Systems of Linear Equations 11/11/12 lntaylor ©

Table of Contents Learning Objectives Solving Systems by Graphing Solving Systems by Substitution Solving Systems by Elimination Word Problems Practice 11/11/12 lntaylor ©

Learning Objectives TOC 11/11/12 lntaylor ©

Solving Systems of Linear Equations In these sections you will learn/review how to: –Meet or exceed proficiency in AF 9.0 –Pass questions regarding these standards on District or State Tests –Solve Systems of Equations using three (3) different methods TOC 11/11/12 lntaylor ©

1 st Method Graphing TOC 11/11/12 lntaylor ©

Systems of Linear Equations Graphing Method –Graph two different linear equations –Determine the point where the lines crosses –Prove that point works with both equations TOC 11/11/12 lntaylor ©

Find the solution to: y = 2x + 1 and y = - x + 4 TOC 11/11/12 lntaylor ©

0,0 Find the solution to y = 2x + 1 and y = - x + 4 Step 1 – Graph the 1 st line Identify slope m and b Locate 0,0 Go up or down b Put a dot for the 1 st point Go up or down Δ y Go right Δ x Put a dot for 2 nd point Connect dots Label the line y = 2x + 1 TOC 11/11/12 lntaylor © Step 2 – Graph the 2 nd line Identify slope m and b Locate 0,0 Go up or down b Put a dot for the 1 st point Go up or down Δ y Go right Δ x Put a dot for 2 nd point Connect dots Label the line Find the point where the lines cross Plug into each equation The point (1,3) is the solution for both equations y = - x + 4 y = 2x + 1 and y = -x = 2(1) + 1 and 3 = = 3 and 3 = 3 (1,3)

Now you try! y = 1/2x + 1 and y = - x + 4 TOC 11/11/12 lntaylor ©

0,0 Find the solution to y = 1/2x + 1 and y = - x + 4 Step 1 – Graph the 1 st line Identify slope m and b Locate 0,0 Go up or down b Put a dot for the 1 st point Go up or down Δ y Go right Δ x Put a dot for 2 nd point Connect dots Label the line y = 1/2x + 1 TOC 11/11/12 lntaylor © Step 2 – Graph the 2 nd line Identify slope m and b Locate 0,0 Go up or down b Put a dot for the 1 st point Go up or down Δ y Go right Δ x Put a dot for 2 nd point Connect dots Label the line Find the point where the lines cross Plug into each equation The point (1,3) is the solution for both equations y = - x + 4 y = 1/2x + 1 and y = - x = 1/2(2) + 1 and 2 = = 2 and 2 = 2 (2,2)

Now you try! y = 1/3x – 4 and y = - x TOC 11/11/12 lntaylor ©

0,0 Find the solution to y = 1/3x - 4 and y = - x Step 1 – Graph the 1 st line Identify slope m and b Locate 0,0 Go up or down b Put a dot for the 1 st point Go up or down Δ y Go right Δ x Put a dot for 2 nd point Connect dots Label the line y = 1/3x - 4 TOC 11/11/12 lntaylor © Step 2 – Graph the 2 nd line Identify slope m and b Locate 0,0 Go up or down b Put a dot for the 1 st point Go up or down Δ y Go right Δ x Put a dot for 2 nd point Connect dots Label the line Find the point where the lines cross Plug into each equation The point (1,3) is the solution for both equations y = - x + 4 y = 1/3x – 4 and y = - x -3 = 1/3(3) – 4 and - 3 = = - 3 and - 3 = - 3 (3, - 3)

2 nd Method Substitution TOC 11/11/12 lntaylor ©

Systems of Linear Equations Substitution Method –Set one equal to y –Substitute the 2 nd equation into the 1 st equation TOC 11/11/12 lntaylor ©

