Solve the System of Equations by the Substitution Method

Video Explanation

Question

Solve the following system of equations:

\[ \frac{44}{x+y} + \frac{30}{x-y} = 10, \\ \frac{55}{x+y} + \frac{40}{x-y} = 13 \]

Solution

Step 1: Make Suitable Substitution

Let

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

Then the given equations become:

\[ 44a + 30b = 10 \quad \text{(1)} \]

\[ 55a + 40b = 13 \quad \text{(2)} \]

Step 2: Express One Variable in Terms of the Other

From equation (1):

\[ 44a = 10 – 30b \]

\[ a = \frac{10 – 30b}{44} \quad \text{(3)} \]

Step 3: Substitute in Equation (2)

Substitute equation (3) into equation (2):

\[ 55\left(\frac{10 – 30b}{44}\right) + 40b = 13 \]

\[ \frac{55}{44}(10 – 30b) + 40b = 13 \]

\[ \frac{5}{4}(10 – 30b) + 40b = 13 \]

\[ 12.5 – 37.5b + 40b = 13 \]

\[ 12.5 + 2.5b = 13 \]

\[ 2.5b = 0.5 \]

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

Step 4: Find the Value of a

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

\[ a = \frac{10 – 30\left(\frac{1}{5}\right)}{44} = \frac{10 – 6}{44} = \frac{4}{44} = \frac{1}{11} \]

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:

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

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