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Solving Systems by Substitution Unit IX, Lesson 2 Online Algebra 1 VHS@pwcs System Review Remember that we can solve a system many ways, but in Algebra 1, we are going to solve them the following ways. Graphing Substitution Elimination. Also recall that there are three types of solutions for systems. One solution: an ordered pair No solution Infinite solutions. Solving Systems by Substitution. Recall that if we have the following expression, 3x – 7 and we know that x = 5, we can substitute 5 in for x. So the value of that expression is: 3(5) – 7 15 – 7 8 We can use substitution to solve systems. Solving Systems by Substitution. y = 2x 6x – y = 8 6x – 2x = 8 4x = 8 x=2 y = 2(2) y=4 1. Since the first equation tells us that y = 2x, we can substitute 2x in for y in the second equation. 2. Solve the second equation for x. Combine the like terms Divide both sides by 4 3. Now sub 2 in for x in one of the equations to solve for y. Using the first would be the easier since it is already solved for y 4. Our solution is (2, 4) Checking our solution. y = 2x 6x – y = 8 We can check by substituting our ordered pair into BOTH equations and simplifying. The ordered pair (2, 4) must work for both equations! Y = 2x 6x – y = 8 4 = 2(2) 6(2) – 4 = 8 4=4 12 – 4 = 8 8=8 The ordered pair (2, 4) works for both equations so we know that it is the solution to the linear system. Let’s try another! 3x – 6y = 30 1. y = -6x + 34 3x – 6(-6x + 34) = 30 2. 3x + 36x – 204 = 30 39x – 204 = 30 39x = 234 x=6 3. Y = -6(6) + 34 4. Get rid of the parentheses using the distributive property Combine like terms Add 204 to both sides Divide by 39 Substitute 6 in for x in the second equation and solve for y. Y = -36 + 34 Y = -2 Since the second equation is solved for y, we can substitute -6x + 24 in for y in the first equation. Solve for x. Again this is easier since it is already solved for y. Our solution is (6, -2) 3x + 5y = 2 Another! This time neither equation is solved for a variable, so we solve one of the equations for a variable. Since the x in the second equation has no coefficient it is easier to solve that for x. Subtract 4y from both sides. Now that the second is solved for x, we can sub -4 – 4y in for x in the first equation. It must be the first since it is the second that we solved. Now follow the steps in the last two slides to solve! Our solution is (4, -2) x + 4y = -4 x + 4y – 4y = -4 – 4y x = -4 – 4y 3(-4 – 4y) + 5y = 2 -12 - 12y + 5y = 2 -12 + -7y = 2 -7y = 14 y = -2 x = -4 – 4(-2) X = -4 + 8 X=4 Try this: Click for the steps! 2x + y = 5 2y = 10 – 4x 2x – 2x + y = 5 – 2x y = 5 – 2x 2(5 – 2x) = 10 – 4x 10 – 4x = 10 – 4x 10 + 4x – 4x = 10 – 4x + 4x 10 = 10 So there are an infinite number of solutions. Let’s solve the first for y; it’s easier! We need to sub 5 – 2 in for y into the second equation and solve for x. Notice that when we try to get the x’s all on one side they cancel out and we get 10 = 10. Remember, from Unit 4 that this means that our solution is all real numbers. If we had gotten a false statement there would be no solution. On your own! Homework page 278: 11 – 27 odd