Find the solution to: y = 2x + 1 and y = - x + 4 TOC 11/11/12 lntaylor ©

Step 4 Step 5 Rewrite without the y and solve Plug the answer into both equations Step 6The 1 st number is x; the 2 nd number is y Step 7The solution is (x,y) or in this equation (1,3) Step 2 Step 1 Write the 1 st equation as y = Write the 2 nd equation as = y Step 3 They are both = y 11/11/12 lntaylor © TOC Find the solution to y = 2x + 1 and y = - x + 4y = 2x + 1- x + 4 = y -x + 4 = 2x – 1 = 2x + x 3 = 3x x = 1 -x + 4 = 2x (1) + 4 = 2(1) = = 3 (1,3)

Now you try! y = 1/2x + 1 and y = - x + 4 TOC 11/11/12 lntaylor ©

Step 4 Step 5 Rewrite without the y and solve Plug the answer into both equations Step 6The 1 st number is x; the 2 nd number is y Step 7The solution is (x,y) or in this equation (2,2) Step 2 Step 1 Write the 1 st equation as y = Write the 2 nd equation as = y Step 3 They are both = y 11/11/12 lntaylor © TOC Find the solution to y = 1/2x + 1 and y = - x + 4y = 1/2x + 1- x + 4 = y -x + 4 = 1/2x – 1 = 1/2x + x 3 = 1½ x 6 = 3x x = 2 -x + 4 = 1/2x (2) + 4 = ½(2) = = 2 (2,2)

Now you try! y = 1/3x – 4 and y = - x TOC 11/11/12 lntaylor ©

Step 4 Step 5 Rewrite without the y and solve Plug the answer into both equations Step 6The 1 st number is x; the 2 nd number is y Step 7The solution is (x,y) or in this equation (3,- 3) Step 2 Step 1 Write the 1 st equation as y = Write the 2 nd equation as = y Step 3 They are both = y 11/11/12 lntaylor © TOC Find the solution to y = 1/3x - 4 and y = - xy = 1/3x - 4- x = y -x = 1/3x = 1/3x + x 4 = 1⅓ x 12 = 4x x = 3 -x = 1/3x (3) = ⅓(3) = = - 3 (3,- 3)

3 rd Method Elimination TOC 11/11/12 lntaylor ©

Systems of Linear Equations Elimination Method –Write the equations underneath each other –Choose a multiplier for one equation that eliminates a variable –Plug in the answer and solve for the 2 nd variable TOC 11/11/12 lntaylor ©

Find the solution to: y = 2x + 1 and y = - x + 4 TOC 11/11/12 lntaylor ©

Step 4 Step 5 Add the equations and solve for the variable Plug the answer into an equation and solve Step 6Make sure you put the numbers in the right order! Step 7The solution is (x,y) or in this equation (1,3) Step 2 Step 1 Write the equations underneath each other Pick a multiplier that will eliminate 1 variable Step 3 Rewrite the equations 11/11/12 lntaylor © TOC Find the solution to y = 2x + 1 and y = - x + 4 y = 2x + 1 y = - x + 4 (2) y = (2)(-x + 4) 2y = - 2x + 8 3y = 9 y = 3 (x,3) y = 2x = 2x = 2x x = 1 (1,3)

Now you try! y = 1/2x + 1 and y = - x + 4 TOC 11/11/12 lntaylor ©

Step 4 Step 5 Add the equations and solve for the variable Plug the answer into an equation and solve Step 6Make sure you put the numbers in the right order! Step 7The solution is (x,y) or in this equation (2,2) Step 2 Step 1 Write the equations underneath each other Pick a multiplier that will eliminate 1 variable Step 3 Rewrite the equations 11/11/12 lntaylor © TOC Find the solution to y = ½x + 1 and y = - x + 4y = ½x + 1y = - x + 4 (2) y = (2)(½x + 1) 2y = x + 2 3y = 6 y = 2 (x,2) y = ½x = ½x = ½x 2 = x (2,2)

