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

equation a: y = 4 - 2z

equation b: 6y = 2 - 4z

step 1: - 6 (y) = - 6 (4 - 2z) [equation a is multiplied by - 6.]

6y = 2 - 4z [equation b]

step 2: 6y = 4 - 2z [equation a in step 1 is simplified.]

6y = 2 - 4z [equation b]

step 3: 0 = 6 - 6z [equations in step 2 are added.]

step 4: 6z = 6

step 5: z = 1

in which step did the student first make an error?

step 4

step 3

step 2

step 1

Question

Dominic Booker

Mathematics

24 July, 07:14

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

equation a: y = 4 - 2z

equation b: 6y = 2 - 4z

step 1: - 6 (y) = - 6 (4 - 2z) [equation a is multiplied by - 6.]

6y = 2 - 4z [equation b]

step 2: 6y = 4 - 2z [equation a in step 1 is simplified.]

6y = 2 - 4z [equation b]

step 3: 0 = 6 - 6z [equations in step 2 are added.]

step 4: 6z = 6

step 5: z = 1

in which step did the student first make an error?

step 4

step 3

step 2

step 1

+4

Answers (1)

Know the Answer?

Willie Bryan · Answer 1

That would be step 2.

step 1 is correct ... multiplying the first equation by - 6 which gives you:

-6y = - 24 + 12z

but in step 2, the simplifying would be : y = 4 - 2z