How do you solve: \[\frac{7-5x}{4}=-2\]

- andriod09

How do you solve: \[\frac{7-5x}{4}=-2\]

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- gorica

multiply both sides by 4, you get
7-5x=-8
7+8=5x
15=5x
x=15/3
x=3

- gorica

* x=15/5

- andriod09

I'm sorry, but that makes no sense. I don't know how to get the 7 down from the \[\frac{7-5x}{4}\] if i could do that, then i know how to solve the problem.

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## More answers

- gorica

\[\frac{ 7-5x }{ 4 } 4 = 7-5x\]

- andriod09

Wheres the reciprocal then?

- andriod09

because you can't do \[\frac{4}{7-5x}\] with out messing up the equation, and you can't just multiply them by 4.

- gorica

you can multiply both sides of equation by the same number and you won't change the result. So, multiply fraction by 4 and multiply -2 by 4 and you will get what I wrote above.

- andriod09

But that's not right. If you had a problem like: \[\frac{9}{4}k=-36\] you would do this: \[(\frac{4}{9})(\frac{9}{4})k=-36(\frac{4}{9})\] then you would divide -36 by 9, to get four, then four times four. How would you do that to this problem?

- gorica

if you multiply
\[\frac{ 9 }{ 4 }k=-36\] by 4, you'll get
9k=-36*4
then you divide this equation by 9 and you get
k=-\[k=-\frac{ 36 }{ 9 } 4\]
k=-4*4
k=-16
You will get the same result if you multiply equation from the start by 4/9. Right?
Ok, if you want reciprocal for your first question, you can write 4 as \[\frac{ 4 }{ 1 }\]. Ok?

- andriod09

But that's not the reciprocal! The reciprocal of the problem would cancel them out, and I can't do it like that!

- gorica

ok, tell me, what is reciprocal for \[\frac{ 1 }{ 4 }\]?

- andriod09

The reciprocal for \[\frac{1}{4}\] is \[\frac{4}{1}\] but the reciprocal for \[\frac{7-5x}{4}\] isn't 4, its \[\frac{4}{7-5x} which would cancel out the x, and just leave you with a normal number.

- gorica

you can't cancel out x because you have to find a value of x

- gorica

that's the aim, to find value of x.

- andriod09

That's what i'm saying!

- gorica

Do you know that \[\frac{ a }{ b }=\frac{ c }{ d }\] is the same as a*d=c*b ?

- andriod09

Yes i do, But you're seriously confusing me.

- gorica

Ok, good! Then let me help you.

- andriod09

Then make it easily understandable!

- gorica

\[\frac{ 7-5x }{ 4 }=-2\] is the same as
\[\frac{ 7-5x }{ 4 }=\frac{ -2 }{ 1 }\] because \[-2=\frac{ -2 }{ 1 }\]
now you have, from
\[\frac{ a }{ b }=\frac{ c }{ d } <=> a*d=c*b\],
(7-5x)*1=4*(-2)
7-5x=-8
7+8=5x
15=5x
x=3

- gorica

ok?

- phi

you misunderstand the "multiply by the reciprocal" rule
in your problem
\[ \frac{1}{4} \cdot (7-5x) = -2\]
one way to solve is to multiply both sides by 4
\[ \frac{4}{4} \cdot (7-5x) = -2\cdot 4\]

- andriod09

That's not my problem though, my problem is this: \[\frac{7-5x}{4}\] that's what it is.

- phi

as you know,
\[ \frac{7-5x}{4} = \frac{1}{4} \cdot (7-5x) \]

- gorica

I'm giving up :D

- phi

another way to write it is
\[ \frac{7-5x}{4} = \frac{7}{4} - \frac{5x}{4} \]

- phi

dividing by 4 is the same as multiplying by 1/4

- andriod09

I don't get it.

- phi

which part ?

- andriod09

The reciprocal part. How do you get 1/4??

