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Algebra I. Algebra I. ~ Chapter 7 ~. Lesson 7-1 Solving Systems by Graphing Lesson 7-2 Solving Systems Using Substitution Lesson 7-3 Solving Systems Using Elimination Lesson 7-4 Applications of Linear Systems Lesson 7-5 Linear Inequalities Lesson 7-6 Systems of Linear Inequalities
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Algebra I Algebra I ~ Chapter 7 ~ Lesson 7-1 Solving Systems by Graphing Lesson 7-2 Solving Systems Using Substitution Lesson 7-3 Solving Systems Using Elimination Lesson 7-4 Applications of Linear Systems Lesson 7-5 Linear Inequalities Lesson 7-6 Systems of Linear Inequalities Chapter Review Systems of Equations & Inequalities
Lesson 7-1 Solving Systems by Graphing Cumulative Review Chap 1-6
Lesson 7-1 Solving Systems by Graphing Cumulative Review Chap 1-10
Lesson 7-1 Solving Systems by Graphing System of linear equations – Two or more linear equations together… One way to solve a system of linear equations is by… Graphing. Solving a System of Equations Step 1: Graph both equations on the same plane. (Hint: Use the slope and the y-intercept or x- & y-intercepts to graph.) Step 2: Find the point of intersection Step 3: Check to see if the point of intersection makes both equations true. Solve by graphing. Check your solution. y = x + 5 y = -4x Your turn… y = -1/2 x + 2 y = -3x - 3 Notes ~ Try another one ~ x + y = 4 x = -1
Lesson 7-1 Solving Systems by Graphing Systems with No Solution When two lines are parallel, there are no points of intersection; therefore, the system has NO SOLUTION! y = -2x + 1 y = -2x – 1 Systems with Infinitely Many Solutions y = 1/5x + 9 5y = x + 45 Since they are graphs of the same line… There are an infinite number of solutions. Notes
Lesson 7-1 Solving Systems by Graphing Homework – Practice 7-1 #1-28 odd Homework
Lesson 7-2 Solving Systems by Substitution Practice 7-1
Lesson 7-2 Solving Systems Using Substitution Using Substitution Step 1: Start with one equation. Step 2: Substitute for y using the other equation. Step 3: Solve the equation for x. Step 4: Substitute solution for x and solve for y Step 5: Your x & y values make the intersection point (x, y). Step 6: Check your solution. y = 2x 7x – y = 15 Your turn… y = 4x – 8 y = 2x + 10 Notes ~ Another example~ c = 3d – 27 4d + 10c = 120
Lesson 7-2 Solving Systems Using Substitution Using Substitution & the Distributive Property 3y + 2x = 4 -6x + y = -7 Step 1: Solve the equation in which y has a coefficient of 1… -6x + y = -7 +6x +6x y = 6x -7 Step 2: Use the other equation (substitute using the equation from Step 1.) 3y + 2x = 4 3(6x – 7) + 2x = 4 18x – 21 + 2x = 4 20x = 25 x = 1 1/4 Notes Step 3: Solve for the other variable Substitute 1 ¼ or 1.25 for x y = 6(1.25) – 7 y = 7.5 -7 y = 0.5 Solution is (1.25, 0.5)
Lesson 7-2 Solving Systems Using Substitution Your turn… 6y + 8x = 28 3 = 2x – y Solution is (2.3, 1.6) or (2 3/10, 1 3/5) A rectangle is 4 times longer than it is wide. The perimeter of the rectangle is 30 cm. Find the dimensions of the rectangle. Let w = width Let l = length l = 4w 2l + 2w = 30 Solve for l… l = 4(3) l = 12 Use substitution to solve. 2(4w) + 2w = 30 8w + 2w = 30 10w = 30 w = 3 Notes
Lesson 7-2 Solving Systems Using Substitution Homework ~ Practice 7-2 even Homework
Lesson 7-3 Solving Systems Using Elimination Practice 7-2
