# Section 5.1 Solving Systems of Linear Equation Using Substitution and Elimination

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1 204 Section 5.1 Solving Systems of Linear Equation Using Substitution and Elimination A system of equations consists of two or more equations. A solution to a system of equations is a point that satisfies all the equations in the system. In this chapter, our focus will be solving systems of linear equations. If we are solving a system of two linear equations in two variables, there are three possible types of systems we could have: Case #1 Case #2 Case #3 Consistent System Inconsistent System Consistent System The equations are The equations are The equations are independent. Independent. dependent The solution is the The lines do not Every point on the where the lines intersect, so there line is a solution. We intersect. In this case, is no solution, }. write the solution as: the solution is (2, 3). The lines are different, (x, y) ax + by = c} The lines have but the lines do have The lines are the different slopes. the same slope. same or coincident. To solve a system of equations, we can graph each equation on the same axes and see where they intersect. The point of intersection will be the solution to the system. We can then check the answer by plugging in the solution into both equations and verify that the point makes both of them true. Solving the following system by graphing: Ex. 1 y = 2 3 x 1 x + 2y = 5

2 205 The first equation is already solved for y. 1) y = 2 3 x 1 m = 2, y-int: (0, 1) 3 To solve the second equation for y, subtract x from both sides and divided both sides by 2: 2) x + 2y = 5 y = 1 2 x m = 1 2, y-int: (0, 5 2 ) The lines intersect at (3, 1) so the solution is (3, 1). This is a consistent system and the equations are independent. You can always check the answer by plugging in the solution into each equation and see if it satisfies all the equations in the system: y = 2 3 x 1 x + 2y = 5 1 = 2 (3) 1 3 (3) + 2(1) = 5 1 = 1 true 5 = 5 true So, the solution checks. Objective #1: Solving Systems of Equations Using Substitution. Using the substitution method, we will solve one equation for one variable and substitute that expression for that variable in the other equation. This will give us one linear equation with one variable. We solve that equation and that will yield the value for one of the coordinates of our solution. We take that answer, substitute it into one of the two original equations and solve to find the value for the other coordinate of our solution. Solve by substitution: Ex x 1 6 y = x y = 3 5

3 It is easiest to clear fractions first. The L.C.D. of equation #1 is 6 and the L.C.D. of equation #2 is 15. 1) 2 3 x 1 6 y = ( 3 x 1 ) 6( 6 y ) = 6( (multiply both sides by 6) 5 6 ) 4x y = 5 2) 1 5 x y = 3 (multiply both sides by 15) 5 15( 1 5 x 1 ) + 15( 3 y 3 ) = 15( 5 ) 3x + 5y = 9 Now, solve equation #1 for y: 4x y = 5 y = 4x + 5 y = 4x 5 Finally, replace y in equation #2 with 4x 5: 3x + 5(4x 5) = 9 (distribute & combine like terms) 17x 25 = 9 (solve) 17x = 34 x = 2 Since y = 4x 5, then y = 4(2) 5 = 8 5 = 3. The solution is (2, 3). Ex. 3 6x 2y = 4 10y + 5x = 0 Solve equation #2 for x: 10y + 5x = 0 5x = 10y x = 2y Replace x in equation #1 with 2y: 6( 2y) 2y = 4 (multiply and combine like terms) 14y = 4 (solve) y = 2 7 The solution is ( Since x = 2y, then x = 2( 2 7 ) = , 2 7 ). 206 Objective #2: Solving Systems of Equations Using Elimination. A variant of the Addition Property of Equality that if you have two equations, you can add the left sides and then add the right sides of the equations to produce an equivalent equation. In other words, If A = B and C = D, then A + C = B + D. We will therefore manipulate one or both equations until the coefficients of one of the variable terms are opposites. This is done by finding the L.C.M. of the coefficients of the term we are trying to eliminate.

