Question

(8.05 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? Step 4 Step 3 Step 2 Step 1

Answer 1

step 2 the other steps after are wrong due to step 2.

Answer 2

A: y = 15 - 2z
B: 3y = 3 - 4z

One
-3y = -3 (15 - 2z)
3y = 3 - 4z Nothing was done here

Step 2
No need to go much further. The brackets are incorrectly removed on the right side of Equation A. It should be
-3y = - 45 + 6z

Answer: Step 2