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LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS
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LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

Dec 24, 2015

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Felicity Short
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Page 1: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

L E S S O N 3 – S O LV I N G S Y S T E M S O F L I N E A R E Q U A T I O N S U S I N G A S U B S T I T U T I O N S T R A T E G Y

SYSTEMS OF LINEAR EQUATIONS

Page 2: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

TODAYS OBJECTIVES

• Solve problems that involve systems of linear equations in two variables, graphically and algebraically, including:• Explain a strategy to solve a system of linear

equations• Solve a problem that involves a system of linear

equations• Determine and verify the solution of a system

of linear equations algebraically

Page 3: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

SOLVING A SYSTEM OF LINEAR EQUATIONS ALGEBRAICALLY

• In the last lesson, you solved linear systems by graphing• This strategy is time consuming and you

can only approximate the solution• We can use algebra to determine an

exact solution• In the next two lessons we will look at two

strategies that use algebra to solve linear systems:• Substitution Strategy• Elimination Strategy

Page 4: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

SUBSTITUTION STRATEGY

• Using a substitution strategy, we transform a system of two linear equations into a single equation in one variable, then we solve for that variable• (2 equations, 2 variables) (1 equation, 1 variable)

Page 5: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE

• Consider this linear system:

• In equation (2), the variable x has a coefficient 1. So, solve equation (2) for x.• • The solution of the system is the point of

intersection of the graphs of the two lines, so the x-coordinate must satisfy both equations.• We substitute the expression for x into equation

(1)•

“expression for x”

Page 6: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE

• • Now collect like terms, solve for y:• • Substitute y = 5 into one of the original equations

and solve for the other variable:• • • To verify the solution, substitute for both variables

in the original equations:• • • Solution is x = -8, y = 5

Page 7: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE (YOU DO)

• Use the substitution method to solve the linear system:

• Solution:• When no equation in a linear system has a variable

with coefficient 1, it is helpful if there are two like terms where one term is a multiple of the other term:• (3)• Now, solve equation (1) for 3y and substitute into

equation (3):•

Page 8: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE (YOU DO)

• Distribute:• • Collect like terms and solve for x:• • Substitute x = 3 into equation (1) or (2):• • Solution is: x = 3, y = -1

Page 9: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE

• Create a linear system to model this situation:• Mr. Heard invested $2000, part at an annual interest rate of

8%, and the rest as an annual interest rate of 10%. After one year, the total interest was $190

• How much money did Mr. Heard invest at each rate?

• Solution: Given: Linear System

2 investments Let x dollars represent the amount invested at 8%Let y dollars represent the amount invested at 10%

Total investment is $2000

x+y = 2000

x dollars at 8% Interest is 8% of x = 0.08x

y dollars at 10% Interest is 10% of y = 0.10y

Total interest is $190 0.08x + 0.10y = 190

The linear system is:

Page 10: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE

• Solve for y in equation (1):• • Substitute y = 2000 – x into equation (2):

• Distribute, collect like terms, solve for x:• • Substitute x = 500 into equation (1):• • Mr. Heard invested $500 at 8% and $1500 at 10%

Page 11: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE (YOU DO)

• Create a Linear System to model this situation:• Mr. Hazel invested $1800, part at an annual

interest rate of 3.5%, the rest at 4.5%. After one year, the total interest was $73.• How much did Mr. Hazel invest at each rate?• Solution:

• Mr. Hazel invested $800 at 3.5% and $1000 at 4.5%

Page 12: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

SUBSTITUTION STRATEGY

• Sometimes you will come across a linear system with fractional coefficients• It is often helpful to multiply both equations in the

system by the LCD to eliminate the fractions• The resulting equations form an equivalent linear

system• Example:

• In this case, the LCD of equation (1) is 6 and the LCD of equation (2) is 12

Page 13: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE

• • • Solve equation (3) for 3x:

• Substitute -6 – 4y for 3x in equation (4):• • Collect like terms, solve for y:• • Substitute into equation (4):•

Page 14: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

EXAMPLE (YOU DO)

• Solve this linear system by substitution:

• Solution:

Page 15: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

WALL QUIZ!

Page 16: LESSON 3 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

HOMEWORK

• For homework, please complete the handout I provide you for next class