Solve the System of Equations by the Substitution Method

Video Explanation

Question

Solve the following system of equations:

\[ \frac{1}{2(x+2y)} + \frac{5}{3(3x-2y)} = -\frac{3}{2}, \\ \frac{5}{4(x+2y)} – \frac{3}{5(3x-2y)} = \frac{61}{60} \]

Solution

Step 1: Make Suitable Substitution

Let

\[ \frac{1}{x+2y} = a,\quad \frac{1}{3x-2y} = b \]

Then the given equations become:

\[ \frac{a}{2} + \frac{5b}{3} = -\frac{3}{2} \quad \text{(1)} \]

\[ \frac{5a}{4} – \frac{3b}{5} = \frac{61}{60} \quad \text{(2)} \]

Step 2: Remove Fractions

Multiply equation (1) by 6:

\[ 3a + 10b = -9 \quad \text{(3)} \]

Multiply equation (2) by 60:

\[ 75a – 36b = 61 \quad \text{(4)} \]

Step 3: Express One Variable in Terms of the Other

From equation (3):

\[ 3a = -9 – 10b \]

\[ a = -3 – \frac{10}{3}b \quad \text{(5)} \]

Step 4: Substitute in Equation (4)

Substitute equation (5) into equation (4):

\[ 75\left(-3 – \frac{10}{3}b\right) – 36b = 61 \]

\[ -225 – 250b – 36b = 61 \]

\[ -225 – 286b = 61 \]

\[ -286b = 286 \]

\[ b = -1 \]

Step 5: Find the Value of a

Substitute \(b = -1\) into equation (5):

\[ a = -3 – \frac{10}{3}(-1) = -3 + \frac{10}{3} = \frac{1}{3} \]

Step 6: Find the Values of x and y

\[ x + 2y = \frac{1}{a} = 3,\quad 3x – 2y = \frac{1}{b} = -1 \]

Adding both equations:

\[ 4x = 2 \Rightarrow x = \frac{1}{2} \]

Substitute \(x = \frac{1}{2}\) into \(x + 2y = 3\):

\[ 2y = 3 – \frac{1}{2} = \frac{5}{2} \Rightarrow y = \frac{5}{4} \]

Conclusion

The solution of the given system of equations is:

\[ x = \frac{1}{2},\quad y = \frac{5}{4} \]

\[ \therefore \quad \text{The solution is } \left(\frac{1}{2},\; \frac{5}{4}\right). \]