4 We will then add the equations together, eliminating the terms with opposite coefficients. This will give us one linear equation with one variable. We solve that equation and that will yield the value for one of the coordinates of our solution. We then take that answer, substitute it into one of the two original equations and solve to find the value for the other coordinate of our solution. Solve by elimination Ex. 4 6x + 5y = 7 x y = 4 We need to manipulate the equations so that the coefficients of one of the variable terms are opposites. Let s work on getting rid of y, the easiest way to do this is to multiply the second equation by 5: x y = 4 5x 5y = 20 Now, the coefficients of the y-terms are opposites, so add the left sides and the right sides. 6x + 5y = 7 5x 5y = 20 11x = 27 (solve) x = Substitute into equation #2: x y = y = y = So, y = Thus, the solution is ( 27 11, ). y = Ex. 5 3x 11 = 2( y + 4) 7(3 y) = 4(x 5) The first step is to simplify the equations and write them in standard form: 3x 11 = 2( y + 4) 7(3 y) = 4(x 5) 3x 11 = 2y y = 4x 20 3x + 2y 11 = 8 4x 20 = 21 7y 1) 3x + 2y = 19 4x + 7y 20 = 21 2) 4x + 7y = 41 Let s eliminate the x-terms. The LCM of 3 and 4 is

5 1) 3x + 2y = 19 (multiply by 4) 12x + 8y = 76 2) 4x + 7y = 41 (multiply by 3) 12x 21y = 123 Now, the coefficients of the x-terms are opposites, so add the left sides and the right sides. 12x + 8y = 76 12x 21y = y = 47 y = To find x, plug y = 47 into equation #1: 13 3x + 2y = 19 3x + 2( ) = 19 3x + 94 = 19 (multiply both sides by 13 to clear fractions) 13 39x + 94 = x = 153 x = = Hence, ( 13, ) is the solution. 208 Objective #3 Systems of equations that are inconsistent. Solve the following: Ex. 6 4x + 2y = 3 10x + 5y = 2 Solve equation #1 for y: 4x + 2y = 3 2y = 4x 3 y = 2x 1.5 In equation #2, replace y with 2x 1.5: 10x + 5y = 2 10x + 5( 2x 1.5) = 2 (distribute) 10x 10x 7.5 = 2 (combine like terms) 7.5 = 2 In the context of solving systems of equations, it means that the lines are parallel. Hence, the system has no solution. Objective #4 System of dependent equations.

6 Solve the following: Ex. 7 12x 6y = 15 4x = 2y + 5 Let s get rid of the x-terms. We will multiply equation #2 by 3: 4x = 2y + 5 (multiply by 3) 12x = 6y + 15 Now, the coefficients of the x-terms are opposites, so add the left sides and the right sides. Be careful to align the equal signs: 12x 6y = 15 12x = 6y y = 6y (add 6y to both sides) 0 = 0 The equations are the same. In this case, we will either solve the equation for x or for y and use our set builder notation to express the answer: (x, y) x = 0.5y 1.25} or (x, y) y = 2x + 2.5} 209 Objectives #5-7: System of three linear equations. The graph of the equation in the form ax + by + cz = d is a plane in three dimensional space. The points in three-dimensional space have three coordinates, x, y, and z. We will list them in alphabetical order: (x, y, z). In solving a system of three linear equations in three variables, we are not looking at intersecting lines, but intersecting planes. Here are some possible types of systems we could have: Type 1: The solution is a point:

7 Type 2: The solution is a line: Two planes are the same. 210 OR Type 3: There is no solution: OR OR Two planes are the same. OR

8 211 Type 4: Solution is a plane: All three planes are the same. The key to solving a system of three equations with three variables is the reduce it to a system of two linear equations with two variables. To accomplish this, pick a variable to eliminate. We pick two equations and use either elimination or substitution to get rid of that variable. We then pick a different pair of equations in the system and do the same thing. After it is reduced to a system of two linear equations in two variables, we then just solve it like before. Solve the following: Ex. 8 1) x + y + z = 3 2) 2x y + 3z = 16 3) x + 2y + 2z = 3 Let s first try to eliminate x. So, take equation #1 plus equation #3: 1) x + y + z = 3 3) x + 2y + 2z = 3 4) 3y + 3z = 6 We will call this equation #4. Now, choose a different pair in the system to eliminate x. So, take equation #2 plus 2 equation #3: 2) 2x y + 3z = ) 2x + 4y + 4z = 6 5) 3y + 7z = 22 We will call this equation #5 Looking at equation #4 and #5, it is easiest to eliminate the y-terms. So, take equation #4 plus 1 equation #5: 4) 3y + 3z = 6 1 5) 3y 7z = 22 4z = 16 (solve) z = 4 Plug in z = 4 into equation #4 (or #5): 4) 3y + 3(4) = 6 (solve) 3y + 12 = 6 3y = 6 y = 2

