Answer two questions about Systems AAA and BBB: System AAA \text{\quad}start text, end text System BBB
\begin{cases}6x-5y=1-2x+2y=-1\end{c ases}
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6x−5y=1
−2x+2y=−1

\begin{cases}4x-3y=0-2x+2y=-1\end{c ases}
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4x−3y=0
−2x+2y=−1


1) How can we get System BBB from System AAA?
Choose 1
Choose 1

(Choice A)
A
Replace one equation with the sum/difference of both equations

(Choice B)
B
Replace only the left-hand side of one equation with the sum/difference of the left-hand sides of both equations

(Choice C)
C
Replace one equation with a multiple of itself

(Choice D)
D
Replace one equation with a multiple of the other equation
2) Based on the previous answer, are the systems equivalent? In other words, do they have the same solution?
Choose 1
Choose 1

(Choice A)
A
Yes

(Choice B)
B
No
from khan

1) In order to get the systems B, we will replace one equation with the sum/difference of both equations

2) The solutions of both systems are different, hence the systems are not equivalent.

Simultaneous equations are equations that involve two or more unknown variables. These equations can either be linear or quadratic in nature

Given the systems of linear equations

6x−5y=1 1

−2x+2y=−1 2

In order to get the systems B as shown. First;

equation 1 + equation 2 (System A)

(6x)+(-2x)+ (-5y+2y) = 1 + (-1)

6x - 2x + (-5y+2y) = 1-1

4x + (-3y) = 0

4x - 3y = 0

As you can see, this equation is the result of equation 1 of system B.

Hence in order to get the systems B, we will replace one equation with the sum/difference of both equations

2) Solving both systems to check if they are equivalent.

For system A:

Multiply equation 1 by 2 and equation 2 by 6

Add the result

-10y + 12y = 2 -12

2y = -10

y = -10/2

y = -5

Substitute y = 5 into (1)

From 1: 6x−5y=1

6x - 5(-5) = 1

6x + 25 = 1

6x = 1-25

6x = -24

x = -24/6

x = -4

Hence the solution to systems A is (-4, -5)

For system B:

Multiply equation 1 by 2 and equation 2 by 4

Add the result

-6y + 8y = 2 -4

2y = -2

y = -2/2

y = -1

Substitute y = -1 into (1)

From 1: 4x−3y=1

4x - 3(-1) = 1

4x + 3 = 1

4x = 1-3

4x = -2

x = -2/4

x = -1/2

Hence the solution to systems B is (-1/2, -1)

Since the solutions of both systems are different, hence the systems are not equaivalent.

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