Sure! Let's solve the system of equations step-by-step:
We are given the following system of equations:
[tex]\[
\left\{\begin{array}{l}
\frac{2}{x} - \frac{1}{y} = 1 \\
\frac{1}{x} + \frac{5}{y} = 6
\end{array}\right.
\][/tex]
First, we will simplify each equation by finding a common denominator for each expression.
Equation 1:
[tex]\[
\frac{2}{x} - \frac{1}{y} = 1
\][/tex]
Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \). Then the first equation becomes:
[tex]\[
2a - b = 1 \quad \text{(1)}
\][/tex]
Equation 2:
[tex]\[
\frac{1}{x} + \frac{5}{y} = 6
\][/tex]
Using the same substitutions, the second equation becomes:
[tex]\[
a + 5b = 6 \quad \text{(2)}
\][/tex]
Now, we have a system of linear equations:
[tex]\[
\left\{\begin{array}{l}
2a - b = 1 \\
a + 5b = 6
\end{array}\right.
\][/tex]
We can solve this system using the substitution or elimination method. We'll use the elimination method here.
First, we can multiply Equation 2 by 2 to align the coefficients of \(a\) in both equations:
[tex]\[
2(a + 5b) = 2 \cdot 6
\][/tex]
[tex]\[
2a + 10b = 12 \quad \text{(3)}
\][/tex]
Now we have:
[tex]\[
\left\{\begin{array}{l}
2a - b = 1 \quad \text{(1)} \\
2a + 10b = 12 \quad \text{(3)}
\end{array}\right.
\][/tex]
Next, subtract Equation 1 from Equation 3:
[tex]\[
(2a + 10b) - (2a - b) = 12 - 1
\][/tex]
[tex]\[
2a + 10b - 2a + b = 11
\][/tex]
[tex]\[
11b = 11
\][/tex]
[tex]\[
b = 1
\][/tex]
Now that we have \( b = 1 \), we can substitute this value back into Equation 2 to find \( a \):
[tex]\[
a + 5b = 6
\][/tex]
[tex]\[
a + 5 \cdot 1 = 6
\][/tex]
[tex]\[
a + 5 = 6
\][/tex]
[tex]\[
a = 1
\][/tex]
Now we have \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \):
[tex]\[
\frac{1}{x} = 1 \quad \Rightarrow \quad x = 1
\][/tex]
[tex]\[
\frac{1}{y} = 1 \quad \Rightarrow \quad y = 1
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
x = 1, \quad y = 1
\][/tex]
