To create a second equation that has an infinite number of solutions with the first equation \(6x + 3y = 9\), we need to ensure that the second equation is a multiple of the first.
First, let's rewrite the first equation in slope-intercept form \(y = mx + b\).
Starting with the equation:
\[ 6x + 3y = 9 \]
We can isolate \(y\):
\[ 3y = -6x + 9 \]
\[ y = -2x + 3 \]
Now, any equation that is a multiple of this equation will also represent the same line, which means it will have an infinite number of solutions.
For example, we can multiply the entire equation by 2, which gives us:
\[ 2y = -4x + 6 \]
Then, dividing everything by 2 to rewrite it again in slope-intercept form:
\[ y = -2x + 3 \]
But we can also use a different multiple, for instance, 0.5:
\[ 0.5y = -3x + 4.5 \]
Then, multiplying by 2 to clear the fraction gives:
\[ y = -6x + 9 \]
So another possible equation in slope-intercept form could be:
\[ y = -2x + 3 \]
or any multiple (e.g., \(y = -4x + 6\), \(y = 0.5x + 1.5\)). However, all these equations will ultimately have the same slope of \(-2\) and the same y-intercept of 3.
Thus, in summary, one valid possible second equation that Cruz could write that has an infinite number of solutions with \(6x + 3y = 9\) is:
\[ y = -2x + 3 \]
or another similar form like:
\[ y = -4x + 6 \]
You can pick one.
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