To find a second equation that has an infinite number of solutions with the first equation \(6x + 3y = 9\), the second equation must be a multiple of the first equation. This means it should have the same slope and thus be equivalent after simplification.
First, let's rewrite the first equation in slope-intercept form (i.e., \(y = mx + b\)):
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Start with the original equation: \[ 6x + 3y = 9 \]
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Subtract \(6x\) from both sides: \[ 3y = -6x + 9 \]
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Divide every term by \(3\) to solve for \(y\): \[ y = -2x + 3 \]
Now we have the slope-intercept form of the first equation: \(y = -2x + 3\).
To create a second equation with an infinite number of solutions, you can multiply the entire first equation by any non-zero constant. For example, let's multiply the first equation by \(2\):
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Start with the original equation: \[ 6x + 3y = 9 \]
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Multiply through by \(2\): \[ 12x + 6y = 18 \]
Now, we can also express this new equation in slope-intercept form by following the same procedure as before. Let's solve for \(y\):
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Subtract \(12x\) from both sides: \[ 6y = -12x + 18 \]
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Divide every term by \(6\): \[ y = -2x + 3 \]
So the second equation, which has an infinite number of solutions with the first equation, can be written as:
\[ y = -2x + 3 \]
In conclusion, both equations \(6x + 3y = 9\) and \(12x + 6y = 18\) (or \(y = -2x + 3\)) represent the same line, leading to an infinite number of solutions.
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