(08.05 HC) The work of a student to solve a set of equations is shown: Equation A: y = 15 βˆ’ 2z Equation B: 3y = 3 βˆ’ 4z Step 1: βˆ’3(y) = βˆ’3(15 βˆ’ 2z) [Equation A is multiplied by βˆ’3.] 3y = 3 βˆ’ 4z [Equation B] Step 2: βˆ’3y = 15 βˆ’ 2z [Equation A in Step 1 is simplified.] 3y = 3 βˆ’ 4z [Equation B] Step 3: 0 = 18 βˆ’ 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?

Respuesta :

The error appears to be in Step 2.

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

Answer:

STEP 2

Step-by-step explanation:Simplifying

Β -3(y) = -3(15 -2z)

should yield

Β -3y = -45 +6z