Mathematics
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OpenStudy (lina17):
solve for x: 2/5 (x-2)=4x help please help!!!!!!
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OpenStudy (lina17):
\[\frac{ 2 }{ 5 } (x-2)=4x\]
OpenStudy (rsadhvika):
As a start, multiply 5 both sides of the equation
OpenStudy (rsadhvika):
That gets rid of the fractions
OpenStudy (lina17):
how would it be set up ?
OpenStudy (lina17):
hello
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OpenStudy (rsadhvika):
I'm here, have you multiplied both sides by 5 ?
OpenStudy (lina17):
ya but what do i multiply 5 from
OpenStudy (rsadhvika):
Good question
before answering that,let me ask you a question
OpenStudy (rsadhvika):
Have you seen a balance used to measure weights before ?
OpenStudy (lina17):
yes
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OpenStudy (rsadhvika):
that one
OpenStudy (rsadhvika):
do you know how it works ?
OpenStudy (lina17):
kind of
OpenStudy (rsadhvika):
how does it work ?
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OpenStudy (lina17):
dont you put them on each side to see which weighs less or more
OpenStudy (rsadhvika):
Yes, what happens if you put 1kg on both sides ?
OpenStudy (lina17):
i would be even
OpenStudy (rsadhvika):
good, what happens if you put 5kg on both sides
OpenStudy (lina17):
it would still be even
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OpenStudy (rsadhvika):
Exactly! As you can see, adding the "same" weight to both sides of the balance won't change the balance. It keeps the balance remain even.
OpenStudy (rsadhvika):
An equation is also like that balance.
OpenStudy (rsadhvika):
If you multiply/add same thing to both sides, the equation will not change. It will remain same.
OpenStudy (rsadhvika):
\[\frac{ 2 }{ 5 } (x-2)=4x\]
OpenStudy (lina17):
so what em i suppose to do
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OpenStudy (rsadhvika):
Since I don't like fractions, and since I am allowed to multiply/add same thing to both sides, I'd like to multiply 5 both sides.
OpenStudy (rsadhvika):
\[\color{red}{5\times }\frac{ 2 }{ 5 } (x-2)= \color{red}{5\times}4x\]
OpenStudy (rsadhvika):
Fine with that step ?
OpenStudy (lina17):
yes
OpenStudy (rsadhvika):
Notice that 5 in the top and bottom eat eachother out
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OpenStudy (rsadhvika):
\[\color{red}{\cancel 5\times }\frac{ 2 }{\cancel 5 } (x-2)= \color{red}{5\times}4x\]
OpenStudy (rsadhvika):
\[2(x-2)= \color{red}{5\times}4x\]
OpenStudy (rsadhvika):
do you know how to simplify further ?
OpenStudy (lina17):
um i think so
OpenStudy (lina17):
let me try
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OpenStudy (rsadhvika):
Awesome, so what would be your next step ?
OpenStudy (lina17):
\[2(x-2)= 2x-4\]
OpenStudy (rsadhvika):
Excellent! that is your left side
OpenStudy (lina17):
5x4x=20x
OpenStudy (lina17):
2x-4=20x
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OpenStudy (lina17):
em i right
OpenStudy (rsadhvika):
Perfect!
OpenStudy (lina17):
so what do i do from there'
OpenStudy (rsadhvika):
Our goal is to solve x
so we need to try and isolate it
OpenStudy (lina17):
do we subtract 2x from both sides
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OpenStudy (rsadhvika):
Yes :
\[2x-4 \color{red}{-2x} = 20x \color{red}{-2x}\]
OpenStudy (lina17):
so it would be 2x=18x
OpenStudy (rsadhvika):
Nope
OpenStudy (rsadhvika):
right hand side is correct,
left hand side is wrong
try again
OpenStudy (lina17):
oh it would be 6x=18x
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OpenStudy (rsadhvika):
looks you're adding apples and oranges
OpenStudy (rsadhvika):
remember, you can only add/subtract like terms
OpenStudy (lina17):
so x = 3
OpenStudy (rsadhvika):
wrong
OpenStudy (lina17):
dang
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OpenStudy (rsadhvika):
\[2x-4 \color{red}{-2x} = 20x \color{red}{-2x}\]
I cannot add -4 and -2x because these are not "like terms"
OpenStudy (rsadhvika):
I can only add 2x and -2x, which gives me 0
OpenStudy (rsadhvika):
\[2x-4 \color{red}{-2x} = 20x \color{red}{-2x}\]
simplifies to
\[-4 = 18x\]
OpenStudy (lina17):
do i divide
OpenStudy (rsadhvika):
Yes, divide 18 both sides so that x gets isolated
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OpenStudy (lina17):
x=2
OpenStudy (lina17):
i mean 9
OpenStudy (rsadhvika):
\[-4 = 18x\]
divide 18 both sides and get
\[-\dfrac{4}{18} = x\]
which is same as
\[-\dfrac{2}{9}=x\]
OpenStudy (rsadhvika):
so \(x = -\dfrac{2}{9}\)
OpenStudy (lina17):
oh ok
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OpenStudy (lina17):
can you help me on one more
OpenStudy (rsadhvika):
I'll try... tag me