Solve the System of Equations by the Substitution Method

Video Explanation

Question

Solve the following system of equations:

\[ \frac{22}{x+y} + \frac{15}{x-y} = 5, \\ \frac{55}{x+y} + \frac{45}{x-y} = 14 \]

Solution

Step 1: Make Suitable Substitution

Let

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

Then the given equations become:

\[ 22a + 15b = 5 \quad \text{(1)} \]

\[ 55a + 45b = 14 \quad \text{(2)} \]

Step 2: Express One Variable in Terms of the Other

From equation (1):

\[ 22a = 5 – 15b \]

\[ a = \frac{5 – 15b}{22} \quad \text{(3)} \]

Step 3: Substitute in Equation (2)

Substitute equation (3) into equation (2):

\[ 55\left(\frac{5 – 15b}{22}\right) + 45b = 14 \]

\[ \frac{55}{22}(5 – 15b) + 45b = 14 \]

\[ \frac{5}{2}(5 – 15b) + 45b = 14 \]

\[ 12.5 – 37.5b + 45b = 14 \]

\[ 12.5 + 7.5b = 14 \]

\[ 7.5b = 1.5 \]

\[ b = \frac{1}{5} \]

Step 4: Find the Value of a

Substitute \(b = \frac{1}{5}\) into equation (3):

\[ a = \frac{5 – 15\left(\frac{1}{5}\right)}{22} = \frac{5 – 3}{22} = \frac{2}{22} = \frac{1}{11} \]

Step 5: Find the Values of x and y

\[ x + y = \frac{1}{a} = 11,\quad x – y = \frac{1}{b} = 5 \]

Adding both equations:

\[ 2x = 16 \Rightarrow x = 8 \]

\[ y = 11 – 8 = 3 \]

Conclusion

The solution of the given system of equations is:

\[ x = 8,\quad y = 3 \]

\[ \therefore \quad \text{The solution is } (8,\; 3). \]