Applications of System of Equations

Applications of System of Equations Example 1) Mary has 36 coins consisting of nickels and dimes worth $2.75. How many of each coin does she have? We have two unknowns in this problem: the number of nickels and the number of dimes. If we choose to use two different variables, then we need two different equations (it is possible to use only one variable by calling the number of nickels x and the number of dimes 36 - x ). One equation is going to relate the number of coins together and the other equation will relate the values of the coins together. But first we need to label our variables as descriptively as possible: Number of dimes: D Number of nickels: N First we know that the two numbers have to add up to 36 (since we have 36 coins): D + N = 36 We also know that if we calculate the value of the coins we get $2.75 (each dime is worth $.10 and each nickel is worth $.05): .10D + .05N = 2.75 So we have a system of two equations with two variables. We can eliminate the decimals by multiplying the equation by a 100. And then multiplying the top equation by -10 allows us to eliminate the D's . D + N = 36 -10D - 10N = -360 .10D + .05N = 2.75 10D + 5N = 275 Now we can add the two equation and get: -5N = -85 And divide both sides by -5: N = 17 And to find D we'll plug in 17 into any of the original equations: D + 17 = 36 Subtract 17 from both sides: D = 19 Mary has 17 nickels and 19 dimes. Example 2) John has invested $3600 in two banks. Part of his money is invested at 4% and the remainder is invested at 6%. If the interest on each investment is the same, how much is each investment? Let A = The amount invested at 4% Let B = The amount invested at 6% We know that the two investments add up to $3600: A + B = 3600 We also know that the interest from the first bank (.04A) and the second bank (.06B) are the same: .04A = .06B or .04A - .06B = 0 We can solve the system by multiplying the second equation by 100 to eliminate the decimals. Then multiplying the top equation by -4 allows us to add the two equations together and eliminate the A's. A + B = 3600 -4A - 4B = -14400 .04A - .06B = 0 4a - 6B = 0 Now we can add the equations together: -10B = -14400 And divide both sides by -10: B = 1440 Now we'll plug in 1440 in the first equation for B: A + 1440 = 3600 Subtracting 1440 from both sides gives us: A = 2160 John has $1440 invested at 6% and $2160 at 4% Example 3) A chemist wants to mix a 20% salt solution with a 40% salt solution. How many liters of each solution does she need in order to get a 40 liter mixture which is 33% salt. Let x = liters of the 20% solution Let y = liters of the 40% solution We know that the mixture has to be 40 liters: x + y = 40 We also know that the amount of salt in the first solution (.20x) mixed with the salt in the second solution (.40y) should equal to the salt in the final solution ( .33(40) ): .20x + .40y = .33(40) or .20x + .40y = 13.2 (Notice that since we need the amount of salt in the final solution, we have to multiply the 33% by the total amount (40). In general, every time we have a value that is per some amount, we have to multiply it by the amount.) We can solve the system by multiplying the bottom equation by a 10 and the top equation by -2: x + y = 40 -2x - 2y = -80 .20x + .40y = 13.2 2x + 4y = 132 Now we can add the two equations together: 2y = 52 Dividing both sides by 2, we get: y = 26 Plugging 26 in the top equation we get: x + 26 = 40 And subtracting 26 from both sides: x = 14 She needs 26 liters of the 20% solution and 14 liters of the 40% solution. Example 4) A rectangle with perimeter 26 has area 40. Find its dimensions of the rectangle. Let L = length of the rectangle Let W = width of the rectangle The rectangle has perimeter 26: 2L + 2W = 26 The area of the rectangle is 40: LW = 40 So we get the following system: 2L + 2W = 26 LW = 40 The second equation in this system is not linear, so we have to solve this system by substitution method. We can solve the top equation for L: 2L = 26 - 2W L = 13 - W And plug in this expression in the second equation for L: (13 - W)W = 40 13W - W2 = 40 This is a quadratic equation so we have to set it equal to zero 0 = W2 - 13 W + 40 0 = (W - 8)(W - 5) So W = 8 or W = 5 And we need to plug in both of these values for W and find L. W=8 L = 13 - 8 = 5 W=5 L = 13 - 5 = 8 The rectangle has dimensions 5 by 8