Graphing Linear Systems

Transcription

1 Graphing Linear Systems Goal Estimate the solution of a system of linear equations by graphing. VOCABULARY System of linear equations A system of linear equations is two or more linear equations in the same variable form. This is also called a linear system. of a linear system A solution of a linear system is an ordered pair {x, y) that makes each equation in the system a true statement. Point of intersection A point (a, b) that lies on the graphs of two or more equations is a point of intersection for the graphs. Example 1 Find the Point of Intersection Use the graph at the right to estimate the solution of the linear system. Then check your solution algebraically. x + 2y = 4 Equation 1 x - 3y = 1 Equation 2-4 The lines appear to intersect once at (-2, -1). x--3y Check Substitute -2 for x and -1 for y in each equation. x + 2y = ( -1 ) =2= -4-4 = -4 x - 3y = ( -1 ) = 1 Answer Because ( -2, -1 ) is a solution of each equation, ( -2, -1 ) is the solution of the system of linear equations. 142 Algebra 1 Concepts and Skills Notetaking Guide Chapter 7

2 SOLVING A LINEAR SYSTEM USING GRAPH-AND-CHECK Step 1 Write each equation in a form that is easy to graph. Step 2 Graph both equations in the same coordinate plane. Step 3 Estimate the coordinates of the point of intersection. Step 4 Check whether the coordinates give a solution by substituting them into each equation of the original linear system. ) ' A line in slopeintercept form, y = mx + b, has a slope of m and a y-intercept of b. Example 2 Graph and Check a Linear System Use the graph-and-check method to solve the linear system. 5x + 4y = -12 Equation 1 3x - 4y = -20 Equation 2 v 1. Write each equation in slope-intercept form. Equation 1 Equation 2 5x + 4y = -12 3x - 4y = -20 4y = -5x y = -3x Graph both equations. 3. Estimate from the graph that the point of intersection is ( -4, 2 ). 4. Check whether (j^4_,_2_) is a solution by substituting -4 for x and 2 for y in each of the original equations. Equation 1 Equation 2 y= x 5x + 4y = -12 3x - 4y = -20 5( -4) + 4(2.) L -12 3( -4) - 4 ( 2 } L = N X, ^!!x - - 4y X > i 5x + IK - j 5 s- S = I 1 y X Answer Because ( -4, 2_) is a solution of each equation in the linear system, ( -4, 2 ) is a solution of the linear system. Lesson 7.1 Algebra 1 Concepts and Skills Notetaking Guide 143

3 Checkpoint Use the graph-and-check method to solve the linear system. 1. 3x - 4y = 4 x + 2y = x + 2y = 4 9x + 2y = 12 XX. - - ~3 _ j ^X^ ^X. S'^ > 7 A ^ Xx? X X- ' ^ (4,2) (2, -3) 3.y = -2x- 3 2x + 5y = y = 3x + 4 7x - 3y = x (-5,7) (-3, -5) 144 Algebra 1 Concepts and Skills Notetaking Guide Chapter 7

4 Solving Linear Systems by Substitution Goal Solve a linear system by substitution. SOLVING A LINEAR SYSTEM BY SUBSTITUTION Step 1 Solve one of the equations for one of its variables. Step 2 Substitute the expression from Step 1 into the other equation and solve for the other variable. Step 3 Substitute the value from Step 2 into the revised equation from Step 1 and solve. " Step 4 Check the solution in each of the original equations. Example 1 Substitution Method: Solve for y First ' When you use the substitution method, you can check the solution by substituting it for xand forin each of the original equations. You can also use a graph to check your solution. Solve the linear system. 4x + y = -5 Equation 1 1. Solve for y in Equation 1. 3x - y = 5 Equation 2 4x + y = -5 Original Equation 1 y = -4x - 5 Revised Equation 1 2. Substitute -4x - 5 for y in Equation 2 and find the value of x. 3x - y = 5 3x - ( -4x - 5 ) = 5 _7x + _5_ = 5 J7x=_0 x=_0_ Write Equation Substitute _0_ for x in the revised Equation 1 and find the value of y. y = -4x - 5 = -4(0) Substitute -4x - 5 for y. Simplify. Subtract 5 from each side. Divide each side by Check that (J3, -5 ) is a solution by substituting _0^ for x and 5 for y in each of the original equations. Lesson 7.2 Algebra 1 Concepts and Skills Notetaking Guide 145

5 When using substitution, you will get the same solution whether you solve for y first or x first. You should begin by solving for the variable that is easier to isolate. Example 2 Substitution Method: Solve for x First Solve the linear system. 2x - 5y = -13 Equation 1 x + 3y = -1 Equation 2 1. Solve for x in Equation 2. x + 3y = -1 Original Equation 2 x = -3y - 1 Revised Equation 2 2. Substitute -3y - 1 for x in Equation 1 and find the value of y. 2x - 5y ( -3y- 1 ) - 5y = -13-6y- 2-5y = y = -11 y = Write Equation 1. Substitute -3y- 1 for x. Use the distributive property. Combine like terms. Add 2 to each side. Divide each side by Substitute 1 for y in the revised Equation 2 and find the value ofx. x = -3(1) - 1 x = -4 Write revised Equation 2. Substitute 1 for y. Simplify. 4. Check that ( -4, _) is a solution by substituting -4 for x ^ for y in each of the original equations. Answer The solution is ( -4, 1 ). 146 Algebra 1 Concepts and Skills Notetaking Guide Chapter 7

