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 of a linear system Point of intersection Example Find the Point of ntersection Use the graph at the right to estimate the solution of the linear system. Then check your solution algebraically. x + 2y = -4 Equation x - 3y = Equation 2 The lines appear to intersect once at (, ) t t--t=t--t--t i!-+--t--+-+3f-----f--t h«+ y - 4, i-"'" 3 x --.--~ x 3Y... 3 t-----"""'... Check Substitute for x and for y in each equation. x + 2y = -4 x - 3y = + 2( ) J: -4-3( ) J: -4 Answer Because (, ) is a solution of each equation, ( ) is the solution of the system of linear equations. =-,' j t-, Algebra. Concepts and Skills Notetaklng Guide. Chapter 7

2 SOLVNG A LNEAR SYSTEM USNG GRAPH AND-CHECK Step Write each equation in a form that is Step 2 Graph both equations in the Step 3 Estimate the coordinates 9f the ~- Step 4 Check whether the coordinates give a solution by them into each equation of the ::-; , 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 = -2 Equation 3x - 4y = -20 Equation 2. Write each equation in slope-intercept form. Equation Equation 2 5x + 4y = -2 3x - 4y = -20 4y = - 2-4y = - 20 y= y= 2. Graph both equations. 3. Estimate from the graph that the point of intersection is (,_). 4. Check whether (_,_) is a solution by substituting for x and for y in each of the original equations. Equation Equation 2 5x + 4y = -2 3x - 4y = -20 5( ) + 4( ) J: -2 3( ) - 4( ) J: Answer Because (, _) is a solution of each equation in the linear system, (, _) is a solution of the linear system.! y 5r- 3 - f x r - r 3 - Lesson 7.. Algebra Concepts and Skills Notetaklng Guide 43

3 o Checkpoint Use the graph-and-check method to solve the linear system..3x - 4y = 4 2.5x + 2y = 4 9x + 2y = 2 x + 2y = 8 -r y y --7 f-f---l--f l ~ 'f---l t f----l --3 t--- --t--+---t---t--t---i, t x 7 x -:-3f-t l 3. y = -2x Y = 3x + 4 7x - 3y = -6 2x + 5y = 25 }' """ -7, t--- y x _ f- ~ x -3 i- 44 Algebra Concepts and Skills Notetaking Guide. Chapter 7

4 So ving Li ear Systems by Subs itution Goal Solve a linear system by substitution. SOLVNG A LNEAR SYSTEM BY SUBSTTUTON Step Solve one of the equations for one"of its ---- Step 2 Substitute the expression from Step into the other equation and solve for the Step 3 Substitute the value from into the revised equation from and sollve. Step 4 Check the solution in each of the equations. Example Substitution Method: Solve for y First When you use the substitution method, you can check the solution by substituting it for x and for y in each of the original equations. You can also use a graph to check your solution. Solve the linear system. 4x + Y = -5 Equation. Solve for y in Equation. 4x + Y = -5 y = 3x - y = 5 Equation 2 Ori'ginal Equation Revised Equation 2. Substitute for y in Equation 2 and find the value of x. 3x - y = 5 Write Equation 2. 3x - ( ) = Substitute for y. x + = 5 Simplify. x= Subtract from each side. x = Divide each side by 3. Substitute for x in the revised Equation and find the value of y. y= = Check that (_, ) is a solution by substituting _ for x and fory in each of the original equations. Lesson 7.2. Algebra Concepts and Skills Notetaklng Guide 45 \

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 = -3 Equation x + 3y = - Equation 2. Solve for x in Equation 2. x + 3y = - Original Equation 2 x= Revised 'Equation 2 2. Substitute for x in Equation and find the ---- value of y Answer The solution is (, ) x - 5y = -3 Write Equation. 2( ) - 5y = -3 Substitute for x. - 5y = -3 Use the distributive property. = -3 Combine like terms. Add to each side. Divide each side by 3. Substitute for y in the revised Equation 2 and find the value of x. x= Write revised Equation 2. x= Substitute for y. x= Simplify. 4. Check that (, ) is a solution by substit,uting for x and for y in eachof the original equations. 46 Algebra Concepts and Skills Notetaking Guide. Chapter 7

6 o Checkpoint Name the variable that you would solve for first. Explain..x - 2y = 0 2.4x + 2y = 0 x - 8y = -5 7x - y = 2 Use substitution to solve the linear system. 3. y = x - 4. Y = -5x + 3, x - 5y = -5 3x + 2y = -8 Lesson 7.2. A'igebra ; Concepts and Skills Notetaking Guide 47

