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

Question

Mathematics

Bryant Wilkins

23 April, 03:36

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

+1

Answers (1)

Know the Answer?

Gage Raymond · Answer 1

Step 1. Equation B could not have simplified out to 3y = 3 - 4z because Equation A's original equation was y = 15 - 2z. That means after the multiplication by - 3, - 3 (y) = - 3 (15 - 2z) it would equal - 3y = - 45 + 6z.