Question

To get system B, the (first or second) equation in system A was replaced by the sum of that equation and the (first or second) equation multiplied by (2, 4, or -3). The solution to system B (is not, or is) the same as the solution to system A?

Answer 1

That's some awkward computer aided education there.

In system B we kept the first equation of system A.  Then apparently we multiplied that equation by 4, giving

4x - 4y = 12

and added that to

-2x + 4y = -2

to give

2x = 10

OK, let's relate that to the boxes.

Answer:

In system B the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 4.  The solution to system B is the same as the solution to system A.

Answer 2

The complete statements are:

• To get system B, the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 4.
• The solution to system B is the same as the solution to system A

The equations in system A are given as:

$x - y=3$

$-2x + 4y = -2$

Multiply the first equation by 4

$4x - 4y = 12$

Add this equation by the second equation in system A

$-2x + 4x +4y - 4y = -2 + 12$

$2x = 10$

So, the equations in system B are:

$x - y=3$

$2x = 10$

Also, the solutions of both systems would remain the same

Read more about systems of equations at:

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