-3y - 12 = 2
-3y = 2 +12
-3y = 14
y = 14/-3
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y = -(14/3)
boomerang285
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wow that was easier than i thought
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You were only stumped by the y^2 - 16
which is simplified as (y-4)(y+4)
boomerang285
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thank you so much for helping me with these!
order
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No probelma :)