1. Why is graphing a beneficial method for solving systems of equations? [OV, page 1]


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1 Student Activity Sheet 1; use with Overview 1. Why is graphing a beneficial method for solving systems of equations? [OV, page 1] (Answers may vary.) Graphing allows you to see quickly whether you have one, no, or many solutions. For example, it allows you to see if lines are parallel, which lets you know that the system has no solution. 2. What are some disadvantages to using tables or graphs a to solve systems of equations? [OV, page 2] (Answers may vary.) It may be harder to find exact solutions when they are not integers or easy fractions. 3. What are the names of two algebraic methods used to solve systems of equations? [OV, page 2] The substitution method and the linear combination method 4. What is the advantage to using an algebraic method, like substitution or linear combination, to solve a system of equations? [OV, page 3] Algebraic methods have the advantage of getting the solution very quickly, even if that answer is an ugly fraction. Page 1 of 1
2 Student Activity Sheet 2; use with Exploring Substitution method Complete the following statement to describe these equations. [EX1, page 1] 2l + 2w = 124 l = 5w 10 system set four variables equations two variables 1. Together, the two equations that model this situation form a system of equations in two variables. 2. Study the completed substitution example above and answer the following questions. a. What substitution was made in the second equation, 2l + 2w = 124? [EX1, page 3] The first equation tells you that l = 5w 10. The expression 5w 10 was substituted for the variable l in the equation 2l + 2w = 124. b. After this substitution was made into the second equation, how many variables were left in that equation? [EX1, page 3] After the substitution was made, the equation became 2(5w 10) + 2w = 124. Only one variable, w, was left in the equation. The resulting equation could then be solved for w. Page 1 of 7
3 Student Activity Sheet 2; use with Exploring Substitution method 3. In this solution, the value for w, 12, was substituted into the equation l = 5w 10 to find the value of l. Verify that you get the same solution if you instead substitute 12 for w in the other equation in the system, 2l + 2w = 124. [EX1, page 4] 4. Check the solution, l = 50 and w = 12, by verifying that these values make both equations in the system true. [EX1, page 4] 5. Summarize the substitution method for solving systems of equations and then write down the steps for using this method to solve a system. [EX1, page 5] Page 2 of 7
4 Student Activity Sheet 2; use with Exploring Substitution method 6. REINFORCE Solve this system of equations using the substitution method. y = 7x 1 5 x+ 2 y= 2 Then, solve the same system using a graph and a table. Discuss how the solutions from the three methods compare. Using the first equation, substitute into the second equation to find the value of x: 5x + 2(7x 1) = 2 5x + 14x 2 = 2 19x 2 = 2 19x = 0 x = 0 Substitute this value into one of the equations to find the value of y: y = 7(0) 1 y = 1 Solving with a table: Solving with a graph: x 7x  1 5x + 2y Both a table and a graph confirm the solution found by substitution. Page 3 of 7
5 Student Activity Sheet 2; use with Exploring Substitution method The table shows that the ordered pair (0,1) causes the value of the expression on the left in the second equation to be 2. The graphical solution shows that the lines that represent each equation intersect at the point (0,1). 7. Which of the following systems would be a good candidate for the substitution method? Explain. [EX1, page 5] System Substitution Explanation 2 x 5 y= 18 y= 2 x 10 x+ 5 y= 4 3 x+ 10 y= 10 2 x+ 3 y= 11 3 x+ 2 y= 4 YES YES NO The second equation is already solved for y, so the first step is already done. You can easily solve the first equation for x by subtracting 5y from both sides: x = 4 5y. No matter which variable you try to solve for, you wind up with an equation with fractions. 8. Use the substitution method to solve the following system of equations: [EX1, page 6] y= 2 x10 2 x 5 y= 18 Step 1 substitute y = 2x 10 into 2x 5y = 18. 2x 5(2x 10) = 18 2x 10x + 50 = 18 8x + 50 = 18 Step 2 solve the equation from step 1 for x and then substitute the value into y = 2x 10 to find y. 8x + 50 = 18 y = 2x = 50 y = 2(4) 108x = 32 y = y = 2 x = 4 Page 4 of 7