9 212 Now, plug in y = 2 & z = 4 in equation #1 (or #2 or #3): 1) x + ( 2) + (4) = 3 (solve) x + 2 = 3 x = 1 Thus, the solution is (1, 2, 4). Ex. 9 1) x 3y + 2z = 1 2) 2x + y z = 3 3) 6x 4y + 2z = 5 First, eliminate z. Take eqn. #1 plus 2 eqn. #2: 1) x 3y + 2z = 1 2 2) 4x + 2y 2z = 6 4) 5x y = 7 Next, take 2 eqn. #2 plus eqn. #3: 2 2) 4x + 2y 2z = 6 3) 6x 4y + 2z = 5 5) 10x 2y = 11 Looking at equations #4 and #5, let s get rid of x-terms. Take 2 eqn. #4 + eqn. #5: 2 4) 10x + 2y = 14 5) 10x 2y = 11 0 = 3 Not possible This system has no solution. Ex. 10 1) 5 2 x 3y + 7z = ) 3 x + y 2 3 z = 2 3 3) 2x y + 4z = 1 First clear fractions by multiply both sides of eqn. #1 by 2 and both sides of eqn. #2 by 3 (we will also list eqn. #3): 2 1) 5x 6y + 14z = 5 call this our new equation #1 3 2) x + 3y 2z = 2 call this our new equation #2 3) 2x y + 4z = 1 Let s eliminate x. Eqn. #1 + 5 eqn. #2: 1) 5x 6y + 14z = 5 5 2) 5x 15y + 10z = 10 4) 21y + 24z = 15

10 213 2 Eqn. #2 + eqn. #3: 2 2) 2x 6y + 4z = 4 3) 2x y + 4z = 1 5) 7y + 8z = 5 Looking at equation #4 and #5, let s eliminate the y-terms. Eqn. #4 + 3 eqn. #5: 4) 21y + 24z = ) 21y 24z = 15 0 = 0 This means that there are an infinite number of solutions. To express our answer, we will need to take two of the variables and rewrite then in terms of the third variable. To do this, we will first take equation #5 and solve it for y: 7y + 8z = 5 7y = 8z 5 y = 8 7 z We will then replace y by 8 7 z and solve for x to get x in terms of z: x + 3( 8 7 z + 5 ) 2z = 2 (distribute and combine like terms) 7 x z = 2 (solve for x) x = 10 7 z 1 7 Our solution is (x, y, z) x = 10 7 z 1 7, y = 8 7 z + 5, z is any real 7 number}. Ex. 11 Maylina buys two packages of Skittles and three bottles of soda for \$5.15. Zoë buys four packages of Skittles and five bottles of soda for \$9.05. Find the cost of one package of Skittles and one bottle of soda. Let S = the cost of one package of Skittles b = the cost of bottle of soda For Maylina, she buys: 2 Skittles + 3 bottles of soda = \$5.15 1) 2S + 3b = 5.15 For Zoë, she buys: 4 Skittles + 5 bottles of soda = \$9.05 2) 4S + 5b = 9.05 So, our system is:

