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Systems of Equations. 10-6. HOMEWORK & Learning Goal. Lesson Presentation. AIMS Prep. Pre-Algebra. PA HOMEWORK Answers. Page 521 #1-11 ALL NO WORK= ZERO CREDIT! NO WORK= ZERO CREDIT !. 3 V. = h. 1. C – S. = l. A. 3. t. P – 2 w. 2.
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Systems of Equations 10-6 HOMEWORK & Learning Goal Lesson Presentation AIMS Prep Pre-Algebra
PA HOMEWORK Answers Page 521 #1-11 ALL NO WORK= ZERO CREDIT! NO WORK= ZERO CREDIT!
3V = h 1 C – S = l A 3 t P – 2w 2 Don’t forget your proper heading! Trade & Grade! 10-5 Lesson Quiz: Part 1 Solve for the indicated variable. 1.P = R – C for C. 2.P = 2l+ 2w for l. 3.V = Ah for h. 4.R = for S. C = R - P C – Rt = S
y 2x 4 7 2 –4 –2 2 4 –2 y = – + 2 –4 Lesson Quiz: Part 2 5. Solve for y and graph 2x + 7y = 14.
Pre-Algebra HOMEWORK Page 526 #17-32 NO WORK= ZERO CREDIT! NO WORK= ZERO CREDIT!
Our Learning Goal Students will be able to solve multi-step equations with multiple variables, solve inequalities and graph the solutions on a number line.
Our Learning Goal Assignments • Learn to solve two-step equations. • Learn to solve multistep equations. • Learn to solve equations with variables on both sides of the equal sign. • Learn to solve two-step inequalities and graph the solutions of an inequality on a number line. • Learn to solve an equation for a variable. • Learn to solve systems of equations.
Today’s Learning Goal Assignment Learn to solve systems of equations.
Vocabulary system of equations solution of a system of equations
A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions of all of the equations. If the system has two variables, the solutions can be written as ordered pairs.
? ? 5(1) + 2 = 7 1 – 3(2) = 11 Additional Example 1A: Identifying Solutions of a System of Equations Determine if the ordered pair is a solution of the system of equations below. 5x + y = 7 x – 3y = 11 A. (1, 2) 5x + y = 7 x – 3y = 11 Substitute for x and y. 7 = 7 –5 11 The ordered pair (1, 2) is not a solution of the system of equations.
? ? 4(1) + 2 = 8 1 – 4(2) = 12 Try This: Example 1A Determine if each ordered pair is a solution of the system of equations below. 4x + y = 8 x – 4y = 12 A. (1, 2) 4x + y = 8 x – 4y = 12 Substitute for x and y. 6 8 –7 12 The ordered pair (1, 2) is not a solution of the system of equations.
? 2 – 3(–3) = 11 ? 5(2) + –3 = 7 Additional Example 1B: Identifying Solutions of a System of Equations Determine if the ordered pair is a solution of the system of equations below. 5x + y = 7 x – 3y = 11 B. (2, –3) 5x + y = 7 x – 3y = 11 Substitute for x and y. 7 = 7 11 = 11 The ordered pair (2, –3) is a solution of the system of equations.
? 2 – 4(–3) = 12 ? 4(2) + –3 = 8 Try This: Example 1B Determine if each ordered pair is a solution of the system of equations below. 4x + y = 8 x – 4y = 12 B. (2, –3) 4x + y = 8 x – 4y = 12 Substitute for x and y. 5 8 14 12 The ordered pair (2, –3) is not a solution of the system of equations.
? ? 5(20) + (3) = 7 20 – 3(3) = 11 Additional Example 1C: Identifying Solutions of a System of Equations Determine if the ordered pair is a solution of the system of equations below. 5x + y = 7 x – 3y = 11 C. (20, 3) 5x + y = 7 x – 3y = 11 Substitute for x and y. 103 7 11 = 11 The ordered pair (20, 3) is not a solution of the system of equations.
? 1 – 4(4) = 12 ? 4(1) + 4 = 8 Try This: Example 1C Determine if each ordered pair is a solution of the system of equations below. 4x + y = 8 x – 4y = 12 C. (1, 4) 4x + y = 8 x – 4y = 12 Substitute for x and y. 8 = 8 –15 12 The ordered pair (1, 4) is not a solution of the system of equations.
Helpful Hint When solving systems of equations, remember to find values for all of the variables.
