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The Substitution Method A method to solve a system of linear equations in 2 variables.

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Presentation on theme: "The Substitution Method A method to solve a system of linear equations in 2 variables."— Presentation transcript:

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2 The Substitution Method A method to solve a system of linear equations in 2 variables

3 When you Solve a system of equations you are looking for a solution that will solve every equation in the system (group).

4 ??? A linear equation Ax + By = C has an infinite number of solution? ???

5 How do we start with two equations, each having an infinite number solutions, and find the common solution (if any)???

6 This Method will create one combined equation with only one variable. This is the kind of equation that you can solve!

7 Substitution Method - Step One Solve one equation for one of the variables Choose either equation and solve for either variable. (Choose the easiest one the solve. )

8 Step One - Solve one equation for one of the variables 2x + y = 53 x + 5y = Choose either equation and solve for either variable. (Choose the easiest one the solve. )

9 Step One - Solve one equation for one of the variables In this problem you could have solved equation #1 for y or solved equation # 2 for x. -5y -2x x = - 5y +139 y = -2x +53 2x + y = 53 1 x + 5y = 139 2

10 Substitution Method - Step Two Step One - Solve one equation for one of the variables Substitute this expression in the other equation and solve.

11 2x + y = 53 1 x + 5y = y = -2x +53 If you solve for y in the first equation take this expression and substitute it in for y in the 2 nd equation x + 5 (-2x +53) = 139 joined

12 This will create one combined equation with only one variable. This is the kind of equation that you can solve!

13 Substitute this expression in the other equation and solve. 2x + y = 53 1 x + 5y = 139 y = -2x +53 Now solve for x x + 5 (-2x +53) = 139 joined x + -10x+265 = x = x = -126 x = -126/-9=14

14 Find the corresponding value of the other variable. After solving the combined equation …. (Substitute the value you found in step 2 to back into the equation)

15 Substitute the value you found for the first variable back into one of the original equations y = -2x +53 2x + y = 53 1 x + 5y = x + 5 (-2x +53) = 139 x + -10x+265 = x = x = -126 x = -126/-9 = 14 From the last step you found that x was 14. Take this value and plug it back into one of the original equations and find y. y = -2x +53 y = -2(14)+53 y = =25 (x, y ) = (14,25)

16 x - 4y = 5 1 3x + 2y = 113 x = 4y +5 3 ( 4y +5 ) + 2y = Substitute the value you found for the first variable back into one of the original equations x = 4y +5 x = 4(7) +5 x = 28+5=33 (x, y ) = (33,7) Solve one of the equations for one of the variables Substitute this expression into the other equation 12y +15+2y = y + 15 = y = 98 y = 98/14 = 7 and solve. 3 ( ) + 2y = 113

17 Step One - Solve one equation for one of the variables Step Two - Substitute this expression in the other equation and solve. Step Three - Find the other variable (Substitute value back into one of the equations)

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19 Im thinking of two numbers. One number is one less then twice the other. The difference of the numbers is 18. Find the numbers.

20 Let x and y represent the numbers. Write two equations to represent the relationships. y=2x-1 y-x=18

21 y = 2x-1 1 y - x = 18 y = 2x - 1 2x x = 18 2 Substitute the value you found for the first variable back into one of the original equations y = 2x - 1 y = 2(19) - 1 x = 38-1=37 (x, y ) = (19,37) Solve one of the equations for one of the variables Substitute this expression into the other equation x-1 = 18 x = 19 and solve.

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23 John had all dimes and quarters worth $5.45 If he had 35 coins in all, find out how many of each coin he had.

24 Let d = the # of dimes and q = the # of quarters Write two equations. d+q=3510d+25q=545 John had all dimes and quarters worth $5.45 If he had 35 coins in all, find out how many of each coin he had.

25 d + q = d + 25q = 545 q = 35 - d 10d + 25( 35 - d ) = Substitute the value you found for the first variable back into one of the original equations q = 35 - d q = q = 13 # of dimes = 22 # of quarters =13 Solve one of the equations for one of the variables Substitute this expression into the other equation 10d d = d = d = -330 d = -330/-15 = 22 and solve.


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