To have an infinite number of solutions, the second equation must be equivalent to the first equation. This means it should be a multiple of the original equation.
Given the equation: \[ 6x + 3y = 9 \]
We can rearrange this into slope-intercept form (y = mx + b):
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Subtract \( 6x \) from both sides: \[ 3y = -6x + 9 \]
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Divide everything by 3: \[ y = -2x + 3 \]
Now, to find an equivalent equation, we can multiply the entire equation by any non-zero scalar. Let's say we multiply it by 2:
\[ 2(y) = 2(-2x + 3) \ 2y = -4x + 6 \]
Rearranging it back to standard form gives us: \[ 4x + 2y = 6 \]
So, an example of a second equation that has an infinite number of solutions with the first equation is:
\[ 4x + 2y = 6 \]
Alternatively, we could also write it in slope-intercept form. Let's say we just want to double the original equation, it would lead us to:
\[ y = -2x + 3 \quad \text{(which is the same as the first equation in slope-intercept form)} \]
Thus, the equation in slope-intercept form could be: \[ y = -2x + 3 \]
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