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Published byCarlos Larsen Modified over 2 years ago

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

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

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

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

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

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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. )

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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. )

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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

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

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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

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

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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

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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)

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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)

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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

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

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

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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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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.

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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.

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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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