Question

(08.05 HC) 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

The error appears to be in Step 2.

Simplifying
  -3(y) = -3(15 -2z)
should yield
  -3y = -45 +6z
not
  -3y = 15 -2z

Answer 2

Answer:

STEP 2

Step-by-step explanation:Simplifying

 -3(y) = -3(15 -2z)

should yield

 -3y = -45 +6z