Solve Linear Systems Using the Substitution Method Key

Solve Linear Systems Using the Substitution Method

Key Concepts of the Substitution Method • Pick a variable to isolate • Substitute the new expression into the other equation and solve for the variable • Substitute the value into either original equation and solve for the remaining variable

Ex 1: Solve the system using the substitution method 1 2 x + 5 y = -5 2 x + 3 y = -3 Pick a variable to isolate. Substitute the new expression into the other equation and solve for the variable Isolate x for equation 2 x + 3 y = -3 -3 y x = -3 y - 3 Substitute x = -3 y – 3 to equation 1 2 x + 5 y = -5 2( -3 y – 3 ) + 5 y = -5 -6 y – 6 + 5 y = -5 -y – 6 = -5 +6 +6 -y = 1 -1 -1 Substitute the value into either original equation and solve for the remaining variable Substitute y = -1 into equation 2 x + 3 y = -3 x + 3( -1 ) = -3 x – 3 = -3 +3 +3 x=0 y = -1 The linear system has one solution (0, -1)

Ex 2: Solve the system using the substitution method 1 3 x + y = 2 2 6 x + 2 y = 5 Pick a variable to isolate Isolate y for equation 1 3 x + y = 2 -3 x y = -3 x + 2 Substitute the new expression into the other equation and solve for the variable Substitute the value into either original equation and solve for the remaining variable Substitute y = -3 x + 2 into equation 2 6 x + 2 y = 5 6 x + 2(-3 x + 2 ) = 5 6 x – 6 x + 4 = 5 STOP 4 = 5 false Since the statement above is false, the linear system has NO SOLUTION

Ex 3: Solve the system using the substitution method 1 2 x + 5 y = 4 2 4 x + 10 y = 8 Pick a variable to isolate Isolate x for equation 1 2 x + 5 y = 4 -5 y 2 x = -5 y + 4 2 2 x = -5/2 y + 2 Substitute the new expression into the other equation and solve for the variable Substitute the value into either original equation and solve for the remaining variable Substitute x = -5/2 y + 2 into equation 2 4 x + 10 y = 8 4( -5/2 y + 2) + 10 y = 8 -10 y + 8 + 10 y = 8 8 = 8 true STOP Since the statement above is true, the linear system has MANY SOLUTIONS
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