Question

Question 6(Multiple Choice Worth 5 points) (8.05 HC) The work of a student to solve a set of equations is shown: Equation A: y = 15 − 2z Equation B: 2y = 3 − 4z Step 1: −2(y) = −2(15 − 2z) [Equation A is multiplied by −2.] 2y = 3 − 4z [Equation B] Step 2: −2y = 15 − 2z [Equation A in Step 1 is simplified.] 2y = 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? Step1 Step 2 Step 3 Step 4

Answer 1

second step

Step-by-step explanation:

because he did not distribute right.

Answer 2

• The studend first made an error in the second setp, since he/she did not apply the distributive property correctly.

Step-by-step explanation:

Here, I copy each step given and review them algebraically, until finding the first error made by the student.

0) Start: System of equations:

• Equation A: y = 15 − 2z

• Equation B: 2y = 3 − 4z

1) Step 1:

−2(y) = −2(15 − 2z) [Equation A is multiplied by −2.]

2y = 3 − 4z [Equation B]

This procedure is algebraically correct, since the student used correctly the multiplicative property of equalities: both sides of the equation A are multiplied by the same number, 2.

Whereas, the equation B is not algebraically manipulated.

2) Step 2:

• −2y = 15 − 2z [Equation A in Step 1 is simplified.]

• 2y = 3 − 4z [Equation B]

This procedure is algebraically wrong, since the student applied wrongly the distributive property of multiplication to simplify the equation A:

• −2(y) = −2(15 − 2z)

• Wrong: -2y = 15 - 2z,

• The error is that the student did not multiply - 2 times 15 and -2 times - 2z

• The correct procedure had been:

-2y = -2(15) - 2( - 2z)

-2y = - 30 + 4z

Conclusion: The studend made an error in the second setp, since he/she did not apply the distributive property correctly.