Question

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

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.

Answer 2

Answer:

Step 2

Step-by-step explanation:

Where did the -3 from the first equation get simplified to?

Step 1: –3(y) = –3(15 – 2z)

Step 2: 3y = 15 – 2z