To find a second equation that has an infinite number of solutions with the equation \( 6x + 3y = 9 \), we need to create an equation that is equivalent to the first.
First, let's rewrite the first equation in slope-intercept form (y = mx + b):
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Start with the equation: \[ 6x + 3y = 9 \]
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Subtract \( 6x \) from both sides: \[ 3y = -6x + 9 \]
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Divide each term by 3: \[ y = -2x + 3 \]
Now, since the second equation must be equivalent to the first, it can be any multiple of this equation. For instance, multiply the entire equation by a constant (let's say 2):
\[ 2y = 2(-2x + 3) \] \[ 2y = -4x + 6 \]
Now, rewrite it in slope-intercept form:
\[ y = -2x + 3 \]
OR you can write it in a different form (let's multiply the original answer by 2):
If we try with a different multiplier, like 1 (which keeps the equation the same) or any other number (such as -1/3), then:
Another equivalent form can be \( y = -2x + 3 \) or you can choose to express it in a different way as follows:
Let's take the original equation and rearrange it into a
- Start from \( 6x + 3y = 9 \)
- Take something like these equivalents: \( 6x + 3y + 0 = 9 \) or
Another possible way can be changing '9' to another value justified by increasing or decrementing the left side equally reserving the slope.
So ultimately, a second equation could be
\[ y = -2x + 3 \]
You can choose whether you take: Using the same slope and y-intercept: You could also write it as:
So one possible answer could be: \[ y = -2x + 3 \]
Or provide numerical change: To ensure linear formation it stays equivalent.
If you take suitable multiplier or subtract/add value with same treatment.
In chance towards version it could appear multiple forms retaining consistency with \( y = m x + b \).
Thus overall the second equation would be:
\[ y = -2x + 3 \]
Feel free to adjust numbers but engage to infinite solutions which relies on that mathematic representation.