7 Solving Linear Systems by Linear ombinat ons Goal Solve a system of linear equations by linear combinations. VOCABULARY Linear combinations SOLVNG A LNEAR SYSTEM BY LNEAR COMBNAT,ONS Step Arrange the equations with terms in columns. Step 2 Multiply, if necessary, the equations by numbers to obtain coefficients that are for one of the variables. Step 3 the equations from Step 2. Combining like terms with opposite coefficents will one variable. Solve for the Step 4 Substitute the obtained from Step 3 into _ and Step 5 Check the solution in each of the equations A'igebra Concepts and Skills Notetaking Guide. Chapter 7

8 Example Add the Equations Solve the linear system. 7x + 2y = -6 Equation 5x - 2y = 6 Equation 2 Add the equations to get an equation in one variable. 7x + 2y = -6 Write Equation. 5x - 2y = 6 Write Equation 2. Add equations. Solve for Substitute for in the first equation and solve for 7( ) + 2y = -6 Substitute for Solve for Check that (_, ) is a solution by substituting for x and for y in each of the original equations. Answer The solution is (_, ). o Checkpoint Use linear combinations to solve the system of linear equations. Then check your solution..4x + Y = x + 3y = 0-4x + 2y = 6 2x- 3y = 6 Lesson 7.3. Algebra Concepts and Skills Notetaking Guide.49

9 Linear Systems and Problem Solving Goal Use linear systems to solve real-ufe problems. Example Choosing a Method Health Food A health food store mixes granola and raisins to malke 20 pounds of raisin granola. Granola costs $4 per pound and raisins cost $5 per pound. How many pounds of each should be included for the mixture to cost a total of $85? Verbal Model Pounds of granola + Pounds of raisins Total pounds Price of granola Pounds of granola + Price of raisins Pounds of raisins = Total cost Labels Pounds of granola =_ (pounds) Pounds of raisins = (pounds) Total pounds = (pounds) Price of granola =_ (dollars per pound) Price of raisins = (dollars per pound) Total cost ~ (dollars) Algebraic Model Equation + = Equation 2 Because the coefficients of x and yare in Equation, is most convenient. Solve Equation for and the result in Equation. Simplify to obtain y = Substitute for y in Equation and solve for x. Answer The solution is pounds of raisins and pounds of granola. Lesson 7.4. Algebra Concepts and Skills Notetaking Guide 5

10 Example 2 Multiply then Add Solve the linear system. 3x - 5y = 5 Equation 2x + 4y = - Equation 2 You can get the coefficients of x to be opposites by multiplying the first equation by and the second equation by 3x - 5y = 5 Multiply by x- y= 2x + 4y = - Multiply by x- y= Add the equations and solve for Substitute -- for in the second equation and solve for 2x + 4y = - Write Equation 2. 2x +- 4( ) = - Substitute for 2x - = - Simplify. Solve for - Answer The solution is ( ). -- o Checkpoint Use linear combinations to solve the system of linear equations. Then check your solution. 3. x - 3y = 8 4.6x + 5y = 23 9x - 2y = -32 3x + 4y = 50 Algebra Concepts and Skills Notetaklng Guide. Chapter 7.._--_.~----

11 Linear Systems and Problem Solving Goal Use linear systems to solve real-life problems. Example Choosing a Method Health Food A health food store mixes granola and raisins to make 20 pounds of raisin granola. Granola costs $4 per pound and raisins cost $5 per pound. How many pounds of each should be included for the mixture to cost a total of $85? Verbal Model Pounds of granola + Pounds of raisins Total pounds Price of granola Pounds of granola + Price of raisins Pounds of raisins i ' - Total cost Labels Pounds of granola =_ (pounds) Pounds of raisins = Total pounds = Price of granola =_ Price of raisins = Total cost ~ (pounds) (pounds) (dollars per pound) (dollars per pound) (dollars) Algebraic + Equation - - Model + Equation 2 Because the coefficients of x and yare in Equation, is most convenient. Solve Equation for and the result in Equation. Simplify to obtain y = Substitute for y in Equation and solve for x. Answer The solution is pounds of raisins and pounds of granola. Lesson 7.4. Algebra Concepts and Skills Notetaking Guide 5

12 WAYS TO SOLVE A SYSTEM OF LNEAR EQUATONS Substitution requires that one of the variables be on one side of the equation. t is especially convenient when one of the variables has a coefficient of or Linear Combinations can be applied to any system, but it is especially convenient when a appears in different equations with that are Graphing can provide a useful method for a solution. o Checkpoint Choose a method to solve the linear system. Explain your choice, and then solve the system.. n Example, suppose the health food store wants to make 30 pounds of raisin granola that will cost a total of $25. How many pounds of granola and raisins do they need? Use the prices given in Example. 52 Algebra Concepts and Skills Notetaking Guide. Chapter 7 --_.--_._ ~------

