supplement for UND s Math 107 precalculus course


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1 Systems of Equations supplement for UND s Math 107 precalculus course by Dave Morstad Introduction A system of equations is any group of equations. A solution to a system of equations is a solution that works in every equation in the group. For example, in the following system of equations: x 5y = 8 x + y the solution is x = 1, y = because these values work in both equations. However, x = 3, y = 0 is not a solution for the system even though it is a solution for the second equation. Three methods of solving sytems of equations will be discussed in this supplement: the graphical method, substitution, and elimination. The Graphical Method The graphical method consists of graphing every equation in the system and then using the graph to find the coordinates of the point(s) where the graphs intersect. The point of intersection is the solution. Example 1. Use the graphical method to solve the following system of equations. x y = x + y Solution: Carefully graph both equations very precisely. If you don t graph neatly, your point of intersection will be way off. A graph is on the next page. The solution is x = 1, y = 1. Substitute these values into both equations to check the solution. page 1
2 Solution to Example 1. Now You Try It #1. Use the graphical method to solve. x 3y = 0 x + 3y = 9 The graphical method also works if the equations are not all linear. Example. Use the graphical method to solve y = x x + y = Solution: Carefully graph both equations very precisely. The graph below reveals two solutions: x = 1, y = 1 and x =, y =. Check both solutions. page
3 Now You Try It #. Use the graphical method to solve x + y = 9 x + y The graphical method is particularly helpful in understanding why a system of equations might have no solution. Example 3. Use the graphical method to solve x + y = 1 x + y = Solution: After carefully graphing the equations, as shown below, it becomes apparent the two lines are parallel. Since parallel lines do not intersect, there is no solution to this system of equation. A system of equations that yields parallel lines is said to be inconsistent: the lines are parallel, the equations are inconsistent. Now You Try It #3. Use the graphical method to solve x 3y = 6 x + 6y = 6 As easy as the graphical method is to use, it is not very helpful when the solution is not integers. Imagine trying to use the graphical method to find the 37 16,. solution if the point of intersection is ( ) Graphical Method Strengths: visualizing possible solutions. Graphical Method Weakness: imprecise page 3
4 The Substitution Method To solve a system of equations using substitution, first solve one of the equations for a variable, then substitute it into the other equation(s). Always substitute into the other equation(s) and always use parentheses. Example. Use substitution to solve the following sytem of equations. + y = 8 x + 5y Solution. Step 1. Solve one of the equations for a variable. Let s solve Eq. 1 for y. + y = 8 y = 8 3 y = Step. Substitute this into the other equation (Eq. ) and simplify. x + 5y x 3 x + 5( x) 15 x + 10 x 15 x x = 7 3 x = 7 x = 7( ) x = Step 3. Backsubstitute this value of x into the equation from Step 1. to solve for y. y = y = y = y = x ( ) The solution is x =, y =. You should put these 3 3 values into both original equations to check your work. page
5 Now You Try It #. Use substitution to solve y = 7 x 3y = 9 If there is no solution to the system of equations (parallel lines), the substitution method will result in something nonsensical, such as =. When you are using the substitution method and end up with an equation that is false, remember that it just means there is no solution. One of the strengths of the substitution method is that it works for systems of equations that are difficult or impossible to graph. For example, the system x + y z = y + z = x y + 3z is not three lines, but rather is three planes. Each equation is a different plane. Solving requires finding the point where the three planes intersect. The diagram below illustrates this. Using substitution allows you to ignore the threedimensional or higher dimensional aspects of the graphical interpretations of such equations. Example 5. Use substitution to solve the following system of equations. x + y z = y + z = x y + 3z page 5
6 Solution: Step 1. Solve an equation for one of the variables. Let s solve Eq. 1 for y. x + y z =, so y = x + z Step. Substitute this value for y into BOTH of the OTHER equations. First into Eq.. When you simplify, it becomes a simple equation with two variables. (Use parentheses to keep the signs straight) : y + z = ( x + z) + z = + ( + x z) + z = 7x + 3z = 7x 3z = Do the same substituting into Eq. 3. x y + 3z x ( x + z) + 3z x + ( + x z) + 3z x + z x z = 1 Step 3. Now use these two resulting equations, 7 x 3z = x z = 1 to solve for x and z. Using substitution all over again on these two equation should give you x = 1, z. Step. Now go back up to Step 1 and backsubstitute these values of x and z into the equation for y. y = x + z y = ( 1) + ( 3) y = The solution is x = 1, y =, z. Use the original three equations to check this solution. page 6
