System of 3 Equations and 3 variables | Mathematics | Page 1

potent solar Dec 4, 2022, 11:02 PM

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This system of equations is keeping me from falling asleep

low spokeBOT Dec 4, 2022, 11:02 PM

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potent solar Dec 4, 2022, 11:03 PM

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I have seriously tried everything i could but i feel like im missing something

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The only logical thing i was able to solve was that b equals c

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Everytime i try to add or substract the equations from each other all i end up with is of no help at all

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I do not know how to approach this problem and the a variable is horrifying

fickle marten Dec 4, 2022, 11:08 PM

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what is after the + sign in each equation

potent solar Dec 4, 2022, 11:08 PM

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1

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I think it's better to just write it down:
(a+1)bc = 12
(b+1)ca = 4
(c+1)ab = 4

fickle marten Dec 4, 2022, 11:10 PM

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okay so

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open all the brackets

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what do you get

potent solar Dec 4, 2022, 11:11 PM

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I get

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abc + bc = 12
abc + ca = 4
abc + ab = 4

fickle marten Dec 4, 2022, 11:12 PM

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now you need to eliminate the abc term

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so you can subtract the second from the first

potent solar Dec 4, 2022, 11:12 PM

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ca - bc = 12 - 4
like this?

fickle marten Dec 4, 2022, 11:13 PM

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the opposite

potent solar Dec 4, 2022, 11:13 PM

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oh

fickle marten Dec 4, 2022, 11:13 PM

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bc - ca = 12 - 4

potent solar Dec 4, 2022, 11:14 PM

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i actually did this before but didn't know what to do after

fickle marten Dec 4, 2022, 11:14 PM

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do the same with the third and the second equations

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second - third

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you get ca = ab

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so that means b = c

potent solar Dec 4, 2022, 11:15 PM

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fickle marten so that means b = c

that was absolutely as far as i got

fickle marten Dec 4, 2022, 11:15 PM

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did you substitute c with b in the first equation?

potent solar Dec 4, 2022, 11:15 PM

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fickle marten bc - ca = 12 - 4

you mean in this one?

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or the original

fickle marten Dec 4, 2022, 11:16 PM

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the original

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and in that one too yes

potent solar Dec 4, 2022, 11:17 PM

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ab² + b² = 12 in the original and the other one is
b² - ab = 8

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i feel like im just thinking about this one blindly

fickle marten Dec 4, 2022, 11:18 PM

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ok one sec

fickle marten Dec 4, 2022, 11:24 PM

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potent solar ab² + b² = 12 in the original and the other one is b² - ab = 8

do you know the answer to this question?

potent solar Dec 4, 2022, 11:24 PM

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no i actually don't

fickle marten Dec 4, 2022, 11:25 PM

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okay so I figured out one

potent solar Dec 4, 2022, 11:25 PM

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oh wow

fickle marten Dec 4, 2022, 11:25 PM

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potent solar ab² + b² = 12 in the original and the other one is b² - ab = 8

substitute b^2 for any variable

potent solar Dec 4, 2022, 11:25 PM

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okay ill let it be X

fickle marten Dec 4, 2022, 11:25 PM

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alright

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so what do you get now

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ax + x = 12
x - ax = 4

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right?

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now if you add them

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what do you get

potent solar Dec 4, 2022, 11:26 PM

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oh....

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0_0

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that's awesome

fickle marten Dec 4, 2022, 11:26 PM

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Did you understand it?

potent solar Dec 4, 2022, 11:26 PM

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we get 2x = 16

fickle marten Dec 4, 2022, 11:26 PM

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no not 2x

potent solar Dec 4, 2022, 11:26 PM

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yeah i did

fickle marten Dec 4, 2022, 11:26 PM

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2ax

potent solar Dec 4, 2022, 11:26 PM

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wait

fickle marten Dec 4, 2022, 11:26 PM

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2ax = 16

potent solar Dec 4, 2022, 11:27 PM

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oh i read it wrong

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hold on

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wait... aren't we supposed to go and find x directly?

fickle marten Dec 4, 2022, 11:27 PM

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no you need to find a and x both

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dont use x it will get you confused

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use p instead

potent solar Dec 4, 2022, 11:28 PM

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okay

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lets use p

fickle marten Dec 4, 2022, 11:28 PM

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wait not 16 one sec

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20*

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cause 12+8 is 20

potent solar Dec 4, 2022, 11:29 PM

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but still, why is it that we don't solve for p which gives us the value of b² directly?