6 Student Activity Sheet 2; use with Exploring Substitution method 9. Consider the following system of equations: [EX1, page 7] x+ 5 y= 4 3 x+ 10 y= 10 Solve the first equation for x. Subtract 5y from each side of the equation to obtain x = 4 5y. 10. Use the substitution method to solve the system. [EX1, page 8] x+ 5 y= 4 3 x+ 10 y= 10 Step 1: substitute x = 4 5y into 3x + 10y = 10. 3(4 5y) + 10y = y + 10y = y = 10 Step 2: solve for y. 12 5y = = 125y = y = 0.4 Step 3: solve for x. x = 4 5y x = 4 5(0.4) x = 4 2 x = 2 Page 5 of 7
7 Student Activity Sheet 2; use with Exploring Substitution method 11. When might you use the substitution method for solving systems? [EX1, page 9] When one equation in the system is already solved for one variable; or When one equation in the system has one of the variables with a coefficient of 1, so you can easily solve it without getting fractions 12. REINFORCE Solve each of the following systems by substitution. a. y= 5x+ 2 x + 3 y = 7 b. x  y = 12 x + y = 3 c. 2x +y = 3 x + y = 5 Substitute for y in the second equation: x + 3(5 x + 2) = 7 x + 15 x + 6 = 7 16 x = 1 y = 5 x x = 16 1 y = y = y = 16 The solution is 1 37, Solve one of the two equations for either x or y, and then substitute into the other equation. For example, solving the first equation for x gives x = 12 + y. Substitute this expression for x in the second equation: 12+ y + y = y =3 2 y =9 9 y = 2 9 x   = x = x = 2 The solution is 15 9, Solve the second equation for either x or y, and then substitute into the first equation, or solve the first equation for y and substitute into the second equation. Using this approach gives y = 3 2x. Substitute this expression for y in the second equation: x +32 x =5 3 x =52 = x y = 32(2) y =7 The solution is (2,7). Page 6 of 7
8 Student Activity Sheet 2; use with Exploring Substitution method d. y= 3x+10 x + 5 y = 2 e. 4x+ 2 y= 165 x+y= 8 f. 3 x+y= 4 3x+ 2 y= 11 Substitute y into the second equation. x + 5(3x + 10) = 2 x + 15x + 50 = 2 16x = 48 x = 3 Substitute this result into the first equation. y = 3( 3) + 10 y = 1 The solution is (3,1). One solution strategy begins with solving for y in the second equation. y = 5x + 8 Use this equation to substitute in for y in the first equation. 4x + 2(5x + 8) = 16 4x + 10x + 16 = 16 14x + 16 = 16 14x = 0 x = 0 Substitute this result into either of the equations in this problem. y = 5(0) + 8 y = 8 One solution strategy begins with solving for y in the first equation. y = 3x + 4 Substitute in for y in the second equation. 3x + 2( 3x + 4) = 11 3x 6x + 8 = 11 3x = 3 x = 1 Substitute this result into either of the equations in the problem. y = 3( 1) + 4 y = 7 The solution is (1,7). The solution is (0,8). Page 7 of 7
9 1. Solve the system x + 2y = 12 3x 2y = 4 using the linear combination method. [EX2, page 1] ( x + 2 y) + (3x  2 y) = x + 2y = y = 12 2y = 10 y = 5 4x + 0 = 8 or 4x 8 = 4 4 x = 2 3x 2y = 4 3(2)  2y = y = 42y = 10 y = 5 2. What property of equality allows you to combine two equations into one, as you do in the linear combination method? [EX2, page 2] The addition property of equality states that if a = b and c = d, then a + b = c + d. This property supports adding the left and right sides of the two equations in the system to obtain a true equation. 3. Why did the system [EX2, page 3] x + 2y = 12 3x 2y = 4 solve so easily using the linear combination method? When you added or subtracted the equations, the coefficients of one of the variables was zero. 4. What if the coefficients don t add or subtract to zero, as in the system 2 x+ 9 y= 12? 4 x+ 3 y = 6 [EX2, page 3] You should multiply the equations (one or both) by numbers that will make the coefficients of one of the variables opposites of each other. Page 1 of 14
10 5. Use linear combination to solve the system 2 x+ 9 y= 12 4 x+ 3 y = 6 by a. first eliminating the xvalues to solve for y, and then finding the value of x [EX2, page 4] b. first eliminating the yvalues to solve for x, and then finding the value of y [EX2, page 4] Page 2 of 14
11 6. REINFORCE Solve each system using the linear combination method. Check your solution. a. x + 3y = 18 x 3y = 24 Solution (3,7). Solution strategies may vary. One possible strategy: Combine the 2 equations in the system and solve for x. 2x = 6 x = 3 Now find y. ( 3) + 3y = 18 3y = 21 y = 7 Alternate strategy: Multiply either equation by 1 to eliminate the xvariables and solve for y. Page 3 of 14
12 b. 2x + y = 5 2x 3y = 7 Solution (1,3). Solution strategies may vary. One possible strategy is: 2x + y = 5 2x  3y = 7 g 1 Multiply the second equation by 1. The new system is: 2x + y = 5 2x + 3y = 7 Combining the equations gives: 4y = 12 y = 3 Substitute this value into either equation. 2x + (3) = 5 2x = 2 x = 1 Alternate strategy: Multiply the first equation by 3 to eliminate the yvariables and solve for x. Page 4 of 14