11 1) 2S + 3b = ) 4S + 5b = 9.05 Let s multiply equation #1 by 2 and add: 2 1) 4S 6b = ) 4S + 5b = 9.05 b = 1.25 So, b = 1.25 Substitute 1.25 for b in equation #1 and solve: 2S + 3(1.25) = S = S = 1.40 S = 0.70 So, a package of Skittles costs \$0.70 & a bottle of soda costs \$ Ex. 12 Juanita decided to divide \$8800 into two investments, one in bonds that ended up earning 8% annual interest and one in a mutual fund that ended up yielding 12% annual interest. If the interest from both was \$924, how much did she invest in each? Let b = the amount invested in bonds. m = the amount invested in a mutual fund Since there was total \$8800 invested, the amount invested in a mutual fund plus the amount invested in bonds was \$8800: 1) b + m = 8800 The interest earned on the bonds was 8% times the amount invested in bonds or 0.08b and the interest earned on the mutual fund was 12% times the amount invested in the mutual fund or 0.12m. The total interest is 2) 0.08b m = 924. So, our system is: 1) b + m = ) 0.08b m = 924. Let s solve equation #1 for m: b + m = 8800 m = 8800 b Now, replace m is equation #2 by 8800 b: 0.08b (8800 b) = 924 (distribute & combine like terms) 0.04b = b = 132 b = 3300 m = 8800 b = 5500 So, \$3300 was invested in bonds and \$5500 in a mutual fund.

12 Ex. 13 Find the equation of the quadratic function in the form y = ax 2 + bx + c whose graph passes through the points ( 2, 29), (3, 14), and (1, 2). The point ( 2, 29) yields: 29 = a( 2) 2 + b( 2) + c or 1) 4a 2b + c = 29 The point (3, 14) yields: 14 = a(3) 2 + b(3) + c or 2) 9a + 3b + c = 14 The point (1, 2) yields: 2 = a(1) 2 + b(1) + c or 3) a + b + c = 2 Thus, we have a system of three equations: 1) 4a 2b + c = 29 2) 9a + 3b + c = 14 3) a + b + c = Let s first try to eliminate c. Take equation #1 plus 1 times equation #3: 1) 4a 2b + c = ) a b c = 2 3a 3b = 27 (divide both sides by 3) 4) a b = 9 We will call this equation #4. Now, choose a different pair in the system to eliminate c. So, take equation #2 plus 1 equation #3: 2) 9a + 3b + c = ) a b c = 2 5) 8a + 2b = 12 We will call this equation #5 Looking at equation #4 and #5, it is easiest to eliminate the b-terms. So, take 2 equation #4 plus equation #5: 2 4) 2a 2b = 18 5) 8a + 2b = 12 10a = 30 (solve) a = 3 Plug in a = 3 into equation #4 4) 3 b = 9 (solve) b = 6 b = 6 Now plug in a = 3, and b = 6 into equation #3 and solve: c = c = 2 c = 5 Thus, the quadratic function is y = 3x 2 + 6x 5.

13 216 Ex. 14 Leroy had a total of \$50,400 invested in three different mutual funds, one that paid 9% interest, another that paid 4% interest, and a third that paid 7% interest. If the total interest was \$3570 and the amount invested at 9% was \$3200 less than twice the amount invested at 4%, how much was invested at each rate? Let n = amount invested at 9% and let f = amount invested at 4% and let s = amount invested at 7% The total amount invested is equal to the amount invested at 9% plus the amount invested at 4% plus the amount invested at 7%: 1) n + f + s = The interest on the 9% investment was 0.09n, the interest on the 4% investment was 0.04f, and the interest on the 7% investment was 0.07s. Since the total interest was \$3570, then 2) 0.09n f s = 3570 The amount invested at 9%, n, was 3200 less than twice the amount invested at 4%, f: n = 2f 3200 So, our system is: 1) n + f + s = ) 0.09n f s = ) n = 2f 3200 n 2f = 3200 Since equation #3 only has n and f in it, we will eliminate s from equations #1 and #2: ) 0.07n 0.07f 0.07s = ) 0.09n f s = ) 0.02n 0.03f = 42 Now, eliminate n from equation #3 and #4: ) 0.02n f = 64 4) 0.02n 0.03f = f = 106 or f = 10,600 Plug in f = 10,600 into equation #3 to find n: n = 2f 3200 = 2(10600) 3200 = 18,000 Plug in f = 10,600 and n = 18,000 into equation #1 to find s: n + f + s = s = s = s = 21,800 Thus, \$18,000 was invested at 9%, \$10,600 was invested at 4% and \$21,800 was invested at 7%.

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