Question

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?


Step 2
Step1
Step 4
Step 3

Answer 1

I will give medals!

Answer 2

Any ideas?

Answer 3

Recall the distributive property of multiplication:
a(b+c)=ab+ac.

Answer 4

Im pretty sure its step 2

Answer 5

Because there he used subtraction instead of division I need to confirm that

Answer 6

Someone Please help!

Answer 7

y = 15 – 2z

–2y = –2•( 15 – 2z )

Answer 8

So im i right with the step 2?

Answer 9

You're right! It is step 2 :-)

Answer 10

okay thank you can u help me with one more?

Answer 11

Variable c is 5 more than variable a. Variable c is also 3 less than variable a. Which pair of equations best models the relationship between c and a?


a = c + 5
a = 3c –3
c = a – 5
c = a + 3
a = c – 5
a = 3c + 3
c = a + 5
c = a – 3

Answer 12

Im pretty sure its the first one

Answer 13

can u confirm?

Answer 14

The -2 was NOT distributed correctly.

Answer 15

okay can u confirm the 2nd one please?

Answer 16

The question asks for which pair.

Answer 17

That means two equations, right?

Answer 18

Variable c is 5 more than variable a:

c = 5 + a

Answer 19

Variable c is 3 less, than variable a:

c = a – 3

Answer 20

You should email your teacher and tell them that there is no relationship between a and c because the equations have no solution.

Answer 21

5+a=a-3 gives 5=-3 which is not true