fickle marten Dec 4, 2022, 11:29 PM

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because to get p we also need a

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after we get a we get p

potent solar Dec 4, 2022, 11:29 PM

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its weird i don't get why

fickle marten Dec 4, 2022, 11:30 PM

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Okay so

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ap + p = 12     ...(1)
p - ap = 8      ...(2)

adding both

2ap = 20
ap = 10 ...(3)
or p = 10/a

substituting this in (1)

a×10/a + 10/a = 12
or, 10 + 10/a = 12
or, 10/a = 2
or, a = 10/2 = 5

if a = 5 then according to (3)
p = 2
and since p = b², b² = 2 or b = √ 2

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and as b = c, c = √ 2 as well

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so a = 5, b = sqrt(2) and c = sqrt(2)

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actually wait this doesnt seem right

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it only works for the first one not the second or third

potent solar Dec 4, 2022, 11:36 PM

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wait a second i feel like something is off

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I hate it when it happens

fickle marten Dec 4, 2022, 11:37 PM

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the reason its problematic is because the equations sketch like this

potent solar Dec 4, 2022, 11:37 PM

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actually when we did the substitution i think we didn't see that there is no squaere on the second term in the second equation

fickle marten Dec 4, 2022, 11:37 PM

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are you allowed to use graphing calculators in this?

potent solar Dec 4, 2022, 11:38 PM

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not really, the thing is, this needs to be solved by hand 😢

fickle marten Dec 4, 2022, 11:38 PM

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I see

potent solar Dec 4, 2022, 11:40 PM

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potent solar actually when we did the substitution i think we didn't see that there is no squ...

this is causing a big problem to me rn

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basically we can't just subtitute with X because that one b will prefer to just stand in there

fickle marten Dec 4, 2022, 11:48 PM

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potent solar this is causing a big problem to me rn

what level math is this btw

potent solar Dec 4, 2022, 11:52 PM

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i don't really get the concept of what "math levels" mean since our schools don't work that way 😦

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can you ask that question in another way?

fickle marten Dec 4, 2022, 11:54 PM

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idk then man

potent solar Dec 4, 2022, 11:56 PM

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sorry about this

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thx for your efforts 🙂

fickle marten Dec 4, 2022, 11:56 PM

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yw

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@potent solar ok i got it finally

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yeah so scratch everything we just did

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we found out b = c right?

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so what we do next is substitute c = b in the first equation and the third equation

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so we get

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ab^2 + b^2 = 12
ab^2 + ab = 4

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now what we do is separate a from these two equations

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so

ab^2 + b^2 = 12 ...(1)
ab^2 + ab = 4 ...(2)

now (1)...
ab^2 + b^2 = 12
or, ab^2 = 12 - b^2
or, a = (12-b^2)/b^2 ...(3)

similarly, (2)...
ab^2 + ab = 4
or, a(b^2 + b) = 4
or, a = 4/(b^2 + b) ...(4)

now since (3) and (4) can be equated, we get

4/(b^2+b) = (12-b^2)/b^2

solve the above equation by cross-mulitplying both sides and you will get b = -2 and b = 3

and therefore a = 2 and a = 0.3333, respectively

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and since b = c therefore, c = -2 and 3

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ping me if you dont get it

potent solar Dec 5, 2022, 12:32 AM

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Thanks a lot, I understood the way you approached it, however im having some trouble solving for b after the cross multiplication

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I end up with b^4 + b^3 -8b^2 + 12b = 0

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factoring the b out will give us b( b^3 + b^2 -8b +12) = 0
Thus b^3 + b^2 -8b +12 = 0 (as b can't be 0)

fickle marten Dec 5, 2022, 12:37 AM

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potent solar I end up with b^4 + b^3 -8b^2 + 12b = 0

yeah you would need the help of a calculator in this one

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you can do it manually but i dont remember the method rn

potent solar Dec 5, 2022, 12:38 AM

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yeah seems like a calculator would be the only hope for me rn

fickle marten Dec 5, 2022, 12:40 AM

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potent solar yeah seems like a calculator would be the only hope for me rn

you can close this by .close

potent solar Dec 5, 2022, 12:41 AM

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okay

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thanks a lot again

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.close

#System of 3 Equations and 3 variables