To solve the system of equations given by: 1. \( 3x - 5y + 5 = 0 \) (Equation 1) 2. \( 4x - 7y + 8 = 0 \) (Equation 2) we will use the method of substitution or elimination. Here, I will demonstrate the elimination method, which is often straightforward for systems of linear equations. ### Step 1: Rearranging the Equations First, let's rearrange both equations to express them in the standard form \( Ax + By = C \). From Equation 1: \[ 3x - 5y = -5 \quad \text{(Rearranging gives us this form)} \] From Equation 2: \[ 4x - 7y = -8 \quad \text{(Rearranging gives us this form)} \] ### Step 2: Elimination of One Variable Next, we will eliminate one of the variables. To do this, we can multiply the equations by suitable coefficients so that the coefficients of \( x \) or \( y \) become the same. Let's eliminate \( x \). We can multiply Equation 1 by 4 and Equation 2 by 3: - Multiply Equation 1 by 4: \[ 4(3x - 5y) = 4(-5) \implies 12x - 20y = -20 \quad \text{(Equation 3)} \] - Multiply Equation 2 by 3: \[ 3(4x - 7y) = 3(-8) \implies 12x - 21y = -24 \quad \text{(Equation 4)} \] ### Step 3: Subtracting the Equations Now, we can subtract Equation 3 from Equation 4 to eliminate \( x \): \[ (12x - 21y) - (12x - 20y) = -24 - (-20) \] This simplifies to: \[ -21y + 20y = -24 + 20 \] \[ -y = -4 \] ### Step 4: Solving for \( y \) Now, we can solve for \( y \): \[ y = 4 \] ### Step 5: Substituting Back to Find \( x \) Now that we have \( y \), we can substitute \( y = 4 \) back into one of the original equations to find \( x \). Let's use Equation 1: \[ 3x - 5(4) + 5 = 0 \] \[ 3x - 20 + 5 = 0 \] \[ 3x - 15 = 0 \] \[ 3x = 15 \] \[ x = 5 \] ### Final Values Thus, we have found: - \( x = 5 \) - \( y = 4 \) ### Conclusion The correct option is **D. 5, 4**. ### Explanation of Other Options - **Option A: -5, -4**: This option suggests both \( x \) and \( y \) are negative, which does not satisfy either of the original equations when substituted back. - **Option B: -4, -5**: Similar to option A, substituting these values into the original equations will not satisfy them. - **Option C: 4, 5**: This option has the values swapped and does not satisfy the equations when checked. ### Revision Summary - Use the elimination method to solve systems of linear equations by making coefficients of one variable equal. - Rearrange equations into standard form \( Ax + By = C \) for clarity. - Substitute back to find the other variable after solving for one. - Always check your solutions by substituting back into the original equations to ensure they hold true.