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