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Mathematics 19 Online
OpenStudy (anonymous):

..

OpenStudy (anonymous):

@priyar Lend a quick hand?!

OpenStudy (priyar):

y not normally by substitution ?

OpenStudy (anonymous):

Thats what they are asking for in my book

OpenStudy (priyar):

but the Q says.."Solve this system by substitution."

OpenStudy (priyar):

okay then just do substitution..as u do usually

OpenStudy (anonymous):

Yes, Im saying thats what they want. Substitution. Im better at elimination.

OpenStudy (priyar):

find r from first eq and substitute in the second..

OpenStudy (anonymous):

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.

OpenStudy (whpalmer4):

take the first equation \[r+s=-12\]solve it for \(r\) in terms of \(s\), what do you get?

OpenStudy (priyar):

subtract s on both sides..in first eq so that u have only r on the left side..

OpenStudy (anonymous):

ok ill show you what I have so far, then just pick up after me..

OpenStudy (priyar):

this is the first step for any substitution problem..(u must have only 1 variable on left side...)

OpenStudy (anonymous):

r+s=-12 r+s-s=-12-s r=-12-s

OpenStudy (whpalmer4):

good. now in the second equation\[4r-6s=12\]replace \(r\) everywhere with \((-12-s)\)

OpenStudy (whpalmer4):

that gives you an equation just in terms of \(r\) which you should be able to easily solve.

OpenStudy (priyar):

so we are making the second eq. to have just "one variable"

OpenStudy (whpalmer4):

then just find \(s\) from your value for \(r\) using your substitution equation\[r=-12-s\]

OpenStudy (whpalmer4):

sorry! I switched the names. you solve for \(s\) then get the value of \(r\)

OpenStudy (priyar):

its like "erasing " r and writing -12-s in its place

OpenStudy (priyar):

so what do u get?

OpenStudy (anonymous):

4r-6s=12 4r-6s-4r=12-4r 6s=12-4r is that correct? for the second equation?

OpenStudy (priyar):

u didn't replace r with "-12-s"..in the second eq..

OpenStudy (anonymous):

I thought we had to do it like I did above before I substituted

OpenStudy (priyar):

ok..@LetsLearn2000 but still u hav emade a mistake in sign..

OpenStudy (anonymous):

Why'd you delete your comment? @whpalmer4

OpenStudy (priyar):

its -6s=12-4r ok?

OpenStudy (anonymous):

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

OpenStudy (priyar):

how did -6s become 6s??

OpenStudy (priyar):

u have made a mistake in last step..

OpenStudy (whpalmer4):

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 \(s\)

OpenStudy (priyar):

@LetsLearn2000 did u understand wht was ur mistake..?

OpenStudy (anonymous):

Yes @priyar Im about to solve @whpalmer4

OpenStudy (priyar):

ok then proceed!

OpenStudy (anonymous):

I got -2 ? @whpalmer4

OpenStudy (whpalmer4):

can you show me the steps?

OpenStudy (anonymous):

Can I just tell you what I did? I wrote it on a piece of paper and it'll take forever to type

OpenStudy (whpalmer4):

well, OK, you made a mistake somewhere, just hoping to find it

OpenStudy (priyar):

did u proceed in ur way or whpalmer4's?

OpenStudy (priyar):

so that i can trace ur steps..

OpenStudy (anonymous):

Im totally confused now.. Whichever way is the substitution way lol I believe thats @whpalmer4 's way

OpenStudy (whpalmer4):

\[4(-12-s)-6s=12\]\[4(-12)+4(-s)-6s=12\]\[-48-4s-6s=12\]agree, disagree?

OpenStudy (anonymous):

Here. I just simplified on a calc so You can see the steps instead of me typing

OpenStudy (whpalmer4):

but you used wrong formula. \[r=-12-s\]NOT\[r=-12-6s\]

OpenStudy (whpalmer4):

I made same typo earlier, and deleted post containing it...

OpenStudy (priyar):

ya..exactly..

OpenStudy (anonymous):

Ok I see, its -6 now correct?

OpenStudy (priyar):

yes!

OpenStudy (whpalmer4):

yes. so value of other variable is what?

OpenStudy (anonymous):

-6 lol

OpenStudy (whpalmer4):

right. and that's the same answer you would get from elimination, of course

OpenStudy (whpalmer4):

\[r+s=-12\]\[4r-6s=12\]multiply first eq. by \(6\) \[6r+6s=-72\]\[4r-6s=12\]add together to get\[6r+4r +6s-6s=-72+12\]\[10r=-60\]etc.

OpenStudy (whpalmer4):

important step for either substitution or elimination: plug answers back into both equations and verify they both work!

OpenStudy (anonymous):

ok, Thanks I appreciate it!

OpenStudy (whpalmer4):

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...

OpenStudy (whpalmer4):

anyhow, you can solve with either approach, useful to know both as one is often a bit easier than the other.

OpenStudy (anonymous):

Yes. thanks once again, I may be back on in a bit, you'll be on?

OpenStudy (whpalmer4):

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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