Answer :
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]
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]
