Yesterday we took our first look at solving a linear system. We learned that a linear system is two or more linear equations taken at the same time. Their solution is the point that all the lines have in common, their intersection. We also learned there are three ways to solve a linear system: 1. Graph the lines, visually locate the intersection: only approximates the solution. 2. Substitution: solve one equation for a variable, substitute into the other. 3. Combination: add the two equations together to eliminate one variable. Today we will work on the latter two; they will give us a precise answer, not an estimation. The basic idea If we have an equation with only one variable, we can solve that equation and find the value of the variable that solves the equation. The problem we have when trying to solve a linear system is that there s more than one variable. The basic idea when solving linear systems is to: 1. Eliminate one of the variables 2. Solve the resulting one variable equation 3. Plug the resulting value back into the original two variable equation to solve for the other variable. The trick is step one: eliminating one of the variables. That s what we ll be learning to day. Using substitution to solve a linear system The first way we can eliminate one of the variables is with substitution. We use substitution to solve algebra problems all the time. Anytime you have an equation and you know the value of x, you use substitution to find y: If y = 3x 9 and x = -2, find y. y = 3x 9 y = 3(-2) 9 y = -6 9 y = -15 Here we substituted -2 for x in order to find the corresponding value of y. We can use the same technique to solve a linear system. Consider the following: y = 2x + 8 3x + y = -2 Page 1 of 6
Here we a linear system. The first equation is in slope-intercept form and the second is in standard form. Can you see a way to use substitution to eliminate one of the variables? Stop for a second and think; the next sentence has a hint. Hint: is there a way you can get rid of y? With the first equation, we know that y is the same thing as 2x + 8. What if in the second equation we substituted 2x + 8 in for y? That makes sense doesn t it? If y equals (or is the same as) 2x + 8 won t everything be the same if we replace y with 2x + 8? Let s try it: y = 2x + 8 3x + y = -2 3x + (2x + 8) = -2 5x + 8 = -2 5x = -10 x = -2 y = 2x + 8 y = 2(-2) + 8 y = -4 + 8 y = 4 here we know what y is in terms of x we ll substitute that in for y in this equation to solve for x here we substituted (2x + 8) for y subtract 8 from both sides to get x alone divide both sides by 5 to get x alone we now know x we ll now find out what y is when x = -2 substitute -2 in for x and we know have the y value too The solution for this linear system is (-2, 4). Is there a way we can double check this? Sure! Actually, there are two ways: 1. Graph the lines 2. Plug this x & y value pair into both equations to verify it works for both. We ll do number 2 here: Equation1 Equation 2 y 2x 8 3x y 2 4 2( 2) 8 3( 2) 4 2 4 4 8 6 4 2 4 4 2 2 Both check out so we can be confident that (-2, 4) is the solution for this linear system. Try substitution again Find the solution for the following linear system; be careful, it s a little different! To get started, take a look at this linear system and compare it to the one we just did. 5x 3y = 2 x + 2y = 3 Page 2 of 6
What is different? Here both equations are in standard form? What enabled us to do substitution in the first example? It was the fact that one of the equations was already solved for one of the variables. In the prior example, the first equation was already solved for y. So what we need to do here is solve one of the equations for a variable. It really doesn t matter if we solve for x or y; pick the easiest one. Which of the two equations do you think will be easiest to solve for a variable, and which variable will you solve it for? I d pick the second equation and solve it for x. Why? Because that is the only equation with a variable all by itself (it has a coefficient of 1). That will make it super easy to solve for x: just subtract 2y from both sides and you re done! This gives us x = -2y + 3. What is our next step? Plug that expression into the first equation in place of x: 5x 3y = 2 start with the first equation 5(-2y + 3) 3y = 2 substitute (-2y + 3) in place of x -10y + 15 3y = 2 use distributive property to get rid of parentheses -13y + 15 = 2-13y = -13 subtract 15 from both sides to get y by itself y = 1 divide both sides by -13 to get what y is equal to 5x 3y = 2 5x 3(1) = 2 5x - 3 = 2 5x = 5 x = 1 go back to the original 2 variable equation substitute the known value for y back into the equation add 3 to both sides to get x alone divide both sides by 5 to solve for x The solution for the linear system is (1, 1). Plug these values into both equations to check the answer. What are the steps for solving using substitution? 1. Solve one equation for one of its variables. 2. Substitute this expression into the other equation and solve for the other variable. 3. Substitute the (now) known value into either of the 2 variable equations & solve. 4. Check the solution in each of the original equations. Page 3 of 6
