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dusya [7]
4 years ago
13

Is my answer correct?

Mathematics
1 answer:
Black_prince [1.1K]4 years ago
4 0

No, we do not multiply the first equation by 2 and then add it to the second equation. We have to multiply the first equation by 4.

To illustrate this point, consider system A:

\left\{ \begin{aligned}x - y &= 3 \\-2x + 4y &= -2\end{aligned}\right.

If we multiply the first equation by 2, we get

2x - 2y = 6

Now, if we replace the second equation with the sum of the second equation and 2x - 2y = 6, we get

\left\{ \begin{aligned}x - y &= 3 \\(-2x+2x) + (4y - 2y) &= -2 + 6\end{aligned}\right.

which simplifies to

\left\{ \begin{aligned}x - y &= 3 \\2y &= 4\end{aligned}\right.

This is not an equivalent system to System B. We can see that we ended up with a 2y = 4 equation.

In order to end up with a 2x = 10 second equation, we have to multiply the first equation of system A by 4 to get

4x - 4y = 12

If we replace the second equation with the sum of the second equation and 4x - 4y = 12, we get

\left\{ \begin{aligned}x - y &= 3 \\(-2x+4x) + (4y-4y) &= -2 + 12\end{aligned}\right.

which simplifies to

\left\{ \begin{aligned}x - y &= 3 \\2x &= 10\end{aligned}\right.

Otherwise, you are correct. The solution to system B is the solution to system A. Adding an equation to another does not change the system.

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