Solving Systems of Equations: Methods and Examples

Math A100 IUSB Sec. 4.1 – 4.3 detailed notes – Week 6 Solving Systems of Equations Systems of Equations are when you have two different lines on the same grid. I am be presenting Sec. 4.1 – 4.3 together (even though section 4.3 is in Week 7), because they give you three different methods for solving these types of questions: 1) By graphing 2) By Substitution 3) By Elimination You WILL have to be able to use Method 1 – Graphing, but you don’t have to learn both Substitution and Elimination – you can use the one you can wrap your head around better. So, for HW, Quizzes, and Tests, if the directions say to either Substitution or Elimination – you can ignore the directions and use the method you are more comfortable with. (How is MML going to know which method you used?? What is a “solution” in a System of Equations? A “solution” is math is a value that makes the equation true/work. In a Systems of Equations, you have two equations (i.e. lines/linear equations), so you are looking for the ONE point that works in both equations. Determining if a point is a “solution” In order for a point/ordered pair to be a solution, it has to work in BOTH equations, i.e. it has to be on both lines. So, to find out, simply plug the x & y coordinates into both equations to see if it works in both. NOTE: If a point/ordered pair works in one but not the other, then it is NOT a solution. Example 1: Is the ordered pair (3, 7) a solution for the following system of equations: y = 2x + 1 y = 5x – 8 Plug (3, 7) in each equations separately to see if it works: For y = 2x + 1  7 = 2(3) + 1  7 = 6 + 1 = 7  (3, 7) works for y = 2x + 1 For y = 5x – 8  7 = 5(3) - 8  7 = 15 – 8 = 7  (3, 7) works for y = 5x – 8 Since (3, 7) works in BOTH equations, then (3, 7) is a solution for this system of equations . Example 2: Is the ordered pair (-2, 5) a solution for the following system of equations: y = -2x + 1 y = 3x - 9 Plug (3, 7) in each equations separately to see if it works: For y = -2x – 1  5 = -2(-2) + 1  5 = 4 + 1 = 5  (-2, 5) works for y = -2x + 1 For y = 3x + 9  5 = 3(-2) - 9  5 = -6 - 9 = -15  Not True  (-2, 5) does N0T works for y = 3x - 9 Since (-4, 5) does not works in BOTH equations, then (-4, 5) is NOT a solution for this system of equations Finding a “solution” to a System of Equations As mentioned above, there are three ways to find a “solution” for a system of equations: 1) By graphing 2) By Substitution 3) By Elimination Let’s look at each. Graphing Believe it or not, you already know how to do this one: 1) Graph each line separately on the same grid. You can use your choice of below to do so: a. Plotting points b. Finding and plotting the intercepts c. Put the equation in y = mx + b form and graph it 2) The solution is the point/ordered pair where the lines cross NOTE: you are required to know how to do this method Example 1 (from above): Find the solution for the following system of equations. y = 2x + 1 y grid. = 5x – 8 1) Graph each line on the same y = 2x + 1 is the orange line & y = 5x – 8 is the blue line 2) The lines cross at (3, 7) so it is your solution. Example 2 (from above): Find the solution for the following system of equations. y = -2x + 1 y = 3x - 9 1) Graph each line on the same grid. y = -2x + 1 is the orange line & y = 3x – 9 is the blue line 2) The lines cross at (2, -3) so it is your solution. NOTE: the solution is a point, i.e. an ordered pair, so make sure you write your answer as an ordered pair. There are a few issues with using graphing: 1) What if the lines cross in the middle of one of those squares? 2) What if the lines cross somewhere off the grid? So, we need another method (or two) to use. Substitution To use Substitution to solve a system of equations, follow the steps below: 1) Get either “x” or “y” by itself in one of the equations. It does not matter whether you pick “x” or “y”. You may have to change one of the equations into y = mx + b form 2) Substitute this equation into the other equation. Why on earth would you do this  to get down to only 1 variable, so now it is Chapter 1 type question! 3) Solve for whichever variable you have 4) Once you have either x = ____ or y = _______, plug it into either of the equations to find the other variable. 