 #  Solve one of the equations for one of the variables.  Isolate one of the variables in one of the equations.  Choose whichever seems easiest.  Substitute.

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 Solve one of the equations for one of the variables.  Isolate one of the variables in one of the equations.  Choose whichever seems easiest.  Substitute the expression for the variable in the other equation.  Use substitution when a system has at least one equation that can be solved quickly for one of the variables.

 Solve the following system:  3y + 4x = 14  -2x + y = -3  The second equation looks easiest to solve for y  So y = 2x – 3  Substitute 2x – 3 for y in the other equation  3(2x – 3) + 4x = 14  Solve for x  x = 2.3  Now substitute 2.3 for x in either equation  y = 1.6  The solution is (2.3, 1.6)

 Solve the following system by substituting:  y = 3x and x + y = -32  (-8, -24)

 Solve the system using substitution  6y + 5x = 10  x + 3y = -7  (8, -5)

 A large snack pack costs \$5 and a small costs \$3. If 60 snack packs are sold, for a total of \$220, How many were large and how many were small?  Let x = large and y = small  Money: 5x + 3y = 220  Amount sold: x + y = 60  Solve: (20, 40)  20 large and 40 small

 Add or subtract two linear equations in order to eliminate one of the variables.  Look for whichever is easiest to cancel by adding or subtracting.  Answers should still be ordered pairs.

 2x + 5y = 17  6x – 5y = -9  Cancel 5y using addition.  Now we have 8x = 8  x = 1  Substitute x = 1 into either equation to find y.  (1, 3)

 x + y = 101  2.5x + y = 164  Use subtraction › Change all the signs of the second equation.  (42, 59)

 Odds p.146 #15-25,31,33,41

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