Question

Cruz wrote a system of equations with an infinite number of solutions. One of his equations is 6x+3y=9
6
𝑥
+
3
𝑦
=
9
. What could be his second equation? (Write the equation in slope-intercept form.)(1 point)
The second equation could be y=
𝑦
=
.

Answer 1

To find a second equation that has an infinite number of solutions with the equation \(6x + 3y = 9\), we can start by rewriting the first equation in slope-intercept form (i.e., \(y = mx + b\)).

1. Start with the original equation:
\[
6x + 3y = 9
\]

2. Isolate \(y\):
\[
3y = -6x + 9
\]
\[
y = -2x + 3
\]

The slope of the equation is \(-2\), and the y-intercept is \(3\).

For the second equation to have an infinite number of solutions, it must be equivalent to the first equation, meaning it can be derived by multiplying the first equation by a non-zero constant.

Let's multiply the entire first equation by \(k\):
\[
k(6x + 3y = 9) \implies 6kx + 3ky = 9k
\]

To express the second equation in slope-intercept form, we can similarly transform it. For example, if we let \(k = 2\):
\[
12x + 6y = 18
\]

Now, we isolate \(y\):
\[
6y = -12x + 18
\]
\[
y = -2x + 3
\]

This is the same as the first equation.

Thus, one possible second equation could be:
\[
y = -2x + 3
\]

However, another example, if we let \(k = 3\):
\[
18x + 9y = 27
\]
leads to:
\[
9y = -18x + 27
\]
and therefore,
\[
y = -2x + 3
\]

In conclusion, the second equation could be:
\[
y = -2x + 3
\]
No matter how we derive it, the result is consistent. An example would be to just choose any multiple of the first equation, such as:
\[
y = -2x + 3 \quad (\text{which is rewritten from } 12x + 6y = 18 \text{ or other multiples})
\]
As long as it is consistent with the slope and intercept, it will work.