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Find whether the following pair of linear equations has a unique solution. If yes, find the solution.
`7x-4y =49 and 5x-6y=57`

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To determine whether the given pair of linear equations has a unique solution, we will follow these steps: ### Given Equations: 1. \( 7x - 4y = 49 \) (Equation 1) 2. \( 5x - 6y = 57 \) (Equation 2) ### Step 1: Identify Coefficients From the equations, we can identify the coefficients: - For Equation 1: - \( A_1 = 7 \) - \( B_1 = -4 \) - \( C_1 = -49 \) - For Equation 2: - \( A_2 = 5 \) - \( B_2 = -6 \) - \( C_2 = -57 \) ### Step 2: Check the Condition for Unique Solution The condition for a unique solution is given by: \[ \frac{A_1}{A_2} \neq \frac{B_1}{B_2} \] Calculating \( \frac{A_1}{A_2} \): \[ \frac{A_1}{A_2} = \frac{7}{5} \] Calculating \( \frac{B_1}{B_2} \): \[ \frac{B_1}{B_2} = \frac{-4}{-6} = \frac{2}{3} \] Now, we compare: \[ \frac{7}{5} \neq \frac{2}{3} \] Since \( \frac{7}{5} \) is not equal to \( \frac{2}{3} \), the condition for a unique solution is satisfied. ### Step 3: Solve the Equations Since we have established that there is a unique solution, we will solve the equations using the elimination method. #### Step 3.1: Multiply the Equations Multiply Equation 1 by 3 and Equation 2 by 2: - \( 3 \times (7x - 4y) = 3 \times 49 \) \[ 21x - 12y = 147 \quad \text{(Equation 3)} \] - \( 2 \times (5x - 6y) = 2 \times 57 \) \[ 10x - 12y = 114 \quad \text{(Equation 4)} \] #### Step 3.2: Subtract the Equations Now, subtract Equation 4 from Equation 3: \[ (21x - 12y) - (10x - 12y) = 147 - 114 \] This simplifies to: \[ 11x = 33 \] #### Step 3.3: Solve for \( x \) Dividing both sides by 11: \[ x = 3 \] ### Step 4: Substitute \( x \) back into one of the original equations Now, substitute \( x = 3 \) back into Equation 1: \[ 7(3) - 4y = 49 \] This simplifies to: \[ 21 - 4y = 49 \] Rearranging gives: \[ -4y = 49 - 21 \] \[ -4y = 28 \] Dividing by -4: \[ y = -7 \] ### Final Solution Thus, the unique solution to the given pair of linear equations is: \[ x = 3, \quad y = -7 \]
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