Spring98
  • Spring98
The work of a student to solve a set of equations is shown: Equation A: y = 15 – 2z Equation B: 3y = 3 – 4z Step 1: –3(y) = –3(15 – 2z) [Equation A is multiplied by –3.] 3y = 3 – 4z [Equation B] Step 2: –3y = 15 – 2z [Equation A in Step 1 is simplified.] 3y = 3 – 4z [Equation B] Step 3: 0 = 18 – 6z [Equations in Step 2 are added.] Step 4: 6z = 18 Step 5: z = 3 In which step did the student first make an error? Step 4 Step 3 Step 2 Step 1
Mathematics
katieb
  • katieb
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Spring98
  • Spring98
pooja195
  • pooja195
Try solving it yourself then see where he made the mistake.
pooja195
  • pooja195
HINT: start with substitution

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Spring98
  • Spring98
i did! is step one wrong?
pooja195
  • pooja195
Yep xD
pooja195
  • pooja195
What did he do wrong?
Spring98
  • Spring98
so step one is the right answer?
pooja195
  • pooja195
Well yeah but what was his mistake?
Spring98
  • Spring98
he put a -3 instead of a 3
pooja195
  • pooja195
yea but one thing i dont understand why is there another y?
Hero
  • Hero
Are you sure Step 1 is wrong?
pooja195
  • pooja195
\[\huge~–3(y) = –3(15 – 2z)\] it says y equals... 15-2z...
Hero
  • Hero
Yes but we're allowed to multiply both sides of an equation by -3
pooja195
  • pooja195
It says y=15-2z the person never plugged it in place of y .
Hero
  • Hero
The problem says the student is working to solve the system of equations. There's more than one way to solve a system.
Hero
  • Hero
Multiplying both sides of an equation by -3 does not count as an "error". Nothing "wrong" was done in that step. The student approached differently than you would have.
pooja195
  • pooja195
Right but it looks like they used substituion.... :/ This is typed out weirdly .-.
Spring98
  • Spring98
so witch one would it be?
Spring98
  • Spring98
i'm confused now!
Spring98
  • Spring98
plz explain this to me!
Spring98
  • Spring98
anyone?
Nnesha
  • Nnesha
y = 15 – 2z 3y = 3 – 4z how would u find the solution ? you need same coefficient but opposite sign to isolate for one variable that's why multiply 1st equation by -3 so -3(y = 15 -2z) = ???
Spring98
  • Spring98
@Hero is the answer step 4
Hero
  • Hero
Guessing is not the best approach to this. How'd you make it through the other steps successfully? The truth is you wouldn't even really need to ask which step is correct, if you had the correct approach. But I understand you're asking for help. So we know step 1 is correct. What are your thoughts on step 2?
Spring98
  • Spring98
i don't see anything wrong with step 2
Hero
  • Hero
So the equations in step 1 are equivalent to the equations in step 2? Because when solving systems, for each step, the equations have to all be equivalent.
Spring98
  • Spring98
im not sure
Hero
  • Hero
In step 1, for the first equation, after multiplying both sides by -3, we have: –3(y) = –3(15 – 2z) In step 2, that equation was supposedly simplified to get: –3(y) = 15 – 2z Tell me, is –3(y) = 15 – 2z equivalent to y = 15 – 2z ?
Spring98
  • Spring98
no
Hero
  • Hero
Why not?
Spring98
  • Spring98
becuase its -3y and the other one is plain y
Hero
  • Hero
So what does this observation tell you?
Spring98
  • Spring98
that step 2 is wrong. so step two is the answer
Hero
  • Hero
Bingo.
Spring98
  • Spring98
thanks for explaning.
Hero
  • Hero
You're most welcome.

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