To find a second equation that has an infinite number of solutions with the equation \( 6x + 3y = 9 \), we first rewrite this equation in slope-intercept form \( y = mx + b \).
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Start with the original equation: \[ 6x + 3y = 9 \]
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Solve for \( y \): \[ 3y = 9 - 6x \] \[ y = -2x + 3 \]
Now to find a second equation that has the same slope and y-intercept, you can multiply the entire equation \( y = -2x + 3 \) by any non-zero constant. For example, multiplying by 2 gives:
\[ 2y = -4x + 6 \]
Rearranging to the y = mx + b form gives:
\[ y = -2x + 3 \]
However, let’s find another equation using a different constant:
Let's multiply by \( \frac{1}{3} \): \[ y = \frac{-2}{3}x + 1 \]
This is not giving the same slope, let’s keep it as same as the original.
Another way to find the same line is just to change it in a way it stays similar:
- Let’s express it again as: \[ 6x + 3y = 9 \]
So if we reduce by the same constant:
Let’s say: \[ 2x + y = 3 \text{ (this is already divided by 3)} \]
This is another form of their equations that gives us the same solution across any scope.
Thus, another example of a second equation could be: \[ y = -2x + 3 \]
To specify, one valid second equation could be: \[ y = -2x + 3 \]
So, your answer can simply be:
\[ y = -2x + 3 \]
This will have an infinite number of solutions as it is equivalent to the original equation.
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