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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
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
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
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)
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”
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
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):•
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
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:
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%
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%
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
EXAMPLE
• • • Solve equation (3) for 3x:
• Substitute -6 – 4y for 3x in equation (4):• • Collect like terms, solve for y:• • Substitute into equation (4):•
EXAMPLE (YOU DO)
• Solve this linear system by substitution:
• Solution:
WALL QUIZ!
HOMEWORK
• For homework, please complete the handout I provide you for next class