# 7.4 – SOLVING SYSTEMS OF LINEAR EQUATIONS USING A SUBSTITUTION STRATEGY SYSTEMS OF LINEAR EQUATIONS.

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7.4 – 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

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

EXAMPLE  “expression for x”

EXAMPLE

EXAMPLE

EXAMPLE

EXAMPLE Create a linear system to model this situation: Mr. Mennie 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. Mennie invest at each rate? Solution: Given:Linear System 2 investmentsLet x dollars represent the amount invested at 8% Let y dollars represent the amount invested at 10% Total investment is \$2000x+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 \$1900.08x + 0.10y = 190

EXAMPLE

EXAMPLE Create a Linear System to model this situation: Mr. Nishi 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. Nishi invest at each rate? Solution: x + y = 1800 (1) 0.035x + 0.045y = 73 (2) Mr. Nishi invested \$800 at 3.5% and \$1000 at 4.5%

SUBSTITUTION STRATEGY

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EXAMPLE 55

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