Question

The work of a student to solve a set of equations is shown:

Equation A: y = 15 − 2z
Equation B: 2y = 3 − 4z


Step 1: −2(y) = −2(15 − 2z) [Equation A is multiplied by −2.]
2y = 3 − 4z [Equation B]
Step 2: −2y = 15 − 2z [Equation A in Step 1 is simplified.]
2y = 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?
Step1
Step 2
Step 3
Step 4

Answer 1

@TheSmartOne

Answer 2

@AriPotta

Answer 3

@campbell_st

Answer 4

@563blackghost

Answer 5

You're still taking the test?

Answer 6

this is a different one

Answer 7

08.09 Module Eight Test Part One

Answer 8

@TheSmartOne

Answer 9

I'd say the teacher made the error in asking for a solution to a set of equations that have no solution...

and then the next problem is asking to use elimination...

the 2 lines are parallel and have no solution...

y = 15 - 2z B: divide by 2 y = 3/2 - 2z

I would have stopped there and said there is no solution.

Answer 10

I'd say the error by the student was attempting this inane question.

Answer 11

so step 1 is wrong ?

Answer 12

the error is in the multiplication of equation A by -2

-2 * y = -2 * 15 - (-2)*2z

it should be

-2y = -30 + 4z

Answer 13

okay