Question

Solution of the equations 3x-4y=7 and 2x+3y=-1 is not equal to

• A. \(\frac{22}{22},\frac{33}{33}\)
• B. \(\frac{33}{33},\frac{-44}{44}\)
• C. \(\frac{44}{44},\frac{-77}{77}\)
• D. \(\frac{77}{77},\frac{-11}{11}\)

Answer 1

The Correct Option is A

To solve this problem, we need to find the solution to the system of linear equations:

\[ \begin{aligned} \text{(1)} \quad 3x - 4y &= 7 \\ \text{(2)} \quad 2x + 3y &= -1 \end{aligned} \]

1. Solve the system using elimination or substitution method.

Step 1: Multiply both equations to eliminate one variable:
Multiply equation (1) by 3 and equation (2) by 4 to make coefficients of \( y \) equal in magnitude: \[ \begin{aligned} 3(3x - 4y) &= 3(7) \Rightarrow 9x - 12y = 21 \quad \text{(3)} \\ 4(2x + 3y) &= 4(-1) \Rightarrow 8x + 12y = -4 \quad \text{(4)} \end{aligned} \]

Step 2: Add equations (3) and (4):

\[ (9x - 12y) + (8x + 12y) = 21 + (-4) \] \[ 17x = 17 \Rightarrow x = 1 \]

Step 3: Substitute \( x = 1 \) into one of the original equations:
Using equation (1):

\[ 3x - 4y = 7 \Rightarrow 3(1) - 4y = 7 \Rightarrow 3 - 4y = 7 \Rightarrow -4y = 4 \Rightarrow y = -1 \]

So the solution is:
\[ x = 1, \quad y = -1 \]

2. Evaluating the options:
Only one option is not equal to (1, -1):

• (1) \( \frac{22}{22}, \frac{33}{33} = 1, 1 \) → Not equal ✔
• (2) \( \frac{33}{33}, \frac{-44}{44} = 1, -1 \) → Equal
• (3) \( \frac{44}{44}, \frac{-77}{77} = 1, -1 \) → Equal
• (4) \( \frac{77}{77}, \frac{-11}{11} = 1, -1 \) → Equal

Final Answer:
The correct answer is (A) as it is not equal to the solution \( (1, -1) \).