andriod09
  • andriod09
How do you solve: \[\frac{7-5x}{4}=-2\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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gorica
  • gorica
multiply both sides by 4, you get 7-5x=-8 7+8=5x 15=5x x=15/3 x=3
gorica
  • gorica
* x=15/5
andriod09
  • 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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gorica
  • gorica
\[\frac{ 7-5x }{ 4 } 4 = 7-5x\]
andriod09
  • andriod09
Wheres the reciprocal then?
andriod09
  • 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
  • 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
  • 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
  • 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
  • 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
  • gorica
ok, tell me, what is reciprocal for \[\frac{ 1 }{ 4 }\]?
andriod09
  • 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
  • gorica
you can't cancel out x because you have to find a value of x
gorica
  • gorica
that's the aim, to find value of x.
andriod09
  • andriod09
That's what i'm saying!
gorica
  • gorica
Do you know that \[\frac{ a }{ b }=\frac{ c }{ d }\] is the same as a*d=c*b ?
andriod09
  • andriod09
Yes i do, But you're seriously confusing me.
gorica
  • gorica
Ok, good! Then let me help you.
andriod09
  • andriod09
Then make it easily understandable!
gorica
  • 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
  • gorica
ok?
phi
  • 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
  • andriod09
That's not my problem though, my problem is this: \[\frac{7-5x}{4}\] that's what it is.
phi
  • phi
as you know, \[ \frac{7-5x}{4} = \frac{1}{4} \cdot (7-5x) \]
gorica
  • gorica
I'm giving up :D
phi
  • phi
another way to write it is \[ \frac{7-5x}{4} = \frac{7}{4} - \frac{5x}{4} \]
phi
  • phi
dividing by 4 is the same as multiplying by 1/4
andriod09
  • andriod09
I don't get it.
phi
  • phi
which part ?
andriod09
  • andriod09
The reciprocal part. How do you get 1/4??
phi
  • 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
  • andriod09
So how do you get 1/4 from 7-5x/4? that's what i need to know...
phi
  • 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
  • phi
that same rule works for complicated expressions like 7-5x (instead of 8)
andriod09
  • andriod09
OKay, but still, how do you separate the 7/4 from the -5x/4?
phi
  • 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
  • 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
  • andriod09
okay, i'm beyond confused. >.< i just need to answer these last few questions...
phi
  • phi
can you solve for x in \[ \frac{4}{4} \cdot (7-5x) = -2\cdot 4 \]?
phi
  • phi
4/4 is 1 -2*4 is -8 so you get \[ 7 -5x= -8\] can you finish ?
andriod09
  • andriod09
\[7-5x=-8\]\[-7=-7\]\[-5x=15\]\[\frac{-5x}{5}=\frac{-15}{-5}\]\[x=3\] right?
phi
  • 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
  • andriod09
Okay, but still, how would you get the four from \[\frac{7-5x}{4} That's what i don't understand.
phi
  • phi
Can you answer this \[ 4 \cdot \frac{x}{4} = \]?
andriod09
  • andriod09
no. not correctly at least.
phi
  • phi
can you do \[ 4 \cdot \frac{2}{4} \]?
andriod09
  • andriod09
uhh. would it be: \[\frac{8}{16}\] or \[\frac{1}{2}\]?
phi
  • 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
  • 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
  • phi
so if you see \[ 4 \cdot \frac{2}{4}= \cancel{4} \cdot \frac{2}{\cancel{4}} =2 \]
phi
  • 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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