(x+1)/(x-1)=-4/(x+3)+8/(x^2+2x-3)
What is the assignment question?
That's it, I have to solve for x out of that
That's what i was looking for the "solve for x" part
Oh, sorry! Would you know how to explain it for me?
Let me see how i can break it down first
Thanks!
Before i start, i wanted to know if this is exactly what you typed up there? \[\frac{ (x+1) }{ (x-1) }=\frac{ -4 }{ (x+3) }+\frac{ 8 }{ (x^2+2x-3) }\]
Yes, that's it
Ok first thing is first look at the denominator and you see (x-1), (x+3) and x^2+2x-3 well all the fractional expressions must have the same denominator x^2+2x-3 x^2+2x-3 is broken down to (x-1)(x+3) you understand this part so far?
yes, I just don't know what to do with the numerators in order to begin solving for x
I'm getting there, i just want you too understand it completely so you aren't lost within my explanation
Now you have to multiply the numerator and denominator of the fractional expressions that aren't in x^2+2x-3 or (x+3)(x-1) form. So on the left hand side you multiply the top and bottom by (x+3) and on the right hand side you multiply the top and bottom by (x-1), but only for the 1st expression the other already has (x+3)(x-1) so you leave it blank This is what it would now look like: \[(x+3)(x+1)=-4(x-1)+8\]
\[(x+3)(x+1)=-4(x-1)+8 \rightarrow x^2+4x+3=-4x+4+8\]
I will simplify it until the expression is set equal to zero \[x^2+4x+3+4x-12=0 \rightarrow x^2+8x-9=0\] Now that you have anew polynomial you have too factor it can you factor it?
before I do, I have gotten down to x^2 +8x-9=0 is that correct?
yes!
(x+9)(x-1)=0 ?
yes!!! now get the two solutions of x
they are -9 & 1 , but when I put them into the system, they tell me that there is only one answer and it is non extraneous
yes that is correct. You plug in the solutions of x into the equation separately and see one that comes out with a non-extraneous answer
Oh, okay! Ill do that now. Sorry if im holding you up
No don't worry, i am relaxing from school
It would be the -9, right?
sqrt{t - 62} - sqrt{t+160} = 140 would you also be able to help me solve this for t?
Yes it yes it is!!! Sure no problem
no solution exists for this problem You can tell that is true by squaring both sides and you know what happens when you square a square root, right?
right, and I also know that it is extraneous.. but im not sure what to put into the system. I tried no solution and it said that it wasn't acccepted
What program or website are you trying to put the answer into?
webwork
if it's not too much to ask can you take a screen shot and post it on here?
Im not able to because im on my universitys desktop, sorry. I have four other problems, would you like me to skip it and if you don't mind you could help me with a different one?
there should probably be a DNE (does not exist) selection as an answer but idk how many attempts you are allowed so let's go on too the next one
(10/8-t)+(2/8+t)+(7/64+t^2)=0
Are there possible answer sets to this problem because this one is a bit tricky? i want to see if i am right or wrong?
no, the only other thing on the screen is solve for t
i'm not sure but try t = -11.684
it says incorrcect
this problem is very long too answer step by step
is there any other possible answer for it?
the others are complex or imaginary numbers.
have you worked with imaginary numbers in your class?
yes
ok put in \[\pm0.2793-8.3933i\]
it still says incorrect
how about sqrt(4x+5) +7=3x
it says find all possible solutions.
hmm idk why it's not working but usually online problems should have example problems for each problem you do so you can check how they want the answer to be inputted ok hold on
x should equal \[\frac{ 1 }{ 9 }(23+\sqrt{133}\] or x equals 3.8370
both answers are the same but in different format
but it asks for the highest and lowest possible solution, and then asks if it is a real solution..
ahh ignore that i did a mistake somewhere the values should be 2.6818, -2.2373
That should definitely be correct
are they both solutions?
both should be solutions but i didn't plug them back in the original equation
did you get them as fractions?
Are they asking for the solutions in fraction form?
I think so because it isn't taking the decimals that you gave me
oh never mind, I got it to work sorry. Are you still up for another?
none of them are solutions to the expression
it's all good and yes i am
sqrt (10-x) +x = -2
it says solve for x, it could be an integer or fraction
if x needs to solve for the equation and make it true then x = -6
that's it!! 1/abs(12-4x) = 10
x = \[\frac{ 119 }{ 40 }\] x = \[\frac{ 121 }{ 40 }\]
Thankyou! You just helped me SOO much!!!!!!
No problem at all!
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