George tells you that when variables are in the denominator, the equation 4/5 + 3/x+1/2 becomes unsolvable. George explains, "There is a value for x that makes the denominator zero, and you can't divide by zero." Demonstrate to George how the equation is still solvable and explain your reasoning.
im a little upset just cos i know this example should be easier than it is but i can't figure out what to do. i know you have to find the LCD, which i guess would be 5? but i dunno. and also it's supposed to be \[\frac{ 4 }{ 5 } + \frac{ 3 }{ x }=\frac{ 1 }{ 2 }\]
@whpalmer4 @tkhunny hey guys any of you guys free right now?
Thank you guys so much for viewing. I'm just a little confused on this one and would love some help
Your LCD would also have \(x\) in it... Why not multiply all the way through by \(5*2*x\), which will "clear" all of the fractions?
ohhh ok
In this particular instance, the left side of your equation is defined for all x other than zero. Why not try the actual solution before attempting to answer the question posted in this problem? Follow whpalmer's suggestion.
so 10x?
Please show all your work at one time, not just tidbits.
^ i tried to do that
but my actual work doesnt make any sense
At least, identify what 10x is supposed to represent. Yes, you could multiply all terms of the given equation by 10x, or by 2*5*x. Do this now, please, and share your result.
\[\frac{4}5*(5*2*x) + \frac{3}{x}*(5*2*x) = \frac{1}{2}*(5*2*x)\] what do you get if you simplify that?
10x/2?
where are you getting a /2 from? what does each term simplify to?
let's find the mistake and fix it so you never do that again :-)
ok so you take the 5*2*x which equals 10x
Again, please show your entire result, not just part of it. whpalmer4 seems to be in agreement with me that we need to see what you are doing.
multiply it by the 1/2
and then you get 10x/2
i think
might be easier to just keep it like I wrote it and cancel out matching things in numerator and denominator first
Let's start over. Multiply EVERY term of the given equation by 10x. share ALL of the result.
umm ok lemme just look at it again
I'm going to let my esteemed colleague finish this off, you're in good hands!
ok lol this really doesnt make any sense to me
Explain what doesn't make sense. I need to know where you're coming from. We have an equation, above. I'm asking you to multiply each term of the equation by the LCD, which is 10x. Questions about that?
#1 Thing to do. Worry about the Domain. 1) What is that 'x' doing in the denominator? 2) We better make sure that thing never takes on the value zero (0). 3) \(Domain:\;x \ne 0\) 4) Whew! That was close. Now that we KNOW x cannot be zero, we can proceed without danger.
wait wait i think i got the answer. is it 10? if i write out my work will you guys tell me if im right?
Please show your work. Otherwise, no feedback from me. sorry, but that's the only way in which i could come to undrstand your thinking.
so after simplifying what whpalmer told me, i did this \[\frac{ 10x }{ 5 } + \frac{ 30x }{ x }=\frac{ 10x }{ 2 }\]
And before you share your work, why not test your result yourself? Merely substitute 10 for x and determine whether or not the equation is true.
that all simplifies to \[2x + 30 = 5x\]
Going to solve that equation?
and then you get 10x just by subtracting an ddividing
isnt that solving it?
I'm still worried about the Domain. If we start with this: 3 = 7 Do we believe that to be correct? Let's multiply both sides by zero (0). 3(0) = 7(0) ==> 0 = 0. There you go. 3 is equal to 7. Of course, that is ridiculous. BEFORE you multiply by x, you MUST state that x is NOT zero. It's not optional.
sorry indeed to sound critical, but your "10x just by subtracting an ddividing" is incorrect. subtracting and dividing are NOT related in any way, other than that both are algebraic operations. Please go back to 4/5 + 3/x = 1/2. Substitute x=10. Is the equation now true?
ok ill check
nope its not
also @tkhunny thank you but im still confused
that means, unfortunately, that x=10 may not be a solution. Looking at this same equation yet again, I see that we have a "1/2" on the right side. Any chance that the left side could be made = to 1/2?
could i just have 1 person at a time help me
umm
tk?
no no i want you to help me
but yea i dunno how to make the left side equal a half
like i tried to keep the 3/x on one side and have the 4/5 be on the 1/2 side
then i tried to isolate 3/x
but idk how to do that
Agreed. However, there are two different questions, here. One is the solution of the equation. The other is the answer to George. The second part is easy. Talk about the Domain and say, "Silly George!".
Here's what I see: If you substitute 10 for x in the equation just mentioned, we get 8/10 + 3/10 = 1/2. Not true. However, if the left side were 8/10 - 3/10, what would that come out to?
it'd be a half
so you multiply 4/5 by 2?
where are you getting 3/10 though?
right: 1/2 on the left, 1/2 on the right. But we supposedly got x=10; that's not a solution. We can't just arbitrarily state that "that 10 is negative." We'd have to get -10 algebraically, or it's "no soap."
by assuming that your answer, x=10, is correct.
m8 im confused. i understand where you got 8/10 but what are you doing to the 3/x to make it 3/10?
Then the2nd term is 3/10.
Here's what I'm seeing:\[\frac{ 8 }{ 10 }+\frac{ 3 }{ x }=\frac{ 1 }{ 2 }\]... for only one value of x. What is that value? Hint: It's not 10.
lol
i mean
could you subtract 8/10 and 1/2?
\[\frac{ 1 }{ 10 }+\frac{ 3 }{ x}=\frac{ 1 }{ 2 }\]
that'd get get you 3/10=3/x
We have already shown that if x=10, the above equation is FALSE. What other choice have you for x?
Hint: is 0.8 - 0.3 = to 0.5?
yea
x=10 is incorrect. Choose another value for x to try. Note that we MUST have the same denominator in each fraction.
dude i know that x=10 is incorrect
but i don't even know what you're asking anymore at this point like ??? ok yes, x=10 is NOT the answer but i don't see how that's helping me get the right answer
the thing is, i don't know what value for x to try
Note that \[\frac{ 3 }{ -10 } \]could be re-written so that the denominator is +10.
Try re-writing it.
where are you getting 3/10 from
if you're still using the x=10 thing i really don't wanna use it, it's just confusing me
First, where did the x=10 come from?
holy crap dude i dont know??? i thought that was the wrong answer so we weren't supposed to use that
can you, or SOMEONE, for the love of God, just give me a straight forward answer of how im supposed to solve this algebraically because i've been here for 45 minutes and all that's happened is that i've become even more confused than before
brriiarr wait wait i think i got the answer. is it 10? if i write out my work will you guys tell me if im right? So, that "x=10" came from you.
All right. x=10 is not correct. Try x=-10.
ok lemme plug it in
well you can't do that
cos the 10 and the -10 cancel out
Never said they should cancel out or that they don't cancel out. Try the following: substitute -10 for x in \[\frac{ 4 }{ 5 }+\frac{ 3 }{ x }=\frac{ 1}{ 2}\]
i know lol that's what i did?
Note that this substitution will result in \[\frac{ 4 }{ 5 }-\frac{ 3 }{ 10 }=\frac{ 1 }{ 2 }\]
Is that true or not?
thats what i got on my paper. so then i did the lcd thing and so 4/5 turns into 8/10 and you subtract it and get 5/10
and 5/10 is equal to 1/2, right?
Which means that ... ??
1/2=1/2
so the answer is 1/2?
wait no
Both whpalmer and tkhunny asked that we pay special attention to the domain of this function. I asked the same in another way: what value can x NOT have? and answered it: x cannot be zero. x can be -10, however.
You are not "finding an answer." You are, actually, determining whether or not the equation is true. Is it true? Is it false?
wait so the answer is /-10
*-10
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