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Mathematics 21 Online
OpenStudy (lina17):

solve for x: 2/5 (x-2)=4x help please help!!!!!!

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

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

OpenStudy (rsadhvika):

that one

OpenStudy (rsadhvika):

do you know how it works ?

OpenStudy (lina17):

kind of

OpenStudy (rsadhvika):

how does it work ?

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

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

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

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

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

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

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

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

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

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

OpenStudy (lina17):

can you help me on one more

OpenStudy (rsadhvika):

I'll try... tag me

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