Lesson 7-3 Solving Systems Using Elimination Adding Equations Step 1: Eliminate the variable which has a coefficient sum of 0 and solve. Step 2: Solve for the eliminated variable. Step 3: Check the solution. 5x – 6y = -32 3x + 6y = 48 8x + 0 = 16 x = 2 Solution is (2, 7) Check 3(2) + 6(7) = 48 6 + 42 = 48 48 = 48 Your turn… 6x – 3y = 3 & -6x + 5y = 3 5x – 6y = - 32 5(2) – 6y = - 32 10 – 6y = -32 -6y = -42 y = 7 Notes
Lesson 7-3 Solving Systems Using Elimination Multiplying One Equation Step 1: Eliminate one variable. -2x + 15y = -32 7x – 5y = 17 Step 2: Multiply one equation by a number that will eliminate a variable. -2x + 15y = -32 3(7x – 5y = 17) Step 3: Solve for the variable 19x = 19 x = 1 Step 4: Solve for the eliminated variable using either original equation. -2(1) + 15y = -32 Solution (1, -2) • -2x + 15y = -32 • 21x - 15y = 51 19x + 0 = 19 Notes -2 + 15y = -32 15y = -30 y = -2
Lesson 7-3 Solving Systems Using Elimination Your turn… 3x – 10y = -25 4x + 40y = 20 Solution (-5, 1) Multiply Both Equations Step 1: Eliminate one variable. 4x + 2y = 14 7x – 3y = -8 Step 2: Solve for the variable 26x = 26 x = 1 Try this one… 15x + 3y = 9 10x + 7y = -4 • 3(4x + 2y = 14) • 2(7x – 3y = -8) • 12x + 6y = 42 • 14x – 6y = -16 26x + 0 = 26 Notes Step 3: Solve for the eliminated variable 4(1) + 2y = 14 2y = 10 y = 5 Solution (1, 5)
Lesson 7-3 Solving Systems Using Elimination Homework – Practice 7-3 odd Homework
Lesson 7-4 Applications of Linear Systems Practice 7-3
Lesson 7-4 Applications of Linear Systems Notes
Lesson 7-4 Applications of Linear Systems Homework – Practice 7-4 #6-10 Homework
Lesson 7-5 Linear Inequalities Practice 7-4
Lesson 7-5 Linear Inequalities • Using inequalities to describe regions of a coordinate plane: • x < 1 • y > x + 1 • y ≤ - 2x + 4 • Steps for graphing inequalities… • (1) First graph the boundary line. • (2) Determine if the boundary line is a dashed or solid line. • Shade above or below the boundary line… (< below or > above) • Graph y ≥ 3x - 1 • Rewriting to Graph an Inequality • Graph 3x – 5y ≤ 10 • Solve for y… (remember if you divide by a negative, the inequality sign changes direction) then apply the steps for graphing an inequality. • Graph 6x + 8y ≥ 12 Notes
Lesson 7-5 Linear Inequalities Homework ~ Practice 7-5 odd Homework
Lesson 7-6 Systems of Linear Inequalities Practice 7-5
Lesson 7-6 Systems of Linear Inequalities Practice 7-5
Lesson 7-6 Systems of Linear Inequalities Practice 7-5
Lesson 7-6 Systems of Linear Inequalities • Solve by graphing… • x ≥ 3 & y < -2 • You can describe each quadrant using inequalities… • Quadrant I? • Quadrant II? • Quadrant III? • Quadrant IV? • Graph a system of Inequalities… • (1) Solve each equation for y… • (2) Graph one inequality and shade. • (3) Graph the second inequality and shade. • The solutions of the system are where the shading overlaps. • Choose a point in the overlapping region and check in each inequality. Notes
Lesson 7-6 Systems of Linear Inequalities Graph to find the solution… y ≥ -x + 2 & 2x + 4y < 4 Writing a System of Inequalities from a Graph Determine the boundary line for the pink region… y = x – 2 The region shaded is above the dashed line… so y > x – 2 Determine the boundary line for the blue region… y = -1/3x + 3 The region shaded is below the solid line… so y ≤ -1/3x + 3 Your turn… Notes
Lesson 7-6 Systems of Linear Inequalities Homework 7-6 odd Practice 7-6
Lesson 7-6 Systems of Linear Inequalities Practice 7-6
Lesson 7-6 Systems of Linear Inequalities Practice 7-5
Lesson 7-6 Systems of Linear Inequalities Practice 7-6
Algebra I Algebra I ~ Chapter 7 ~ Chapter Review
Algebra I Algebra I ~ Chapter 7 ~ Chapter Review