- phi

maybe it makes more sense if we use numbers as examples:
Example:
\[ \frac{8}{4}=2 \\ \text{ and } \\
\frac{1}{4} \cdot 8= 2\]
dividing 8 by 4 is the same as multiplying 8 by 1/4

- andriod09

So how do you get 1/4 from 7-5x/4? that's what i need to know...

- phi

if you know how to multiply fractions (top times top over bottom times bottom) we can say
\[ \frac{1}{4} \cdot 8 = \frac{1}{4} \cdot\frac{8}{1}=\frac{1 \cdot 8}{4 \cdot 1}= \frac{8}{4} = 2 \]

- phi

that same rule works for complicated expressions like 7-5x (instead of 8)

- andriod09

OKay, but still, how do you separate the 7/4 from the -5x/4?

- phi

Here is an example of how to add two fractions with the same denominator,
\[ \frac{ 1}{4} + \frac{3}{4} \]
the answer is : keep the same denominator and add the numerators
\[ \frac{ 1}{4} + \frac{3}{4}= \frac{1+3}{4}\]
notice you can switch between those two forms
example
\[ \frac{7-5x}{4} = \frac{7}{4} + \frac{-5x}{4} \]

- phi

another way to solve your problem (though this is a bit messy)
\[ \frac{7}{4} + \frac{-5x}{4}= -2 \\ \frac{-5x}{4}= -2-\frac{7}{4} \]
simplify, then multiply both sides by -4/5

- andriod09

okay, i'm beyond confused. >.< i just need to answer these last few questions...

- phi

can you solve for x in
\[ \frac{4}{4} \cdot (7-5x) = -2\cdot 4 \]?

- phi

4/4 is 1
-2*4 is -8
so you get
\[ 7 -5x= -8\]
can you finish ?

- andriod09

\[7-5x=-8\]\[-7=-7\]\[-5x=15\]\[\frac{-5x}{5}=\frac{-15}{-5}\]\[x=3\] right?

- phi

yes, though you have a typo on the 2nd to last line: you mean -5x/ -5
but that is the answer.
Now to do the original problem you have to convince your self that multiplying both sides by 4 is a good thing to do.

- andriod09

Okay, but still, how would you get the four from \[\frac{7-5x}{4} That's what i don't understand.

- phi

Can you answer this
\[ 4 \cdot \frac{x}{4} = \]?

- andriod09

no. not correctly at least.

- phi

can you do
\[ 4 \cdot \frac{2}{4} \]?

- andriod09

uhh. would it be: \[\frac{8}{16}\] or \[\frac{1}{2}\]?

- phi

a calculator would tell you. But if you don't understand fractions you will get stuck doing algebra.
when you get time, see
http://www.khanacademy.org/math/arithmetic/fractions/multiplying_and_dividing_frac/v/multiplying-fractions
(and any of the videos about fractions that you don't know)
but the rule for multiplying fractions is multiply top times top and bottom times bottom
if one of the numbers is a whole number, set its "bottom" to 1
\[ 4 \cdot \frac{2}{4}= \frac{4}{1}\cdot \frac{2}{4} = \frac{4\cdot 2}{1\cdot 4}\]

- phi

one of the rules you know about multiplying is you can swap the order. so
\[ \frac{4\cdot 2}{1\cdot 4} = \frac{2\cdot 4}{1\cdot 4}\]
we can think of that as multiplying
\[ \frac{2}{1}\cdot \frac{4}{4}\]
of course 4/4 is 1
\[ \frac{2}{1}\cdot 1 \]
and that simplifies to 2
if you practice, you will learn that if you have the same thing in the top and bottom (like the 4 in this example), you can cancel it. and get 2 immediately

- phi

so if you see
\[ 4 \cdot \frac{2}{4}= \cancel{4} \cdot \frac{2}{\cancel{4}} =2 \]

- phi

\[ \frac{7-5x}{4} = -2 \\ 4 \cdot \frac{7-5x}{4} = -2 \cdot 4 \]
now use the rule "cancel" the 4, because you have a 4 in the top and bottom

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