Solve the System of Equations by the Substitution Method

Video Explanation

Question

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

\[ \frac{6}{x+y} = \frac{7}{x-y} + 3, \\ \frac{1}{2(x+y)} = \frac{1}{3(x-y)} \]

Solution

Step 1: Solve the Second Equation

\[ \frac{1}{2(x+y)} = \frac{1}{3(x-y)} \]

Cross-multiplying,

\[ 3(x-y) = 2(x+y) \]

\[ 3x – 3y = 2x + 2y \]

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

Step 2: Substitute in the First Equation

Substitute \(x = 5y\) into \(\displaystyle \frac{6}{x+y} = \frac{7}{x-y} + 3\):

\[ \frac{6}{5y + y} = \frac{7}{5y – y} + 3 \]

\[ \frac{6}{6y} = \frac{7}{4y} + 3 \]

\[ \frac{1}{y} = \frac{7}{4y} + 3 \]

Multiply both sides by \(y\):

\[ 1 = \frac{7}{4} + 3y \]

\[ 3y = 1 – \frac{7}{4} = -\frac{3}{4} \]

\[ y = -\frac{1}{4} \]

Step 3: Find the Value of x

Substitute \(y = -\frac{1}{4}\) into equation (1):

\[ x = 5\left(-\frac{1}{4}\right) = -\frac{5}{4} \]

Conclusion

The solution of the given system of equations is:

\[ x = -\frac{5}{4},\quad y = -\frac{1}{4} \]

\[ \therefore \quad \text{The solution is } \left(-\frac{5}{4},\; -\frac{1}{4}\right). \]