A. Use Substitution to solve \( x \) and \( y \) : 1. \( x+2 y=5 \) and \( 4 x+2 y=-10 \) 2. \( 2 x-4 y=10 \) and \( 3 x+5 y=-29 \) 3. \( 2 x-3 y=4 \) and \( 4 x-3 y=20 \)

1. Given the system \[ x + 2y = 5, \] \[ 4x + 2y = -10. \] Solve the first equation for \( x \): \[ x = 5 - 2y. \] Substitute \( x = 5 - 2y \) into the second equation: \[ 4(5 - 2y) + 2y = -10. \] Simplify: \[ 20 - 8y + 2y = -10, \] \[ 20 - 6y = -10. \] Subtract 20 from both sides: \[ -6y = -30. \] Divide by \(-6\): \[ y = 5. \] Substitute \( y = 5 \) back into \( x = 5 - 2y \): \[ x = 5 - 2(5) = 5 - 10 = -5. \] **Solution for the first system:** \( x = -5 \), \( y = 5 \). 2. Given the system \[ 2x - 4y = 10, \] \[ 3x + 5y = -29. \] Solve the first equation for \( x \): \[ x = \frac{10 + 4y}{2} = 5 + 2y. \] Substitute \( x = 5 + 2y \) into the second equation: \[ 3(5 + 2y) + 5y = -29. \] Simplify: \[ 15 + 6y + 5y = -29, \] \[ 15 + 11y = -29. \] Subtract 15 from both sides: \[ 11y = -44. \] Divide by 11: \[ y = -4. \] Substitute \( y = -4 \) back into \( x = 5 + 2y \): \[ x = 5 + 2(-4) = 5 - 8 = -3. \] **Solution for the second system:** \( x = -3 \), \( y = -4 \). 3. Given the system \[ 2x - 3y = 4, \] \[ 4x - 3y = 20. \] Notice that both equations have the term \(-3y\). Subtract the first equation from the second to eliminate \(y\): \[ (4x - 3y) - (2x - 3y) = 20 - 4. \] Simplify: \[ 4x - 3y - 2x + 3y = 16, \] \[ 2x = 16. \] Divide by 2: \[ x = 8. \] Substitute \( x = 8 \) into the first equation: \[ 2(8) - 3y = 4, \] \[ 16 - 3y = 4. \] Subtract 16 from both sides: \[ -3y = 4 - 16 = -12. \] Divide by \(-3\): \[ y = 4. \] **Solution for the third system:** \( x = 8 \), \( y = 4 \).