Question

solve the system of equations $x - 3y = 0$ and $3x - 6y = 9$ by combining the equations.
$(x - 3y = 0)$
$(3x - 6y = 9)$
\

$$\begin{align*} x - 3y &= 0 \\\\ 3x - 6y &= 9 \\\\ \\hline 0x + 0y &= \\end{align*}$$

answer attempt 1 out of 2
you must answer all questions above in order to submit.

Explanation:

Step1: Multiply first equation

To eliminate \(x\) or \(y\), multiply the first equation \(x - 3y = 0\) by \(-3\) to get \(-3x + 9y = 0\).

Step2: Add to second equation

Now add this new equation \(-3x + 9y = 0\) to the second equation \(3x - 6y = 9\).
\((-3x + 9y)+(3x - 6y)=0 + 9\)
Simplify: \((-3x+3x)+(9y - 6y)=9\), so \(3y = 9\), but wait, the problem's format is about combining the original equations as given (maybe using elimination by making coefficients of \(x\) or \(y\) opposites). Wait, original equations: first is \(x - 3y = 0\), second is \(3x - 6y = 9\). Let's try to eliminate \(x\). Multiply first equation by \(-3\): \(-3x + 9y = 0\). Now add to second equation: \((-3x + 9y)+(3x - 6y)=0 + 9\) → \(3y = 9\), but the problem's box is for the combination result. Wait, maybe the problem expects to multiply the first equation by -2? Wait, first equation: \(x - 3y = 0\) (multiply by -2: \(-2x + 6y = 0\)), second equation: \(3x - 6y = 9\). Now add: \((-2x + 6y)+(3x - 6y)=0 + 9\) → \(x = 9\). Wait, maybe I misread. The given format is:
\(\begin{align} x - 3y &= 0\\ 3x - 6y &= 9\\ \hline 0x + 0y &= \text{?} \end{align}\)
No, that can't be. Wait, maybe the user made a typo, but actually, to solve, let's do elimination properly. From first equation, \(x = 3y\). Substitute into second equation: \(3(3y)-6y = 9\) → \(9y - 6y = 9\) → \(3y = 9\) → \(y = 3\), then \(x = 9\). But the problem's combination step: let's see, if we multiply first equation by -3: \(-3x + 9y = 0\), add to second equation \(3x - 6y = 9\): \((-3x + 9y)+(3x - 6y)=0 + 9\) → \(3y = 9\), so \(y = 3\), then \(x = 9\). But the box is for the sum of the left sides and right sides. Wait, maybe the problem's intended operation is to multiply the first equation by -3, then add to the second:
Left side: \(-3(x - 3y)+(3x - 6y)= -3x + 9y + 3x - 6y = 3y\)
Right side: \(-3(0)+9 = 9\)
But the problem's format has \(0x + 0y=\), which is wrong. Wait, maybe the user's problem has a different approach. Wait, original equations:
1. \(x - 3y = 0\)
2. \(3x - 6y = 9\)

Let's solve for \(x\) from first equation: \(x = 3y\). Substitute into second equation: \(3(3y)-6y = 9\) → \(9y - 6y = 9\) → \(3y = 9\) → \(y = 3\), then \(x = 9\). But the combination step in the given format: maybe the problem expects to write the result of adding the two equations after multiplying the first by -3: \(-3x + 9y = 0\) + \(3x - 6y = 9\) gives \(3y = 9\), so \(y = 3\), then \(x = 9\). But the box is for the sum of the left sides equaling the sum of the right sides. Wait, maybe I messed up. Let's do it as per the problem's format:

First equation: \(x - 3y = 0\)

Second equation: \(3x - 6y = 9\)

Let's multiply the first equation by -3: \(-3x + 9y = 0\)

Now add to second equation:

\((-3x + 9y)+(3x - 6y)=0 + 9\)

Simplify: \(0x + 3y = 9\)

Ah, maybe the problem's format had a typo, and the coefficients of \(x\) and \(y\) are not both zero. So the correct combination (after multiplying first by -3) is \(0x + 3y = 9\), but the problem's box is \(0x + 0y=\), which is incorrect. Wait, maybe the user intended to multiply the first equation by -2:

First equation: \(x - 3y = 0\) (multiply by -2: \(-2x + 6y = 0\))

Second equation: \(3x - 6y = 9\)

Add: \((-2x + 6y)+(3x - 6y)=0 + 9\) → \(x = 9\), so \(1x + 0y = 9\), but the problem's format is \(0x + 0y=\). I think there's a mistake, but assuming we follow the elimination to find the value, but the problem's answer for the combination (maybe the sum of the equations as is, but that would be \(4x - 9y = 9\), which is not helpful). Wait, no, the correct way is:

From \(x - 3y = 0\) → \(x = 3y\)

Substitute into \(3x - 6y = 9\):

\(3(3y) - 6y = 9\)

\(9y - 6y = 9\)

\(3y = 9\)

\(y = 3\)

Then \(x = 3*3 = 9\)

So the solution is \(x = 9\), \(y = 3\). But the problem's box is for the combination result. Maybe the problem expects to write the result of adding the two equations after multiplying the first by -3, which is \(3y = 9\), so \(y = 3\), then \(x = 9\). So the final answer for the system is \(x = 9\), \(y = 3\).

Answer:

The solution to the system is \(x = 9\) and \(y = 3\), so when combining the equations (after appropriate multiplication), we find \(x = 9\) and \(y = 3\). If we strictly follow the combination step in the given format (correcting the coefficient error), the result of combining (after multiplying first equation by -3 and adding to second) is \(3y = 9\) (or \(x = 9\) if multiplying first by -2), but the correct solution is \(x = 9\), \(y = 3\).