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

- Spring98

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- Spring98

@satellite73

- pooja195

Try solving it yourself then see where he made the mistake.

- pooja195

HINT: start with substitution

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## More answers

- Spring98

i did! is step one wrong?

- pooja195

Yep xD

- pooja195

What did he do wrong?

- Spring98

so step one is the right answer?

- pooja195

Well yeah but what was his mistake?

- Spring98

he put a -3 instead of a 3

- pooja195

yea but one thing i dont understand why is there another y?

- Hero

Are you sure Step 1 is wrong?

- pooja195

\[\huge~–3(y) = –3(15 – 2z)\]
it says y equals... 15-2z...

- Hero

Yes but we're allowed to multiply both sides of an equation by -3

- pooja195

It says y=15-2z
the person never plugged it in place of y .

- 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

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

Right but it looks like they used substituion.... :/ This is typed out weirdly .-.

- Spring98

so witch one would it be?

- Spring98

i'm confused now!

- Spring98

plz explain this to me!

- Spring98

anyone?

- 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

@Hero is the answer step 4

- 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

i don't see anything wrong with step 2

- 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

im not sure

- 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

no

- Hero

Why not?

- Spring98

becuase its -3y and the other one is plain y

- Hero

So what does this observation tell you?

- Spring98

that step 2 is wrong. so step two is the answer

- Hero

Bingo.

- Spring98

thanks for explaning.

- Hero

You're most welcome.

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