7 Now You Try It #5. Use substitution to solve y + z = x + 3y + z = 5 + y + 3z = 8 The Elimination Method Elimination consists of adding equations together to eliminate variables. Sometimes you have to multiply equations by a number before you add them. The goal is to end up with one equation that has just one variable. Then you can use backsubstitution to solve for the other variable(s). When using elimination, eliminate one variable at a time. It is also important to write down instructions that indicate how you are manipulating the equations going from step to step. Example 6. Use elimination to solve 5x + 3y = 7 x + 5y = 3 Solution Method1. Let s get rid of the y variable. Multiplying the first equation by 5 and the second equation by ( 3) will make the y terms cancel. 5 ( eq. 1) 3 (. ) 5x + 3y = 7 eq 5x + 15y = 35 x + 5y = 3 1x 15y = 9 Now add the two new equations: 5x + 15y = 35 + ( 1x 15y = 9) y = 6 So x =. Use backsubstitution to find that y = 1. This works, but Method below is a little quicker and it generalizes to larger systems of equations much more easily. Method is what you ll want to use most of the time. page 7
8 Solution Method. In this method, you leave one equation the same and replace the other equation(s). Add 5 times the first equation to 3 times the second equation. Put the result in for the second equation. 5x + 3y = 7 x + 5y = 3 ( 5)( eq. 1) + ( 3)( eq. ) ( eq. ) 5 x + 3 y = y = 6 Next multipy the second equation so x has a coefficient of 1. Put it in for the second equation. 1 ( 13) ( eq. ) ( eq. ) 5 x + 3 y = 7 x + 0 y = So x =. Once again, use backsubstituion to find that y = 1. The solution is x =, y = 1. Now You Try It #6. Use elimination to solve 6 y 7 x 5y = 11 If there is no solution to the system of equations, the elimination method will yield a false statement, just like substitution does. The real power of the elimination method becomes apparent in larger systems of equations. Make sure you eliminate one variable at a time and be very systematic. Also, once you ve got zeros in front of the variable you re eliminating, avoid doing anything later that will remove those zeros. Example 7. Use elimination to solve x + y z = y + z = 6x y + z = Solution. First let s knock out the y in the second and third equations. x + y z = x + y z = y + z = 6x y + z = ( )( eq. 1) + ( eq. ) ( eq. ) ( eq. 1) + ( eq. 3) ( eq. 3) 7 x + 0 y 3z = x + 0 y + z = page 8
9 Now we can multiply eq. 3 by 1 to give the z term a coefficient of 1. Coefficients of 1 are nice to work with. Trying to make any other coefficient 1 would introduce fractions, which is all right, but they can be messy to work with. x + y z = 7 x + 0 y 3z = x + 0 y + z = ( 1 ) ( eq. 3) ( eq. 3) x + y z = 7 x + 0y 3z = x + 0 y + z = 1 Next add 3 times eq. 3 to eq. and put it in for eq.. x + y z = ( 3)( eq. 3) + ( eq. ) ( eq. ) x + 0y + 0z = 1 x + 0y + z = 1 Equation reveals x = 1. Backsubstitute into eq. 3 to find z. Backsubstitute both x and z into eq. 1 to get y =. The solution is x = 1, y =, z. Now You Try It #7. Use elimination to solve y + 5z = 3 x + 6 y z = 0 x 3y + 5z = 5 This technique of elimination is used extensively in other areas of mathematics, including matrices and linear programming. Systematically and neatly rewriting all the equations in each step is very important in preventing errors and loosing big points on exams. It also is very important in the way this technique is generalized to other problems. page 9
10 Exercises In exercises 11, solve the systems of equations graphically. 1. y = x + y =. x 3y = 0 x + y = y = 1 6 x + y = 0 c + 5d = 8 3c d = x 9 y = 1 x + 3y = 1 3c + d = 7 5c 6d = 1 3. y = x x + y = 0. y y = x = 1 1. x x + y = 3 x y. x + x + = y x y = 5. x + y = + 6 y = 6 6. x y = 1 6 x + 3y = s + t = s s + 7 = t. s s t = 3 + t = y = 6 x + y = 8 x + y = 16 x = x 3y = 11 x + y = 1 x + y = 8 x = y In exercises 5, solve the systems of equations using elimination. Be sure to write out your work neatly with directions between each step. 5. x + 3y = 1 x y = y = 6 x 5y = y = x x + y + y = x 3y = 6 x + 7 y = x + 9 y = 7 x 5y = 7 1. y = x x + y y + 3 = 0 9. y = 5 6 x + y = x 3y = 0 1x 9 y = In exercises 13, solve the systems of equations using substitution.. Don t forget to use parentheses. 31. x 3y = 8 x + 1 y = x y 15x 6 y = y = 5 x + y = 9 1. x + y = 7 5x + y = y = 11 + y = x + 3y x y = 15. a + b = 8 a = b 16. a + b = 0 a b = x y + 3z = 7 x + y 3z = 8 y + z = 7 page 10
11 36. x + y 3z = 5 3y + z = 10 5x + y 3z = y 9z 5x 7 y + 10z = 11x + y 6z = x 3y + 5z = 8 6x + 8 y z 0 x + 5y 9z = x 3y + z = 1 x + 3y z = x 5y + z = 0. x + y 3z = 3 3y + z = 3 x y + 5z = 1. x + y z + 3w = 11 x + 3 y + 3 z w = 5 x + y 5z w = 1 y + z + 3 w = 1. x y + z + w = + 5y 3z + w = x y + z w = 16 x 3y + z 3w = page 11
12 Solutions to Now You Try It. NYTI #1. NYTI #. NYTI #3. NYTI #. x = 15, y = 13 NYTI #5. x = 0, y = 1, z = NYTI #6. x = 3, y = NYTI #7. x = 1, y, z = 0 page 1
13 Solutions to odd numbered exercises page 13
14 x =, y = 15. a =, b = and a =, b = no solution 19. c =, d = x = 0, y = 3 and x, y = 0 3. no solution 5. x =, y = 1 7. x =, y = 9. no solution 31. if x = any number, then y = x 33. x =, y = 1 and x =, y = x = 1, y = 0, z = 37. x = 0, y =, z = no solution 1. x = 1, y =, z, w = 3 3 page 1
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