13 c. 5x 2y = 18 x + 3y = 10 Solution (2,4). Solution strategies may vary. One possible strategy is: 5x 2y = 18 x + 3y = 10 g  5 Multiply the second equation by 5. Now the system is: 5x 2y = 18 5x 15y = 50 Combining the equations gives 17y = 68 y = 4 Substitute this value into either equation. 5x 2(4) = 18 5x 8 = 18 5x = 10 x = 2 Alternate strategies require transforming both equations. Page 5 of 14
14 d. 2x + 3y = 3 4x 9y = 4 Solution 1 2,. Solution strategies may vary. One possible strategy: 2 3 2x + 3y = 3 4x  9y = 4 g 2 Multiply the first equation by 2. 4x 6y = 6 4x 9y = 4 Combining the new equations gives: 15y = 10 y = 2 3. Substitute this value into either equation. 2x = 3 3 2x + 2 = 3 2x = 1 x = 1 2 Alternate strategy: Multiply the first equation by 3 to eliminate the y variables and solve for x. Page 6 of 14
15 7. Describe how to use the coefficients of the x terms in the system 2 x+ 3 y= 11 5 x+ 4 y= 10 to solve for y, and then solve for y. Show your work in the Process column. [EX2, page 5] Steps Multiply the first equation by the coefficient of the x term of the other equation; then multiply the second equation by the opposite of the coefficient of the x term of the other equation. Process Add the two left sides, and add the two right sides. 8. Finish solving the system from problem 7 by substituting the value you found for y into either equation, and then solving for x. [EX2, page 5] 2x + 3(5) = 11 2x + 15 = 11 2x = 4 x = 2 Page 7 of 14
16 9. What do you notice about the solution to the system 2 x+ 3 y= 11 REINFORCE 5 x+ 4 y= 10 if you solve it by using the coefficients of the y term first? Use the coefficients to eliminate the y terms to solve for x, and then use your result to find the value for y. 2x + 3y = 11 5x + 4y = 10 4 ( 3) 8x + 12y = 4415x  12y = 307x = 14 x = 2 2x + 3y = 11 2( 2) + 3y = y = 11 3y = 15 y = 5 The solution is the still the same, x = 2 and y = 5. Page 8 of 14
17 10. Compare four different ways of solving the following system. [EX2, page 6] 8a+ 3b = 7 2a+ 9b = 23 a. Eliminate a by multiplying each term in the first equation by 2 and each term in the second equation by 8. b. Eliminate b by multiplying each term in the first equation by 9 and each term in the second equation by 3. c. Eliminate a by multiplying each term in the first equation by 1 and each term in the second equation by 4. d. Eliminate b by multiplying each term in the first equation by 3 and each term in the second equation by 1. Page 9 of 14
18 11. What could you do first to begin solving the system 3 x 8 y= 12? [EX2, page 7] 4 x 7 y= 8 Answers will vary. Answers could include: Multiply the first equation by 4 and the second by 3, and then add. Multiply the first equation by 7 and the second by 8, and then subtract. Multiply the first equation by 4 and the second by 3, and then subtract. Multiply the first equation by 7 and the second by 8, and then add. 12. Why is the following plan a good way to solve the following system? [EX2, page 8] 3x  8y = 12 4x  7y = 8 4 ( 3) Because it kept the numbers small and adding negative values causes most people fewer problems than subtracting negative values. Page 10 of 14
19 13. Solve the following system for y using the given plan. [EX2, page 8] 3x  8y = 12 4x  7y = 8 4 ( 3) 12x  32y = 4812x + 21y = y = 24 y = Solve the following system for x using the given plan. [EX2, page 9] 3x  8y = 12 4x  7y = 8 7 ( 8) 21x  56y = 8432x + 56y = x = 20 x = Page 11 of 14
20 15. REINFORCE Solve each system using the linear combination method. Check your solution. a. 2 x+7 y=5 3x y=4 One possible solution strategy is to multiply each term in the second equation by 7: 21x 7y = 28 Then add to the first equation to get: 23x = 23 x = 1 Substitute this answer into either equation, or eliminate, to find that y = 1. Solution (1,1). b. x+2 y=9 x2 y=4 2 x =13 13 x = 2 Substitute this answer into one of the equations, or eliminate to find that y is equal to 5 4. The solution is 13 5, 2 4. c. 2 x+5 y=1 3x2 y=8 One possible solution strategy: Multiply each term in the first equation by 3 and each term in the second equation by 2. 6x + 15y = 36x + 4y = 16 Then combine the equations to get 19y = 19 y = 1. Substitute this result into either of the equations, or eliminate, to find that x = 2. Solution (2,1) Page 12 of 14