13 Special Types of Linear Systems al dentify how many solutions a linear system has. NUMBER OF SOLUTONS OF A LNEAR SYSTEM f the two sollutions have slopes, then the system has ---- one solution. Lines intersect: solution f the two solutions have the slope but _ y-intercepts, then the system has no solution. Lines are parallel: solution. f the two equa.tions have the slope and the _ y-intercepts, then the system has infinitely many solutions. Lines coincide: solutions. x Lesson 7.5. Algebra Concepts and Skills Notetaking Guide 53

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

15 Example 2 A Linear System with nfinitely Many s Show that the linear system has many solutions. 3x + y = - Equation - 6x -' 2y = 2 Equation 2 Method : Graphing Rewrite each equation in slope-intercept form. Then graph the linear system. y = y = Revised Equation Revised Equation 2 Answer From these equations you can see that the equations represent the same line. point on the line is a solution. --- Method 2: Linear Combinations You can multiply Equation by Ox + Oy = 0 Multiply Equation by. - 6x - 2y = 2 Write Equation 2. o = 0 Add equations. statement Answer The var'iables are and you are left with a statement that is regardless of the values of x and y. This tells you that the system has Y f- -, : x o Checkpoint Solve the linear system and tell how many solutions the system has.. x - 2y = x + 3y = 4-5x + 0y = -5-4x + 6y = 0 Lesson 7.5. Algebra Concepts and Skills Notetaking Guide 55

16 Example 2 A Linear System with nfinitely Many s Show that the linear system has many solutions. 3x + y = - Equation - 6x - 2y = 2 Equation 2 Method : Graphing Rewrite each equation in slope-intercept form. Then graph the linear system. y = y = Revised Equation Revised Equation 2 Answer From these equations you can see that the equations represent the same line. point on the line is a solution. --- Method 2: Linear Combinations You can multiply Equation by Ox + Oy = 0 Multiply Equation by - 6x - o 2y = 2 Write Equation 2. = 0 Add equations. statement Answer The variables are and you are left with a statement that is regardless of the values of x and y. This tells you that the system has i y x.~ 3 o Checkpoint Solve the linear system and tell how many solutions the system has..x - 2y = x + 3y = 4-5x + 0y = -5-4x + 6y = 0 Lesson 7.5. Algebra Concepts and Skills Notetaking Guide 55

17 Systems of Linear nequalities Goal Graph a system of linear i'nequalities. VOCABULARY System of linear inequalities of a system of linear inequalities GRAPHNG A SYSTEM OF LNEAR NEQUALTES Step the boundary lines of each inequality. Use a line if the inequality is < or > and a line if -- the inequality is :::; or z. Step 2 the appropriate half-plane for each inequality. Step 3 the solution of the system of inequalities as the intersection of the half-planes from Step Algebra Concepts and Skills Notetaking Guide. Chapter ~ _.. _-_..._.. _--

18 Example Graph a System of Two Linear nequalities \ To check your graph, choose a point in the overlap of the half-planes. Then substitute the coordinates into each inequality. f each inequality is true, then the point is a solution. Graph the system of linear inequalities. y - x ::::: - nequality x + 2y < nequality 2 Graph both inequalities in the same coordinate plane. The graph of the system is the overlap, or of the two half-planes. Example 2 J 3 i y r~;--l + ~ i -3-3 xl T Graph a System of Three Linear nequalities Graph the system of linear inequalities. y::::: - 3 nequality x < 2 nequality 2 y < x + nequality 3 3 J The graph of y ::::: -3 is the half-plane and the line,. y 3 i The graph of x < 2 is the half-plane to the of the line The graph of y < x + is the half-plane the line i _- t3 -i 3, i X Finally, the graph of the system is the, or, of the :-:----:---:-: three half-planes. Lesson 7.6. Algebra Concepts and Skills Notetaking Guide 57

19 Example 3 Write a System of Linear nequalities Write a system of inequalities that defines the shaded region at the right. The graph of one inequality is the. half-plane to the left of --- The graph of the other inequality is the half-plane to the right of _ The shaded region of the graph is the vertical band that lies ---- the two vertical lines, and, but not the Hnes. 3 Y -il i r t- ~~! Answer The system of linear inequalities below defines the shaded region. nequality nequality 2 o Checkpoint Complete the following exercises.. Graph the system of linear inequalities. y<2x+2 y> --x - 2 y 3-3 -,- 3 x Write a system of linear inequalities that defines the shaded region y -3 - i 3 x 58 Algebra Concepts and Skills Notetaklng Guide. Chapter 7

20 Words to Review Give an example of the vocabulary word. System of linear equations of a linear system Point of intersection Linear combination System of linear inequalities of a system of Unear inequalities Review your notes and Chapter 7 by using the Chapter Review on pages of your textbook. Words to Review. Algebra Concepts and Skills Notetaking Guide 59