Presentation on theme: " 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."— Presentation transcript:
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)