For a system of equations to have an infinite number of solutions, the two equations must represent the same line. This means that the second equation must be a scalar multiple of the first equation.
Let's start with the first equation:
\[ 6x + 3y = 9. \]
First, we can convert this equation into slope-intercept form (y = mx + b).
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Subtract \( 6x \) from both sides to isolate the \( y \) term:
\[ 3y = -6x + 9 \]
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Divide every term by 3:
\[ y = -2x + 3. \]
Now that we have it in slope-intercept form, we can create a second equation that is equivalent to the first by multiplying the entire equation by a non-zero constant. Let's choose a multiplier of \( 2 \):
\[ 2(y) = 2(-2x + 3) \]
Distributing the \( 2 \):
\[ 2y = -4x + 6. \]
Now, rearranging this into slope-intercept form, we can write it as:
\[ 2y = -4x + 6 \]
or by dividing by \( 2 \):
\[ y = -2x + 3. \]
Since this is equivalent to the first equation, you can also multiply it by \( -1 \):
This gives us another form:
\[ 2y + 4x = 6, \]
To provide a different valid form, let's also multiply our original by \( \frac{1}{2} \):
This leads us to:
\[ 3y = \frac{3}{2} - 3x. \]
So, any equation that is a scalar multiple of \( 6x + 3y = 9 \) will be valid. Another example could be something along the lines of:
\[ 4y = -8x + 12. \]
Including multiples such as:
- \( 9x + 6y = 27 \)
- Many such equations yield infinite solutions.
Thus, you may say explicitly:
A possible second equation is:
\[ 6x + 3y = 9. \]
But to be completely clear, a different scalar multiple leading to a newly represented equation could be:
\[ y + 2x = 3. \]
This will preserve the properties of there being infinitely many solutions.
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