To solve the system of equations:
[tex]\[
\frac{3}{x} + \frac{2}{y} = 1
\][/tex]
and
[tex]\[
\frac{4}{x} + \frac{3}{y} = \frac{17}{6}
\][/tex]
we will transform the equations and solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
Step 1: Transform the equations
First, we will make substitutions to simplify these equations. Let:
[tex]\[
u = \frac{1}{x} \quad \text{and} \quad v = \frac{1}{y}
\][/tex]
This gives us the following system of linear equations:
[tex]\[
3u + 2v = 1 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
4u + 3v = \frac{17}{6} \quad \text{(Equation 2)}
\][/tex]
Step 2: Solve the system of linear equations
To solve this system, we can use the elimination or substitution method. Here, we will use the elimination method.
Multiply Equation 1 by 3:
[tex]\[
9u + 6v = 3 \quad \text{(Equation 3)}
\][/tex]
Multiply Equation 2 by 2:
[tex]\[
8u + 6v = \frac{34}{6} \quad \text{(Equation 4)}
\][/tex]
Now, subtract Equation 4 from Equation 3 to eliminate [tex]\( v \)[/tex]:
[tex]\[
(9u + 6v) - (8u + 6v) = 3 - \frac{34}{6}
\][/tex]
Simplifying this, we get:
[tex]\[
u = 3 - \frac{34}{6} = \frac{18}{6} - \frac{34}{6} = \frac{-16}{6} = -\frac{8}{3}
\][/tex]
So,
[tex]\[
u = -\frac{8}{3}
\][/tex]
Since [tex]\( u = \frac{1}{x} \)[/tex],
[tex]\[
\frac{1}{x} = -\frac{8}{3} \Rightarrow x = -\frac{3}{8}
\][/tex]
Step 3: Substitute [tex]\( u \)[/tex] back to find [tex]\( v \)[/tex]
Using Equation 1:
[tex]\[
3u + 2v = 1
\][/tex]
Substitute [tex]\( u = -\frac{8}{3} \)[/tex]:
[tex]\[
3 \left( -\frac{8}{3} \right) + 2v = 1
\][/tex]
[tex]\[
-8 + 2v = 1
\][/tex]
Solve for [tex]\( v \)[/tex]:
[tex]\[
2v = 1 + 8 = 9
\][/tex]
[tex]\[
v = \frac{9}{2}
\][/tex]
Since [tex]\( v = \frac{1}{y} \)[/tex],
[tex]\[
\frac{1}{y} = \frac{9}{2} \Rightarrow y = \frac{2}{9}
\][/tex]
So, the solution to the system of equations is:
[tex]\[
x = -\frac{3}{8} \quad \text{and} \quad y = \frac{2}{9}
\][/tex]
Conclusion
Hence, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations are:
[tex]\[
x = -0.375 \quad \text{and} \quad y = 0.222222 \ldots
\][/tex]
