Question

Can someone please help me, this is my last question I have been stuck on it for 30 min

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

The mistake was made in step 2. In simplifying the equation, he has to distribute the -2 through the parentheses. Step 2 simplification should be

-2y = -30 + 4z

When the 2 equations are added, the resultant equation is 0 = -27 therefore it has no solution.

Answer 2

step 2
he didn't distribute
what he did:
-2(15-2z)=15-2z
that is obviously false

what should have happened is
-2(15-2z)=-30+4z
then add

-2y=-30+4z
2y=4-4z
0=-26
false
no solution
inconsitient, these lines ar paralell


anyway, step 2 is error