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3.2 Solve Linear Systems Algebraically Algebra II.

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Presentation on theme: "3.2 Solve Linear Systems Algebraically Algebra II."— Presentation transcript:

1 3.2 Solve Linear Systems Algebraically Algebra II

2 Substitution Method 1)Solve one of the equations for one of its variables. 2)Substitute the expression from Step 1 into the other equation and solve for the other variable. 3)Substitute the value from Step 2 into the revised equation from Step 2 and solve.

3 Example – Substitution Method Solve this system:2x + 5y = -5  equation 1 x + 3y = 3  equation 2 Step 1)Solve equation 2 for x x = -3y + 3 Step 2)Substitute into equation 1 and solve for y 2 (-3y + 3) + 5y = -5 y = 11 Step 3)Substitute the value of y into revised equation 2 and solve for x. x = -3 (11) +3 x = -30 Solution: (-30, 11) CHECK!

4 Elimination Method 1)Multiply one or both of the equations by a constant to obtain coefficients that differ only in sign for one of the variables. 2)Add the revised equations from Step 1. Combining like terms will eliminate one of the variables. Solve for the remaining variable. 3)Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable.

5 Example - Elimination Method Solve this system:8x + 2y = 4  equation 1 -2x + 3y = 13  equation 2 Step 1)Multiply equation 2 by 4 so coefficients of x differ only in sign. -8x + 12 y = 52 Step 2)Add the revised equations for x and solve. 8x + 2y = 4 -8x + 12 y = 52 14y = 56 y = 4 Step 3)Substitute the value of y into one of the original equations and solve for x. 8x + 2(4) = 4 x = -.5Solution: (-.5, 4)

6 Example 3 Solve this linear system:x – 2y = 4  equation 1 3x – 6y = 8  equation 2 Because the coefficient of x in the first equation is, use the substitution method. Step 1)Solve for x:x = 2y + 4 Step 2) Substitute:3(2y + 4) – 6y = 8 6y + 12 – 6y = 8 12 = 8 Because the statement is never true, there is no solution

7 Example 4 Solve this linear system:4x – 10y = 8  equation 1 -14x + 35y = -28  equation 2 Step 1) 4x – 10y = 8  x7  28x – 70y = 56 -14x + 35y = -28  x2  -28x + 70y = -56 Step 2) Add0 = 0 Because the statement 0 = 0 is always true, there are infinitely many solutions

8 Homework Page 164-167/ 4-14 even, 56, 72


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