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@priyar Lend a quick hand?!
y not normally by substitution ?
Thats what they are asking for in my book
but the Q says.."Solve this system by substitution."
okay then just do substitution..as u do usually
Yes, Im saying thats what they want. Substitution. Im better at elimination.
find r from first eq and substitute in the second..
Take me through the steps of substitution for this problem if you don't mind? I think I have an idea of how to do it, but I can't afford to get this wrong.
take the first equation \[r+s=-12\]solve it for \(r\) in terms of \(s\), what do you get?
subtract s on both sides..in first eq so that u have only r on the left side..
ok ill show you what I have so far, then just pick up after me..
this is the first step for any substitution problem..(u must have only 1 variable on left side...)
r+s=-12 r+s-s=-12-s r=-12-s
good. now in the second equation\[4r-6s=12\]replace \(r\) everywhere with \((-12-s)\)
that gives you an equation just in terms of \(r\) which you should be able to easily solve.
so we are making the second eq. to have just "one variable"
then just find \(s\) from your value for \(r\) using your substitution equation\[r=-12-s\]
sorry! I switched the names. you solve for \(s\) then get the value of \(r\)
its like "erasing " r and writing -12-s in its place
so what do u get?
4r-6s=12 4r-6s-4r=12-4r 6s=12-4r is that correct? for the second equation?
u didn't replace r with "-12-s"..in the second eq..
I thought we had to do it like I did above before I substituted
ok..@LetsLearn2000 but still u hav emade a mistake in sign..
Why'd you delete your comment? @whpalmer4
its -6s=12-4r ok?
4r-6s=12 4r-6s-4r=12-4r 6s=12-4r yeah thats what I did but you guys are saying I made a mistake
how did -6s become 6s??
u have made a mistake in last step..
I had a typo in it \[r+s=-12\]\[r=-12-s\]that is the substitution equation second equation is \[4r-6s=12\]replace \(r\) with \((-12-s)\) to get\[4(-12-s)-6s=12\] simplify and solve for value of \(s\)
@LetsLearn2000 did u understand wht was ur mistake..?
Yes @priyar Im about to solve @whpalmer4
ok then proceed!
I got -2 ? @whpalmer4
can you show me the steps?
Can I just tell you what I did? I wrote it on a piece of paper and it'll take forever to type
well, OK, you made a mistake somewhere, just hoping to find it
did u proceed in ur way or whpalmer4's?
so that i can trace ur steps..
Im totally confused now.. Whichever way is the substitution way lol I believe thats @whpalmer4 's way
\[4(-12-s)-6s=12\]\[4(-12)+4(-s)-6s=12\]\[-48-4s-6s=12\]agree, disagree?
Here. I just simplified on a calc so You can see the steps instead of me typing
but you used wrong formula. \[r=-12-s\]NOT\[r=-12-6s\]
I made same typo earlier, and deleted post containing it...
ya..exactly..
Ok I see, its -6 now correct?
yes!
yes. so value of other variable is what?
-6 lol
right. and that's the same answer you would get from elimination, of course
\[r+s=-12\]\[4r-6s=12\]multiply first eq. by \(6\) \[6r+6s=-72\]\[4r-6s=12\]add together to get\[6r+4r +6s-6s=-72+12\]\[10r=-60\]etc.
important step for either substitution or elimination: plug answers back into both equations and verify they both work!
ok, Thanks I appreciate it!
easy to come up with values that satisfy one equation but not the other. for example, you had \(s=-2\) earlier, and that would give you \(r=-10\) as other half of solution. \[r+s=-12\]\[(-10)+(-2)=-12\]\[-12=-12\]looks good so far, right? \[4r-6s=12\]\[4(-10)-6(-2)=12\]\[-40+12=12\]\[-28=12\]uh, nope...
anyhow, you can solve with either approach, useful to know both as one is often a bit easier than the other.
Yes. thanks once again, I may be back on in a bit, you'll be on?
if you tag me on a question, I will have a look when able. if someone else helps you first, I will check it and add any further points I think useful.
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