Additional Example 2: Solving Systems of Equations y = x – 4 Solve the system of equations. y = 2x – 9 y=y y= x – 4 y =2x – 9 x – 4 = 2x – 9 Solve the equation to find x. x – 4 = 2x – 9 – x– x Subtract x from both sides. –4 = x – 9 + 9+ 9 Add 9 to both sides. 5 = x
? ? 1 = 5 – 4 1 = 2(5) – 9 Additional Example 2 Continued To find y, substitute 5 for x in one of the original equations. y = x – 4 = 5 – 4 = 1 The solution is (5, 1). Check: Substitute 5 for x and 1 for y in each equation. y = x – 4 y = 2x – 9 1 = 1 1 = 1
Try This: Example 2 y = x – 5 Solve the system of equations. y = 2x – 8 y=y y= x – 5 y =2x – 8 x – 5 = 2x – 8 Solve the equation to find x. x – 5 = 2x – 8 – x– x Subtract x from both sides. –5 = x – 8 + 8+ 8 Add 8 to both sides. 3 = x
? ? –2 = 3 – 5 –2 = 2(3) – 8 Try This: Example 2 Continued To find y, substitute 3 for x in one of the original equations. y = x – 5 = 3 – 5 = –2 The solution is (3, –2). Check: Substitute 3 for x and –2 for y in each equation. y = x – 5 y = 2x – 8 –2 = –2 –2 = –2
To solve a general system of two equations with two variables, you can solve both equations for x or both for y.
Additional Example 3A: Solving Systems of Equations Solve the system of equations. A. x + 2y = 8 x – 3y = 13 Solve both equations for x. x + 2y = 8 x – 3y = 13 –2y–2y+ 3y+ 3y x = 8 – 2y x = 13 + 3y 8 – 2y = 13 + 3y Add 2y to both sides. + 2y+ 2y 8 = 13 + 5y
= –5 5 5y 5 Additional Example 3A Continued 8 = 13 + 5y Subtract 13 from both sides. –13–13 –5 = 5y Divide both sides by 5. –1 = y x = 8 – 2y = 8 – 2(–1)Substitute –1 for y. = 8 + 2 = 10 The solution is (10, –1).
Try This: Example 3A Solve the system of equations. A. x + y = 5 3x + y = –1 Solve both equations for y. x + y = 5 3x + y = –1 –x–x– 3x– 3x y = 5 – x y = –1 – 3x 5 – x = –1 – 3x Add x to both sides. + x+ x 5 = –1 – 2x
Try This: Example 3A Continued 5 = –1 – 2x + 1+ 1 Add 1 to both sides. 6 = –2x –3 = x Divide both sides by –2. y = 5 – x = 5 – (–3)Substitute –3 for x. = 5 + 3 = 8 The solution is (–3, 8).
Helpful Hint You can choose either variable to solve for. It is usually easiest to solve for a variable that has a coefficient of 1.
= – –3 –3y –3 3x –3 –3 Additional Example 3B: Solving Systems of Equations Solve the system of equations. B. 3x – 3y = -3 2x + y = -5 Solve both equations for y. 3x – 3y = –3 2x + y = –5 –3x–3x–2x–2x –3y = –3 – 3xy = –5 – 2x y = 1 + x 1 + x = –5 – 2x
–6 3x 3 3 = Additional Example 3B Continued 1 + x = –5 – 2x Add 2x to both sides. + 2x+ 2x 1 + 3x = –5 Subtract 1 from both sides. –1–1 3x = –6 Divide both sides by 3. x = –2 y = 1 + x Substitute –2 for x. = 1 + –2 = –1 The solution is (–2, –1).
Try This: Example 3B Solve the system of equations. B. x + y = –2 –3x + y = 2 Solve both equations for y. x + y = –2 –3x + y = 2 – x– x+ 3x+ 3x y = –2 – xy = 2 + 3x –2 – x = 2 + 3x
Try This: Example 3B Continued –2 – x = 2 + 3x Add x to both sides. + x+ x –2 = 2 + 4x Subtract 2 from both sides. –2–2 –4 = 4x Divide both sides by 4. –1 = x y = 2 + 3x Substitute –1 for x. = 2 + 3(–1) = –1 The solution is (–1, –1).
1 2 ( , 2) Don’t forget your proper heading! Trade & Grade! 10-6 Lesson Quiz 1. Determine if the ordered pair (2, 4) is a solution of the system. y = 2x; y = –4x + 12 Solve each system of equations. 2.y = 2x + 1; y = 4x 3. 6x – y = –15; 2x + 3y = 5 4. Two numbers have a some of 23 and a difference of 7. Find the two numbers. yes (–2,3) 15 and 8