Answer :
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]
[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]