1. 5.2 Solving Systems of Equations by the Substitution Method

2. Solving Using Substitution 1.Solve either eqn. for x or y (may be done already). 2.Substitute the expression for the variable obtained in step 1 into the other eqn. and solve it. 3.Substitute the value for the variable from step 2 into any eqn. in the system that contains both variables and solve for the other variable. 4.Check soln. in BOTH eqns., if necessary.

3. Ex. Solve by the substitution method: x + y = 11 y = 2x – 1 1.y = 2x – 1 2. x + y = 11 x + (2x – 1) = 11 x + 2x – 1 = 11 3x – 1 = 11 3x – 1 + 1 = 11 + 1 3x = 12 3 3 x = 4 3.y = 2x – 1 y = 2(4) – 1 sub 4 for x y = 8 – 1 y = 7 Soln: {(4, 7)} 4.Check: x + y = 11 y = 2x – 1 4 + 7 = 11 7 = 2(4) – 1 11 = 11 7 = 8 – 1 7 = 7

4. Ex. Solve by the substitution method: 2x – 2y = 2 x – 5y = -7 1. x – 5y = -7 x – 5y + 5y = -7 + 5y x = 5y – 7 2. 2x – 2y = 2 2(5y – 7) – 2y = 2 10y – 14 – 2y = 2 8y – 14 = 2 8y + 14 – 14 = 2 + 14 8y = 16 8 8 y = 2 3.x = 5y – 7 x = 5(2) – 7 sub 2 for y x = 10 – 7 x = 3 Soln: {(3, 2)} 4.Check: 2x – 2y = 2 x – 5y = -7 2(3) – 2(2) = 2 3 – 5(2)= -7 6 – 4 = 2 3 – 10 = -7 2 = 2 -7 = -7

5. Ex. Solve by the substitution method: x + 2y = -3 First: eliminate fractions by mult. by LCD 1.x – 3y = 2 x – 3y + 3y = 2 + 3y x = 3y + 2 2. x + 2y = -3 3y + 2 + 2y = -3 5y + 2 = -3 5y + 2 – 2 = -3 – 2 5y = -5 5 5 y = -1 3.x = 3y + 2 x = 3(-1) + 2 sub -1 for y x = -3 + 2 x = -1 Soln: {(-1, -1)}

6. Ex. Solve by the substitution method: -4x + 4y = -8 -x + y = -2 1. -x + y = -2 -x + y + x = -2 + x y = x – 2 2. -4x + 4y = -8 -4x + 4(x – 2) = -8 -4x + 4x – 8 = -8 -8 = -8 No variables remain and a TRUE stmt.  Lines coincide, so there are infinitely many solns. Soln: {(x, y)| -x + y = -2} or {(x, y)| -4x + 4y = -8}

7. Ex. Solve by the substitution method: 6x + 2y = 7 y = 2 – 3x 1.y = 2 – 3x 2. 6x + 2y = 7 6x + 2(2 – 3x) = 7 6x + 4 – 6x = 7 4 = 7 No variables remain and a FALSE stmt.  lines are parallel, so there is NO SOLN. (empty set) no soln. or ø

8. Groups Page 307 – 308: 33, 41, 45, 49