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HELPPP ME STUDY! solve algebraically for a and b and check: 3a+5b=4 4a+3b=-2
You can use Matrix method for solving these two equations simultaneously
i only know substitution, graphing, and elimination method and those are the only methods i can use.
try eliminating either a or b
If you're looking for the point of intersection or the solution to those two then that's helpful if not, sorry, i have no clue what to do next ..
i need to find the value of a and b, not find the solution. like get a or b by itself.
how about let a = 0 or something like that...
Here's the plan; try to "get a in terms of b" (or "b in terms of a"). Really, I mean try to solve for a, so that it equals something that involves b (or solve for b so that it equals something that involves a).
yeah, i know. but how do i do that?
You can look at either equation, and solve for either variable. It's up to you!
It's all fair game in this problem. So pick and equation and pick a variable to solve for!
ok, 3a+5b=4. let's solve for a.
Okay, you start! I'll watch.
Unless you want to workon it together now; we can do that.
yeah, let's work on it together.
Alright. When equations aren't too complex, a great starting point is getting the terms containing variable you're solving for, a, on the opposite side of all the other terms. So look at 3a+5b=4. Which term (containing a) do we want to it's own side?
I agree! And it's with 5b, which we need to get rid of. And it appears you know how to do that! Good! So now a is still stuck with something; it's being multiplied by 3! So can you fix that, too?
\[a= \frac{ 5b+4 }{ 3 }\]
close! You forgot your sign of 5b! You had it right before! I'm guessing you were distracted by Open Study's fancy equation thing, but I don't know.
\[a=\frac{-5b+4}{3}\]Haha, that's fine! So we solved for a in that one equation, 3a+5b=4. a is always a is always (-5b+4)/3. So we can substitute it into another a's place. Try it in the other equation, 4a+3b=-2. the a will be replaced and the only variable in the equation will be b. The we can solve for b!
So, start by substitution!
Make sure you type (-20b+16)/3 + 3b = -2 ! I see you factored correctly, but the way you wrote it it didn't look like it was all divided by 3, though it really was! :)
So\[\frac{-20b+16}{3}+3b = -2\]
Just so that left-most b isn't divided by it.
So\[3(\frac{-20b+16}{3}+3b)=3(-2)\]
Common mistake! Don't worry! You just have to remember to multiply each entire side by 3. That way you're doing the same thing to each side, so that the two new sides will still be equal.
i multiplied by 3 to -20b + 16 so that canceled the 3 out, and then i multiplied the -2 by 3, giving -6. what i do wrong?
You have to multiply 3b by 3 as well!
I'm sure you'll see it soon.
-20b+16+9b=-6 -20b+9b=-22 -11b=-22 22=11b 22/11=b 2=b Yeah!
I had to work it out to check, but we got the same thing.
yay :D thanks eric :D yur awesome :D
Sorry for the wait, Open Study glitched :P Recap, we solved for "a in terms of b" and we solved for b. Now we know b and can now solve for a!
My pleasure! Thank you for doing your part! You did great!
\[a=\frac{-5b+4}{3}=\frac{-5(2)+4}{3}\] and compute... finale!
i just hope i do good on the quiz tomorrow...
My pleasure. Did you get a = 2?
Best of luck! Tag me in any other questions you have, or you can ask them here. But the most important thing in these an dmany other math problems is planning! Knowing what you will do. This technique we used is good for this situation.
Haha!! Caught me! -2 ! :D
Serves me right for trying to do it in my head.
My responses are out of order. Open Stud is scrambled, it seems.
yeah. open study is being really laggy today. if i can just finish up my science homework, ill bring up some more questions to study with. thanks again :D and bye for now.
I'm glad it's not just me! No problem. Good luck with everything!