6 Solving Linear Systems by Linear Combinations w Solve a system of linear equations by linear combinations. VOCABULARY Linear combinations A linear combination of two equations is an equation obtained by (1) multiplying one or both equations by a constant and (2) adding the resulting equations. SOLVING A LINEAR SYSTEM BY LINEAR COMBINATIONS Step 1 Arrange the equations with like terms in columns. Step 2 Multiply, if necessary, the equations by numbers to obtain coefficients that are opposites for one of the variables. Step 3 Add the equations from Step 2. Combining like terms with opposite coefficents will eliminate one variable. Solve for the remaining variable. Step 4 Substitute the value obtained from Step 3 into either of the original equations and solve for the other variable. Step 5 Check the solution in each of the original equations. 148 Algebra 1 Concepts and Skills Notetaking Guide Chapter 7

7 Add the Equations Solve the linear system. 7x + 2y = -6 Equation 1 5x - 2y = 6 Equation 2 Add the equations to get an equation in one variable. 7x + 2y = -6 Write Equation 1. 5x - 2y = 6 Write Equation 2. 12x = 0 Add equations. x = 0 Solve for x. i Substitute 0 for x in the first equation and solve for y. 7(_0_) + 2y = -6 Substitute _0_ for _x_. y = -3 Solve for y. Check that (_0_, -3 ) is a solution by substituting _0_ for x and -3 for y in each of the original equations. Answer The solution is ( 0, -3 ). Checkpoint Use linear combinations to solve the system of linear equations. Then check your solution. 1. 4x + y = -4-4x + 2y = x + 3y = 10 12x - 3y = 6 (-2, 4) (1,2) Lesson 7.3 Algebra 1 Concepts and Skills Notetaking Guide 149

8 Multiply then Add Solve the linear system. 3x - 5y = 15 Equation 1 2x + 4y = -1 Equation 2 ( You can get the coefficients of x to be opposites by multiplying the first equation by _2_and the second equation by -3. 3x - 5y = 15 Multiply by _2...; '.' S 6 x- 10 y = 30 2x + 4y = -1 Multiply by (.-3). Add the equations and solve for y. -6 x- 12 y = 3-22y = 33 y = -1.5 Substitute -1.5 for y in the second equation and solve for x. 2x + 4y = -1 Write Equation 2. 2x + 4( -1.5) = -1 Substitute -1.5 for. 2x-_6_ = -1 Simplify. x = 2.5 Solve for X. Answer The solution is ( 2.5, 1.5 ). Checkpoint Use linear combinations to solve the system of linear equations. Then check your solution. 3. x - 3y = 8 3x + 4y = x + 5y = 23 9x - 2y = -32 (5,-1) (-2, 7) 150 Algebra 1 Concepts and Skills Notetaking Guide Chapter 7

9 Special Types of Linear Systems ) Goal Identify how many solutions a linear system has. NUMBER OF SOLUTIONS OF A LINEAR SYSTEM If the two solutions have different slopes, then the system has one solution. Lines intersect: Exactly one solution. ) If the two solutions have the same slope but different y-intercepts, then the system has no solution. X Lines are parallel: No solution. If the two equations have the same slope and the same y-intercepts, then the system has infinitely many solutions. Lines coincide: Infinitely many solutions. Lesson 7.5 Algebra 1 Concepts and Skills Notetaking Guide 153

10 Example 1 A Linear System with No Show that the linear system has no solution. -x + y = -3 Equation 1 -x + y = 2 Equation 2 Method 1: Graphing Rewrite each equation in slope-intercept form. Then graph the linear system. y = x - 3 Revised Equation 1 y = x + 2 Revised Equation 2 -x + y- 2 i 3, -x -5 y + = z X 3 Answer Because the lines have the same slope but different y-intercepts, they are parallel. Parallel lines do not intersect, so the system has no solution. Method 2: Substitution Because Equation 2 can be rewritten as y = x + 2, you can substitute x + 2 for y in Equation 1. -x + y = -3 Write Equation 1. -x + x+ 2 = -3 Substitute x+ 2 fory. 2^-3 Combine like terms. ( Answer The variables are eliminated and you are left with a statement that is false regardless of the values of x and y. This tells you that the system has no solution. 154 Algebra 1 Concepts and Skills Notetaking Guide Chapter 7

11 Example 2 A Linear System with Infinitely Many s Show that the linear system has many solutions. 3x + y = -1 Equation 1-6x - 2y = 2 Equation 2 Method 1: Graphing Rewrite each equation in slope-intercept form. Then graph the linear system. y = ~3x - 1 Revised Equation 1 y ~ ~3x - 1 Revised Equation 2 Answer From these equations you can see that the equations represent the same line. Every point on the line is a solution. 3* + Y = -4 * x A - 2 y = 5 3. y 1 2 J 2 X Method 2: Linear Combinations You can multiply Equation 1 by 2. X + -6x - 2y = Multiply Equation 1 by _2_. Write Equation 2. Add equations. True statement Answer The variables are eliminated and you are left with a statement that is true regardless of the values of x and y. This tells you that the system has infinitely many solutions. Checkpoint Solve the linear system and tell how many solutions the system has. 1. x - 2y = 3-5x + loy = -15 infinitely many solutions 2. -2x + 3y = 4-4x + 6y = 10 no solution Lesson 7.5 Algebra 1 Concepts and Skills Notetaking Guide 155