To create a second equation that has the same infinite solutions as the first equation \(6x + 3y = 9\), we need to ensure that the second equation is a multiple of the first equation.
First, we can simplify the first equation:
- Divide the entire equation by 3:
\[ 2x + y = 3 \]
- Now, convert this equation to slope-intercept form \(y = mx + b\):
\[ y = -2x + 3 \]
Now, we can create a second equation by multiplying the entire equation by any non-zero constant \(k\). For example, let's choose \(k = 2\):
\[ 2(2x + y) = 2(3) \]
This leads to:
\[ 4x + 2y = 6 \]
Now, we can convert it into slope-intercept form:
\[ 2y = -4x + 6 \] \[ y = -2x + 3 \]
We can choose any constant multiple. For instance, if \(k = 1/2\):
\[ \frac{1}{2}(6x + 3y) = \frac{1}{2}(9) \]
This simplifies to:
\[ 3x + \frac{3}{2}y = \frac{9}{2} \]
Converting this equation to slope-intercept form gives:
\[ \frac{3}{2}y = -3x + \frac{9}{2} \] \[ y = -2x + 3 \]
Thus, one example for Cruz's second equation could be:
\[ y = -2x + 3 \]
Or he could choose any other multiple of the first equation, for example:
\[ y = -4x + 12 \]
So, a possible answer (which maintains an infinite number of solutions) could also be:
\[ y = -4x + 12 \]
You can express this as:
\[ y = -2x + 3 \]
as already mentioned, or in other equivalent forms.
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