Answer:
When \(y=1\), \(x=\frac{11}{18}\)
When \(x=1\), \(y=\frac{1}{30}\)
Step-by-step explanation:
\(2 + \frac{1}{6y} = 3x + 4\)
\(2 -4+ \frac{1}{6y} = 3x\)
\(-2= 3x-\frac{1}{6y}\)
The only way to find the values for this equation is find values of \(x\) and \(y\) relative to each other.
That means we substitute a value for \(x\) and find the value of \(y\) for that particular \(x\) value.
Let's find \(x\) when \(y=1\):
\(-2= 3x-\frac{1}{6y}\)
\(-2= 3x-\frac{1}{6\times1}\)
\(-2= 3x-\frac{1}{6}\)
\(-2+\frac{1}{6}= 3x\)
\(\frac{11}{18}= x\)
Therefore, when \(y=1\), \(x=\frac{11}{18}\)
Let's find \(y\) when \(x=1\):
\(-2= 3x-\frac{1}{6y}\)
\(-2= 3\times1-\frac{1}{6y}\)
\(-2= 3-\frac{1}{6y}\)
\(\frac{1}{6y}= 3+2\)
\(\frac{1}{6y}= 5\)
\(\frac{1}{6\times5}= y\)
\(\frac{1}{30}= y\)
Therefore, when \(x=1\), \(y=\frac{1}{30}\)