5) Check your solution in the other equation. Example 1 (from above): Find the solution for the following system of equations. y = 2x + 1 y = 5x – 8 Steps: 1) You already have y by itself in both equations 2) Substitute it into the other equation 2x + 1 = 5x – 8 3) Solve for whichever variable you have 2x + 1 = 5x – 8 subtract 2x from both sides to get: 1 = 3x – 8 add 8 to both sides to get: 9 = 3x divide by 3 to get: 3=x 4) Plug into the either equation to find y: y = 2x + 1 plug in 3 for x y = 2(3) + 1 do the math y=6+1=7 y=7 Your solution is (3, 7) 5) Checking: y = 5x – 8  7 = 5(3) – 8  7 = 15 – 8 = 7 it works! Example 2: Find the solution for the following system of equations. 2x + y = 1 -3x + y = - 9 Steps: 1) Get either “x” or “y” by itself in one of the equations. Using 2x + y = 1 Subtract 2x from both sides, giving you: y = -2x + 1 2) Substitute it into the other equation -3x + y = - 9 becomes -3x + (-2x + 1) = - 9 3) Solve for whichever variable you have -3x + (-2x + 1) = - 9 clean it up to get: combine like terms to get: -3x + -2x + 1 = - 9 -5x + 1 = - 9 subtract 1 from both sides to get: -5x = - 10 divide by -5 to get: x=2 4) Plug into the either equation to find y: 2x + y = 1 plug in 2 for x 2(2) + y = 1 do the math 4 +y=1 y = -3 Your solution is (2, -3) 5) Checking: -3x + y = - 9  -3(2) + -3 = -6 +-3 = - 9 it works! Elimination Another way you can solve a System of Equations is by using Elimination. To use Elimination to solve a system of equations, follow the steps below: 1) Get all 4 parts of both equations lined up: “x”, “y”, “=”, extra number 2) Change one or both equations (if necessary) to get either the x’s or y’s opposite of each other in the two equations. 3) Add the equations. Why on earth would you do this  to get down to only 1 variable, so now it is Chapter 1 type question! NOTE: you should have only x’s or y’s left after adding. If you still have both left, you have done something wrong! 4) Solve for whichever variable you have 5) Once you have either x = ____ or y = _______, plug it into either of the equations to find the other variable. 6) Check your solution in the other equation. Example 1 (from above): Find the solution for the following system of equations. y = 2x + 1 y = 5x – 8 Steps: 1) You already have all four parts lined up 2) Multiply the 2nd equation by -1 to make the y’s opposite: -1(y) = -1(5x – 8)  -1y = -5x + 8 So your “new” equations are: y = 2x + 1 -1y = -5x + 8 3) Add the equations + 1y = 2x + 1 -1y = -5x + 8 0 = -3x + 9 4) Solve for the variable you have: 0 = -3x + 9 subtract 9 from both sides to get: -9= -3x divide by -3 to get: 3=x 5) Plug in 3 for x to find y y = 2x + 1 y = 2(3) + 1  y = 7 Your solution is (3, 7) 6) Checking: y = 5x – 8  7 = 5(3) – 8  7 = 15 – 8 = 7 it works! Example 2: Find the solution for the following system of equations. 2x = 1 - y -3x + y = - 9 Steps: 1) Get all four parts lined up. Add “y’ to both sides on the top equation to do so. 2x + y = 1 -3x + y = - 9 2) Multiply the 2nd equation by -1 to make the y’s opposite: -1(-3x + y) =-1(- 9) becomes 3x - y = 9 3) Add the equations + 2x + y = 1 3x - y = 9 5x = 10 4) Solve for the variable you have : 5x = 10 do the math x=2 5) Plug in 2 for x to find y 2x = 1 - y 2(2) = 1 - y 4=1–y 3 = -y -3 = y Your solution is (2, -3) 6) Checking: -3x + y = - 9  -3(2) + -3 = -6 +-3 = - 9 it works! NOTE: Systems of Equations that have one solution are also referred to as Consistent and Independent. NOTE (again): You do NOT have to learn both Substitution AND Elimination. Pick the one you like best and use it! Special Cases We have, unfortunately, 2 special cases to deal with: 1) One is where there is No Solution. 