Now you try! y = 1/3x – 4 and y = - x TOC 11/11/12 lntaylor ©

Step 4 Step 5 Add the equations and solve for the variable Plug the answer into an equation and solve Step 6Make sure you put the numbers in the right order! Step 7The solution is (x,y) or in this equation (3,- 3) Step 2 Step 1 Write the equations underneath each other Pick a multiplier that will eliminate 1 variable Step 3 Rewrite the equations 11/11/12 lntaylor © TOC Find the solution to y = ⅓x - 4 and y = - x y = ⅓x - 4 y = - x (3) y = (3)(⅓x - 4) 3y = x y = - 12 y = - 3 (x,- 3) y = ⅓x = ⅓x = ⅓x 3 = x (3,- 3)

Word Problems TOC 11/11/12 lntaylor ©

Step 4 Step 5 Pick a multiplier that will eliminate 1 variable Add the equations and solve Step 6Plug in the answer and solve Step 2 Step 1 Highlight the first equation Highlight the second equation Step 3 Write the equations underneath each other 11/11/12 lntaylor © TOC 4A + 9C = $82 4A + 1C = $34 – 1(4A + 1C) = – 1($34) – 4A – 1C = - $34 4A + 9C = $82 – 4A – 1C = – $34 8C = $48 C = $6 4A + 9C = $82 4A + 9($6) = $82 4A + $54 = $82 4A = $28 A = $7 Adult Tickets were $7 Child Tickets were $6

Step 4 Step 5 Pick multiplier(s) that will eliminate 1 variable Add the equations and solve Step 6Plug into the original equation and solve Step 2 Step 1 Highlight the first equation Highlight the second equation Step 3 Write the equations underneath each other 11/11/12 lntaylor © TOC 5A + 7C = $129 6A + 10C = $174 6(5A + 7C) = 6($129) or 30A + 42C = $774 30A + 42C = $774 – 30A – 50C = – $870 – 8C = – $96 C = $12 5A + 7C = $129 5A + 7($12) = $129 5A + $84 = $129 5A = $45 A = $9 Adult Tickets were $9 Child Tickets were $12 – 5(6A + 10C) = – 5($174) or – 30A – 50C = – $870

Step 4 Step 5 Pick multiplier(s) that will eliminate 1 variable Add the equations and solve Step 6Plug into the original equation and solve Step 2 Step 1 Highlight the first equation Highlight the second equation Step 3 Write the equations underneath each other 11/11/12 lntaylor © TOC 3V + 6B = V + 5B = (3V + 6B) = 10(192) or 30V + 60B = V + 60B = 1920 – 30V – 15B = – B = 1305 B = 29 3V + 6B = 192 3V + 6(29) = 192 3V = 192 3V = 18 V = 6 6 students in each van 29 students in each bus – 3(10V + 5B) = – 3(205) or – 30V – 15B = – 615

Step 4 Step 5 No multiplier(s) is needed here! Add the equations and solve Step 6Plug into the original equation and solve Step 2 Step 1 Highlight the first equation Highlight the second equation Step 3 Write the equations underneath each other 11/11/12 lntaylor © TOC A + B = 16 A – B = 2 2A = 18 A = 9 A + B = B = 16 B = 7 A is 9 B is 7

Practice TOC 11/11/12 lntaylor ©

TOC ProblemAnswer y = 2x + 5 and y = – 5x – 16 y = – 4x + 5 and y = – 2x + 3 y = 5x + 10 and 7x – 7y = - 14 y = – 6x + 20 and – x – y = - 10 y = 5x – 5y = – 20 and x + 2y = 17 x – 4y = 16 and – 3x – 3y= 12 – 5x – 4y = – 22 and – 3x – y = – 16 – x – 5y = – 3 and – 8x – 6y = 10 2x – 7y = 6 and – 7x + 5y = – 21 > (-3, -1) > > > > > (1, 1) (- 2, 0) (2, 8) (3, 7) (0, - 4) > (6, - 2) > (- 2, 1) > (3, 0) clear answers