21 d. x+ y=9 x2 y=4 e. 3 x+ y= x+ y=8 f. 3 x+2 y=9 x5 y= 4 3 y =5 5 y = 3 Substitute this answer into one of the equations, or eliminate to find that x is equal to The solution is 22 5, 3 3. One possible solution strategy: Multiply each term in the second equation by 1 to get 2x y = 8 and add this to the first equation to get 5x = 20 x = 4 Substitute this into either of the equations, or eliminate, to find that y = 0. Solution (4,0) To eliminate x, multiply each term in the bottom equation by 3: 3x + 15y = 12 Then add to obtain 17 y = 3 3 y = 17 To eliminate y, multiply each term in the top equation by 5 and each term in the bottom equation by 2: 15x + 10y = 45 and 2x 10y = 8 17x = 53 x = The solution is 53 3, Page 13 of 14
22 g. 3 x+4 y=0 2 x+5 y=7 One possible solution strategy: Multiply each term in the first equation by 2 and each term in the second equation by 3. 6x + 8y = 06x 15y = 21 Add these results together to get. 7y = 21 y = 3 Substitute or eliminate to find that x = 4. Solution (4,3) h. 8x4 y= 1 x+ y=1 One possible solution strategy: Multiply each term in the second equation by 8 to get 8x 8y = 8, and then add to get: 12y = 9 y = ¾ Substitute or eliminate to find that x = ¼. Solution 1 3, 4 4 Page 14 of 14
23 Student Activity Sheet 4; use with Exploring Why linear combination works 1. Consider the following system. Complete the puzzle to describe two different correct strategies for solving this system. [EX3, page 1] x + y = 4 2x 3y = Strategy 1 Strategy 2 Multiply the bottom equation by 1. Multiply the bottom equation by 1. Multiply the top equation by 2 to eliminate x. Add the two equations and solve to find y = 1. Substitute this value into one of the equations to find x = 3. Multiply the top equation by 3 to eliminate y. Add the two equations and solve to find x = 3. Substitute this value into one of the equations to find y = To begin to understand why the method of linear combination works, first rewrite each equation in the system so that it is equal to 0. Then, add the two equations together. Why can you do this? [EX3, page 2] Since each expression equals zero, you can add the two expressions together and still have zero for an answer. Page 1 of 4
24 Student Activity Sheet 4; use with Exploring Why linear combination works 3. Graph all three equations the system and its linear combination on the same axes. What do you see when you look closely at the graph? [EX3, page 3] The graph of the linear combination is a line, and the line goes through the point of intersection of the graphs of the two equations in the original system of equations. The coordinates of this point of intersection are the solution to the system. 4. Why does the graph of the linear combination equation pass through the point that gives the solution to the system? [EX3, page 4] The solution to the system is given by the one pair of values of x and y that make both x + y 4 = 0 and 2x 3y 3 = 0 true. Because these values make each expression equal to 0, they make the sum of the expressions equal to 0 as well, so they make the linear combination equation true. This means the graph of the linear combination equation contains the point with those coordinates. Page 2 of 4
25 Student Activity Sheet 4; use with Exploring Why linear combination works 5. Study the three equations that are equal to 0. Why is the equation 2(x + y 4) 3(2x 3y 3) = 0 also true? [EX3, page 5] Because x + y 4 and 2x 3y 3 are both equal to 0, multiplying each expression by some nonzero number gives two new expressions that each still equal 0. This means the sum of these two new expressions will also be Simplify the new equation found in question 5 and graph it on the same axes with the graphs of the other three equations: x + y 4 = 0, 2x 3y 3 = 0, and (x + y 4) + (2x 3y 3) = 0. What does the graph of the new equation have in common with the other three graphs? [EX3, page 5] The graph of the new linear combination you get after multiplying also passes through the intersection point of the original two equations. Page 3 of 4
26 Student Activity Sheet 4; use with Exploring Why linear combination works 7. Simplify and solve the equation 2(x + y 4) + 1(2x 3y 3) = 0. What is true about the graph of your answer? [EX3, page 6] 2(x + y 4) + 1(2x 3y 3) = 02x 2y x 3y 3 = 02y + 8 3y 3 = 05y = 5 y = 1 The graph of y = 1 is a horizontal line that contains the solution to the system, (3,1). 8. Simplify and solve the equation 3(x + y 4) + 1(2x 3y 3) = 0. What is true about the graph of your answer? [EX3, page 6] 3(x + y 4) + 1(2x 3y 3) = 0 3x + 3y x 3y 3 = 0 3x x 3 = 0 5x = 15 x = 3 The graph of x = 3 is a vertical line that contains the solution to the system, (3,1). Page 4 of 4
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