2) The other is where we have Infinitely Many solutions. No Solution Recall that the “solution” to these types of questions is the one point what works in BOTH equations. In order for there to be NO Solution, we have to have a situation where no points work. NOTE: Systems of Equations that have No Solution are also referred to as Inconsistent. Example: Solve the following System of Equations y = 1/2x – 4 the blue line 2y = x + 6 the orange line 1) By Graphing – following the steps above, when you graph both lines on the same grid, you get: As you can see, the lines DON’T cross, so there is no point that works  No Solution NOTE: Parallel lines have No Solution 2) By Substitution y = 1/2x – 4 2y = x + 6 Substituting in gives you: 2(1/2x – 4) = x + 6 Distributing gives you: 2(1/2x) – 2(4) = x + 6 Cleaning it up: x-8=x+6 Subtract “x” from both sides gets you: -8 = 6 3) By Elimination y = 1/2x – 4 2y = x + 6 Multiply the 1st equation by -2: Gives you: Now add the equations together: -2(y) = -2(1/2x – 4) -2y = -1x + 8 -2y = -1x + 8 NOT TRUE  No Solution + 2y = x + 6 0 = 14 NOT TRUE  No Solution Infinitely Many solutions. Recall that the “solution” to these types of questions is the one point what works in BOTH equations. What if you have a whole bunch of points that are on both lines??? The only way for that to happen is if the lines are on top of each other!! NOTE: Systems of Equations that have Infinitely Many solutions are also referred to as Consistent and Dependent Example: Solve the following System of Equations y = 1/2x – 4 the blue line 2y = x - 8 the orange line 1) By Graphing – following the steps above, when you graph both lines on the same grid, you get: As you can see, the lines are on top of one another, so you have Infinitely many Solutions. 2) By Substitution y = 1/2x – 4 2y = x - 8 Substituting in gives you: 2(1/2x – 4) = x - 8 Distributing gives you: 2(1/2x) – 2(4) = x - 8 Cleaning it up: x-8=x-8 Subtract “x” from both sides gets you: -8 = - 8 ALWAYS TRUE  Infinitely many Solutions 3) By Elimination y = 1/2x – 4 2y = x - 8 -2(y) = -2(1/2x – 4) Multiply the 1st equation by -2: Gives you: -2y = -1x + 8 Now add the equations together: -2y = -1x + 8 + 2y = x - 8 0=0 ALWAYS TRUE  Infinitely many Solutions Word Questions In real life, you are not given equations straight out. Instead, you are given a word question situation. In these cases, you have to write the equations yourself, then solve them using Substitution or Elimination. First decide what variables you should use, then see if there are two “topics” the word question is talking about to make your equations. Example 1: 25 people go to a play, adults & kids. Adult’s tickets are $8 each. Kid’s tickets are $5 each. The total receipts for the play is $176. How many adults and how many kids attended? Let’s call the number of adults = a and the number of children = c The questions talks about 1) number of people attending, and 2) $$ If 25 people go to the play and they are adults, “a”, and children, “c”, then # of adults + # of children = 25 people attending, so a + c = 25  your 1st equation total $$ was 176. Adult tickets are $8 and children’s are $5. $8(# of adults) + $5(# of children) = $176, so 8a + 5c = 176  your 2nd equation Your 2 equations are: a + c = 25 8a + 5c = 176 Use Substitution or Elimination to get: Answer: a = 17 adults , c = 8 children Example 2: At a clothing store, all items of the same type are a fixed price (i.e. all shirts are the same price). If one pair of pants and three shirts cost $60, and three pairs of pants and four shirts cost $110, then find the price of one shirt and one pair of pants. Let’s call the number of price of pants = p and the price of shirts = s The questions talks about 2 different shoppers and how much each spent. Shopper 1: buys one pair of pants and three shirts at a cost of $60  1p + 3s = 60  1st equation Shopper 2: buys three pairs of pants and four shirts at a cost of $110  2nd equation  3p + 4s = 110 Your 2 equations are: 1p + 3s = 60 3p + 4s = 110 Use Substitution or Elimination to get: Answer: p = $18 , s = $14 Videos that may help: Solve Linear systems by Graphing http://www.youtube.com/watch?v=cauaqqwAWCE Solve a System of Linear Equations by the Substitution Method http://www.youtube.com/watch?v=oUC931TcBuw http://www.youtube.com/watch?v=MIXL35YRzRw http://www.youtube.com/watch?v=j_BsjiDF8_4 Solve a System of Linear Equations by the Addition Method (Elimination by Addition) http://www.youtube.com/watch?v=ova8GSmPV4o No Solutions http://www.youtube.com/watch?v=kTtKfh5gFUc http://www.youtube.com/watch?v=z5_ACYtzW98 Infinitely Many Solutions http://www.youtube.com/watch?v=Pcqb109yK5Q http://www.youtube.com/watch?v=NRxh9Q16Ulk