Solve the System of Equations by the Substitution Method

Video Explanation

Question

Solve the following system of equations, where \(x+y \ne 0\) and \(x-y \ne 0\):

\[ \frac{xy}{x+y} = \frac{6}{5}, \\ \frac{xy}{y-x} = 6 \]

Solution

Step 1: Remove Denominators

From the first equation:

\[ \frac{xy}{x+y} = \frac{6}{5} \]

\[ 5xy = 6(x+y) \quad \text{(1)} \]

From the second equation:

\[ \frac{xy}{y-x} = 6 \]

\[ xy = 6(y-x) \quad \text{(2)} \]

Step 2: Express One Expression in Terms of the Other

From equation (2):

\[ xy = 6y – 6x \quad \text{(3)} \]

Step 3: Substitute in Equation (1)

Substitute equation (3) into equation (1):

\[ 5(6y – 6x) = 6(x+y) \]

\[ 30y – 30x = 6x + 6y \]

\[ 24y = 36x \]

\[ y = \frac{3}{2}x \quad \text{(4)} \]

Step 4: Find the Value of x

Substitute equation (4) into equation (2):

\[ x\left(\frac{3}{2}x\right) = 6\left(\frac{3}{2}x – x\right) \]

\[ \frac{3}{2}x^2 = 6\left(\frac{1}{2}x\right) \]

\[ \frac{3}{2}x^2 = 3x \]

\[ x = 2 \]

Step 5: Find the Value of y

Substitute \(x = 2\) into equation (4):

\[ y = \frac{3}{2}(2) = 3 \]

Conclusion

The solution of the given system of equations is:

\[ x = 2,\quad y = 3 \]

\[ \therefore \quad \text{The solution is } (2,\; 3). \]