Solve the System of Equations by the Substitution Method

Video Explanation

Question

Solve the following system of equations:

\[ \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}, \\ \frac{1}{2}(3x+y) – \frac{1}{2}(3x-y) = -\frac{1}{8} \]

Solution

Step 1: Simplify the Second Equation

\[ \frac{1}{2}\big[(3x+y) – (3x-y)\big] = -\frac{1}{8} \]

\[ \frac{1}{2}(2y) = -\frac{1}{8} \]

\[ y = -\frac{1}{8} \quad \text{(1)} \]

Step 2: Substitute in the First Equation

Substitute \(y = -\frac{1}{8}\) into \(\displaystyle \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}\):

\[ \frac{1}{3x-\frac{1}{8}} + \frac{1}{3x+\frac{1}{8}} = \frac{3}{4} \]

\[ \frac{(3x+\frac{1}{8}) + (3x-\frac{1}{8})}{(3x)^2 – \left(\frac{1}{8}\right)^2} = \frac{3}{4} \]

\[ \frac{6x}{9x^2 – \frac{1}{64}} = \frac{3}{4} \]

Step 3: Solve for x

Cross-multiplying,

\[ 24x = 27x^2 – \frac{3}{64} \]

Multiply both sides by 64:

\[ 1536x = 1728x^2 – 3 \]

\[ 1728x^2 – 1536x – 3 = 0 \]

Using the quadratic formula,

\[ x = \frac{1536 \pm \sqrt{1536^2 + 4(1728)(3)}}{2(1728)} \]

\[ x = \frac{1536 \pm \sqrt{264448}}{3456} \]

Step 4: Write the Solutions

\[ x = \frac{1536 + \sqrt{264448}}{3456} \quad \text{or} \quad x = \frac{1536 – \sqrt{264448}}{3456} \]

and

\[ y = -\frac{1}{8} \]

Conclusion

The solutions of the given system of equations are:

\[ \left(\frac{1536 + \sqrt{264448}}{3456},\; -\frac{1}{8}\right) \quad \text{and} \quad \left(\frac{1536 – \sqrt{264448}}{3456},\; -\frac{1}{8}\right) \]

\[ \therefore \quad \text{These are the required solutions.} \]