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

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

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.

The answer to this is pi/3