Determine how much was invested at each rate Let x represent the amount

Determine how much was invested at each rate let x

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Determine how much was invested at each rate. Let x represent the amount invested at 4% Let y represent the amount invested at 5% Total amount invested is $5 000 so, x + y = 5 000 Recall that I = prt or interest = principal x rate x time For 1 year: Interest earned on x dollars invested at 4% is 0.04x Interest earned on y dollars invested at 5% is 0.05y Total interest earned is $230 so, 0.04x + 0.05y = 230 For this problem, the linear system is: x + y = 5 000 0.04x + 0.05y = 230 This system can be solved by isolating either x or y in the first equation then substituting into the second equation. On your own , complete the solution to find that $2 000 will be invested at 4% and $3 000 will be invested at 5%. The correct solution is: (2000,3000), or: $2000.00 was invested at 4%, and $3000.00 was invested at 5%. Time for an exercise! In your textbook, read pages 417 424. On pages 425 426, do questions 4 22. Example 2 on page 420 can also be viewed as an animation in the e-book. Look for Lesson 7.4 Using a Substitution Strategy to Solve a System of Linear Equations Example 2a and Example 2b .
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10C C5-6 LG Systems of Linear Equations Page 19 of 42 Solving linear systems of equations by substitution may require us to multiply an equation by a non-zero number in order to eliminate any fractional coefficients. It is important to note the following. SOLVING LINEAR SYSTEMS OF EQUATIONS USING ELIMINATION There is one final algebraic strategy for solving linear systems of equations that we will study. The elimination strategy requires us to eliminate one variable (temporarily) in order to solve for the variable that remains. We then substitute the known value into one of the original equations to solve for the variable that was originally eliminated. When using the elimination strategy, the two equations are either added or subtracted in order to eliminate one variable. Carefully study the next example. Example Solve the following system using the elimination strategy . 3x 5y = -9 4x + 5y = 23 Do you notice that the coefficients of y in the two equations are opposites or additive inverses of one another? Since opposites or additive inverses add to 0 , adding the two equations will eliminate y . 3x 5y = -9 4x + 5y = 23 7x + 0y = 14 7x = 14 one equation in one variable x = 2 We now substitute x = 2 into either equation to solve for y. Let‟s substitute into the second equation because there are no negative signs! Multiplying or dividing both sides of an equation by a non-zero number does not change the solution but simply produces an equivalent equation. Line up like terms vertically then add! y is eliminated!
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10C C5-6 LG Systems of Linear Equations Page 20 of 42 4x + 5y = 23 4(2) + 5y = 23 8 + 5y = 23 5y = 15 y = 3 The solution is (2, 3) . Let‟s check by substituting the point into both equations: Let‟s check by substituting the point into both equations: 3x 5y = -9 4x + 5y = 23 3(2) 5(3) -9 4(2) + 5(3) 23 6 15 8 + 15 -9 = -9 23 = 23 LS = RS LS = RS Therefore, (2, 3) is the solution. You can also check by graphing but remember to first rewrite each equation in the form y = mx + b .
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