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Original Question
Find any possible values for x and y if the point (x,y) lies on the line y=3x+7Asked by Jennie
Find any possible values for x and y if the point (x,y) lies on the line 4y-16x+12=0
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Answered by
GPT-4o mini
To find the possible values for \(x\) and \(y\) that lie on the line described by the equation \(4y - 16x + 12 = 0\), we can start by rearranging this equation into the slope-intercept form \(y = mx + b\).
1. Begin with the original equation:
\[
4y - 16x + 12 = 0
\]
2. Isolate \(4y\):
\[
4y = 16x - 12
\]
3. Divide everything by 4 to solve for \(y\):
\[
y = 4x - 3
\]
Now, we have expressed \(y\) in terms of \(x\). This means that for any value of \(x\), we can find a corresponding value of \(y\).
### Example values:
- If \(x = 0\):
\[
y = 4(0) - 3 = -3 \quad \Rightarrow \quad (0, -3)
\]
- If \(x = 1\):
\[
y = 4(1) - 3 = 1 \quad \Rightarrow \quad (1, 1)
\]
- If \(x = 2\):
\[
y = 4(2) - 3 = 5 \quad \Rightarrow \quad (2, 5)
\]
- If \(x = -1\):
\[
y = 4(-1) - 3 = -7 \quad \Rightarrow \quad (-1, -7)
\]
### Conclusion:
The point \((x, y)\) can take on infinitely many values of \(x\) and the corresponding \(y\) can be calculated using the equation \(y = 4x - 3\). Thus, any point \((x, 4x - 3)\) will lie on the line defined by the equation \(4y - 16x + 12 = 0\).
1. Begin with the original equation:
\[
4y - 16x + 12 = 0
\]
2. Isolate \(4y\):
\[
4y = 16x - 12
\]
3. Divide everything by 4 to solve for \(y\):
\[
y = 4x - 3
\]
Now, we have expressed \(y\) in terms of \(x\). This means that for any value of \(x\), we can find a corresponding value of \(y\).
### Example values:
- If \(x = 0\):
\[
y = 4(0) - 3 = -3 \quad \Rightarrow \quad (0, -3)
\]
- If \(x = 1\):
\[
y = 4(1) - 3 = 1 \quad \Rightarrow \quad (1, 1)
\]
- If \(x = 2\):
\[
y = 4(2) - 3 = 5 \quad \Rightarrow \quad (2, 5)
\]
- If \(x = -1\):
\[
y = 4(-1) - 3 = -7 \quad \Rightarrow \quad (-1, -7)
\]
### Conclusion:
The point \((x, y)\) can take on infinitely many values of \(x\) and the corresponding \(y\) can be calculated using the equation \(y = 4x - 3\). Thus, any point \((x, 4x - 3)\) will lie on the line defined by the equation \(4y - 16x + 12 = 0\).
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