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 \( 4 x-3 y=20 \)
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To solve these systems of equations using substitution, let’s first isolate one variable in one equation and substitute it into the other. 1. From the first equation \( x + 2y = 5 \), we can isolate \( x \): \[ x = 5 - 2y \] Substitute this expression for \( x \) in the second equation \( 4x + 2y = -10 \): \[ 4(5 - 2y) + 2y = -10 \] Simplifying this gives: \[ 20 - 8y + 2y = -10 \implies -6y = -30 \implies y = 5 \] Now substitute \( y = 5 \) back into \( x = 5 - 2y \): \[ x = 5 - 2(5) = -5 \] So, the solution is \( x = -5 \), \( y = 5 \). 2. In the second system, starting with \( 2x - 4y = 10 \), we can isolate \( x \): \[ 2x = 10 + 4y \implies x = 5 + 2y \] Substitute this into the second equation \( 3x + 5y = -29 \): \[ 3(5 + 2y) + 5y = -29 \] Simplifying gives: \[ 15 + 6y + 5y = -29 \implies 11y = -44 \implies y = -4 \] Now substituting \( y = -4 \) back into \( x = 5 + 2y \): \[ x = 5 + 2(-4) = 5 - 8 = -3 \] So, the solution is \( x = -3 \), \( y = -4 \). 3. For the third system starting from \( 2x - 3y = 4 \), isolate \( x \): \[ 2x = 4 + 3y \implies x = 2 + \frac{3}{2}y \] Substitute this into the second equation \( 4x - 3y = 20 \): \[ 4(2 + \frac{3}{2}y) - 3y = 20 \] Simplifying gives: \[ 8 + 6y - 3y = 20 \implies 3y = 12 \implies y = 4 \] Now substitute \( y = 4 \) back: \[ x = 2 + \frac{3}{2}(4) = 2 + 6 = 8 \] So, the solution is \( x = 8 \), \( y = 4 \). Final Solutions: 1. \( (x, y) = (-5, 5) \) 2. \( (x, y) = (-3, -4) \) 3. \( (x, y) = (8, 4) \)