Using combination to solve linear systems The second method for solving linear systems is the combination method. In the combination method, we arrange the equations so that if we add them together one of the variables is cancelled out/eliminated. To do this we: 1. Arrange the two equations so like terms are aligned in columns. 2. Multiply one (or both) equations by a constant so that one of the variable sets differs only in sign. In other words, when you add them together they cancel out. 3. Add the equations and solve for the remaining variable. 4. Substitute the now known value of the solved variable back into either of the original equations and solve for the other variable. 5. Check your solution in each of the original equations. Let s try this out. Solve this linear system: 11x + 7y = 9 6x + 7y = 24 Step one: Arrange the two equations so like terms are aligned in columns. We already there the x s and the y s are aligned above each other. Step two: Multiply one (or both) equations by a constant so that one of the variable sest differs only in sign. We re close. In both equations y has a coefficient of 7. All we need to do is multiply one of the equations (both sides) by -1. I ll pick the first one: -1(11x + 7y) = -1(9) -11x 7y = -9 mult both sides by -1-11x 7y = -9 here s the adjusted 1 st equation 6x + 7y = 24 and here s the original 2 nd equation Notice how the y s have the same coefficient (7) but differ only in sign. Step three: Add the equations and solve for the remaining variable. To add the equations, just add like terms. This is easy since we ve arranged the equations so that like terms are aligned in columns. Page 4 of 6
11x 7 y 9 add thesetwo... 6x 7 y 24... equations together 5x 0 y 15 this basically cancels the y out 5x 15 x 3 divideboth sides by 5to solve for x Step four: Substitute the now known value for the solved variable back into either of the original equations and solve for the other variable. I ll pick the second original equation because it has simpler coefficients: 6x 7 y 24 6( 3) 7 y 24 18 7 y 24 7 y 42 y 6 Step five: Check your solution in each of the original equations: 11x 7y 9 6x 7y 24 11( 3) 7(6) 9 6( 3) 7(6) 24 33 42 9 18 42 24 9 9 24 24 Both check out so the solution is (-3, 6). Parallel or same lines when using substitution or combination If when you substitute or combine: 1. The resulting equation is false (for example 2 = 3), the lines are parallel. 2. The resulting equation is true (for example 3 = 3), the lines are the same. Examples: 9x 12y 3 2x 4y 2 ( by 5) 3x 4y 2 ( by 3) 10x 20y 10 9x 12y 3 10x 20y 10 9x 12y 6 ( combine) 10x 20y 10 ( combine) 0 6 ( false; no solution) 0 0 ( true; number of solutions) Page 5 of 6
Cookbook for solving linear systems algebraically General guidelines 1. Use substitution if one of the equations has a variable with a coefficient of 1 or -1. 2. Use combination otherwise. Combination is easier if the equations already have a variable set with the same coefficient. 3. System has infinite number of solutions (lines are the same) if when you eliminate, both variables are canceled and you have a true statement. 4. System has no solution (lines are parallel) if when you eliminate, both variables are canceled and you have a false statement. Substitution cookbook 1. Solve one equation for one of its variables. 2. Substitute this expression into the other equation and solve for the other variable. 3. Substitute the (now) known value into either of the 2 variable equations & solve. 4. Check the solution in each of the original equations. Combination cookbook 1. Arrange the two equations so like terms are aligned in columns. 2. Multiply one (or both) equations by a constant so that one of the variable sets differs only in sign. In other words, when you add them together they cancel out. 3. Add the equations and solve for the remaining variable. 4. Substitute the now known value for the solved variable back into either of the original equations and solve for the other variable. 5. Check your solution in each of the original equations. Page 6 of 6