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Learn how to solve systems of equations by substitution method with examples and practice problems.
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Name:Date:Period:Topic: Solving Systems of Equations by Substitution Essential Question: When is using substitution a better option to find a solution to systems of equations? A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were:The fifth number plus the third number equals fourteen. The fourth number is one more than the second number. The first number is one less than twice the second number. The second number plus the third number equals ten. The sum of all five numbers is 30.What were the five numbers and in what order? Warm-Up:
Name:Date:Period:Topic: Solving Systems of Equations by Substitution Essential Question: When is using substitution a better option to find a solution to systems of equations? Find the solution to the following system of equations by GRAPHING:y = 2x + 22x + 4y = 8 Warm-Up: Period 1, 5
GIVEN EXAMPLE: y= 4x 3x+y=-21 STEP1: y=4x (Already solved for y) STEP 2: Substitute into second equation: 3x + y = -21 becomes:
GIVEN EXAMPLE: y= 4x 3x+y=-21 STEP 3: Solve for the variable 3x + y = -21 becomes 3x + 4x = - 21
GIVEN EXAMPLE: y= 4x 3x+y=-21 STEP 4: Solve for the other variable use x=-3 and y=4x
y = 4x 3x + y = -21 Step 5: Check the solution in both equations. Solution to the system is (-3,-12). 3x + y = -21 3(-3) + (-12) = -21 -9 + (-12) = -21 -21= -21 y = 4x -12 = 4(-3) -12 = -12
Solving a system of equations by substitution Pick the easier equation. The goal is to get y= ; x= ; a= ; etc. Step 1: Solve an equation for one variable. Step 2: Substitute Put the equation solved in Step 1 into the other equation. Step 3: Solve the equation. Get the variable by itself. Step 4: Plug back in to find the other variable. Substitute the value of the variable into the equation. Step 5: Check your solution. Substitute your ordered pair into BOTH equations.
Solve the system using substitution x + y = 5 y = 3 + x Step 1: Solve an equation for one variable. The second equation is already solved for y! Step 2: Substitute x + y = 5x + (3 + x) = 5 2x + 3 = 5 2x = 2 x = 1 Step 3: Solve the equation.
Solve the system using substitution x + y = 5 y = 3 + x x + y = 5 (1) + y = 5 y = 4 Step 4: Plug back in to find the other variable. (1, 4) (1) + (4) = 5 (4) = 3 + (1) Step 5: Check your solution. The solution is (1, 4). What do you think the answer would be if you graphed the two equations?
What about this one?: x + y = 10 5x – y = 2 Step 1: Solve one equation for one variable. Step 2: Substitute the expression from step one into the other equation.
x + y = 10 5x – y = 2 Step 3: Simplify and solve the equation.
x + y = 10 5x – y = 2 Step 4: Substitute back into either original equation to find the value of the other variable.
x + y = 10 5x – y = 2 Step 5: Check the solution in both equations. Solution to the system is (2, 8).
Solve by substitution: Your Turn!
Which answer checks correctly? 3x – y = 4 x = 4y - 17 • (2, 2) • (5, 3) • (3, 5) • (3, -5)
Additional Practice: • Page 371 (1 – 4)
Wrap-Up: • Determine what problems from previous class could you easily solve by utilizing the substitution strategy • Vocabulary Review • Summary • Home-Learning # 7 • Page 371 – 373 (12, 17, 28, 35, 44)