7.1: Systems of Linear Equations (2 Variables)

1. y y y x x x 7.1: Systems of Linear Equations (2 Variables) y = ½ x + 3 2y = x + 6 Infinite solutions (x,y) y = ½ x - 3 y = ½ x + 2 No Solutions y = ½ x –3 y = (-2/3)x + 2 One Solution (x, y)

2. y x Solving Systems: Graphical Method Step 1: Graph both equations Step 2: Find the point of intersection x + 2y = 7 x – y = 4 x y x y 0 3 ½ 0 -4 1 3 1 -3 Solution: (5, 1)

3. y x Solving Systems: Substitution Method Step 1: Solve for x or y in 1 equation Step 2: Substitute into the 2nd equation Step 3: Solve algebraically to find 1 variable Step 4: Plug the ‘found variable’ back in and solve for the second variable. x + 2y = 7 x - y = 4 Step 1: x – y = 4 [add y to both sides] x = y + 4 Step 2:x + 2y = 7 (y + 4) + 2y = 7 Step 3: 3y + 4 = 7 3y = 3 y = 1 Step 4: x = y + 4 x = (1) + 4 x = 5 Solution (5, 1)

4. 5 -3 2 2x = 18 x + 2y = 7 2x - 2y = 8 • 2 • x = 9 3x = 15 3 3 x = 5 Solving Systems:Addition Method x + y = 10 x – y = 8 x + 2y = 7 x - y = 4 2A + 3B = -1 3A + 5B = -2 10A + 15B = -5 -9A - 15B = 6 A = 1 • 2A + 3B = -1 • 2(1) + 3B = -1 • + 3B = -1 • 3B = -3 • B = -1 • x + y = 10 • + y = 10 • y = 1 • Solution (9, 1) x – y = 4 5 - y = 4 y = 1 Solution (5, 1)

5. Applications Write a system of equations for each scenario and solve: 1 Bank Account Problem: Carlos has 2 bank accounts. He has seven times as much in his savings account as in his checking account. In all, he has $3,200 in the bank. Find out how much Carlos has in each account. 2. Coins in a Jar: There are 93 coins in a jar. The coins are quarters and dimes All together the coins total $18.60. How many quarters and dimes are in the jar? 3. Dimensions of a Rectangle: A soccer field has a perimeter of 320 yards. The length measures 40 yards more than its width. What are the field dimensions? 4. Acid Mixture (Revisited): How many ounces of a 5% hydrochloric acid and 20% hydrochloric acid must be combined to get 10 oz of solution that is 12.5% acid?

6. Finding Equilibrium Find the equilibrium point if the supply and demand functions for a new brand of digital video recorder (DVR) are given by the system: P = 60 + .0012x (Supply Equation) x = number of units P = 80 - .0008x (Demand Equation) P = price in dollars As the price of a product increases, demand for it decreases As the price of a product increases, supply of the product increases The equilibrium point occurs when the supply and demand is equal. 60 + .0012x = 80 - .0008x => Solving gives x = 10,000 P = 60 + .0012 (10000) = 72

7. 7.2 Three Equations & Three Unknowns Step 1: Eliminate 1 variable (the same variable) from two sets of equations Step 2: Solve the resulting 2 equations/2 unknowns w/substitution or addition Step 3: Solve for the 3rd unknown using found variables and the 3rd equation. Find solution: {5, -3, -2} Find solution: {1, 1, 2} x + y + z = 0 x + 2y – 3z = 5 3x + 4y + 2z = -1 x + z = 3 x + 2y – z = 1 2x – y + z = 3

8. Application: Investments • Kelly has $20,000 to invest As her financial planner, • you recommend that she diversify into three investments: • Treasury bills that yield 5% simple interest • Treasure bonds that yield 7% simple interest, and • Corporate bonds that yield 10% simple interest. • Kelly wishes to earn $1390 per year in income. Also, Kelly wants her investment • In Treasury bills to be $3000 more than her investment in corporate bonds. • How much money should Kelly place in each investment?

9. NonSquare Systems Linear Systems where the number of equations is unequal to the number of variables. Example X + 5y + 3z = 7  -2x - 10y - 6z = -14 2x + 11y -4z = 6  2x + 11y - 4z = 6 Y -10z = -8 Now we have: X + 5y + 3z = 7 y -10z = -8 => Y = 10z -8 Subtitute: X + 5(10z -8) + 3z = 7 X +53z -40 = 7  X = -53z +47 The system has infinitely many solutions, one for each z The solution set is {-53z +47, 10z -8, z)

10. y y x x Review: Graphing Systems of Linear Inequalities Graph: x + y < 4 x y 0 4 1 3 Graph the system of inequalities 2x – y < 2 x + 2y > 6 x y x y 0 -2 0 3 1 0 2 2 Step 1: Graph the line Step 2: Decide: Dashed or Solid Step 3: Choose a test point on 1 side Step 4: Plug in test point & check Step 5: Shade the TRUE side. Test and Shade See in-class example Test Point (0,0)  0 + 0 < 4 --- TRUE

11. 7.4 Non-Linear Systems of Equations For non-linear equations, there is no general methodology. Only experience using the techniques in your ‘mathematical toolbox’ can help you. Graphing the equations, this can give you quick insight into howmany intersection points / solutions the system has. 3x – y = -2 3(-1/2) – y = -2 2x2 – y = 0 y = 1/2 Using substitution, or Y = 2x2 3x - 2x2 = -2 3(2) –y = -2 2x2 -3x -2 = 0 y = 8 ( 2x + 1)(x – 2) = 0 X = -1/2 or x = 2

12. Graphing Systems of Non-Linear Inequalities

13. Try these Systems: Y ≥ x2 + 1 Y ≤ -x + 13 Y < 4x + 13

14. 7.5 Partial Fraction Decomposition (4 Cases) (Decomposing a rational expression (P/Q) into a sum of fractions) Case 1:Q has only nonrepeated linear factors If Q can be factored as Q(x) = (x – a1)(x – a2)…(x – an) A1…An are constants 3x + 26 (x – 3)(x + 4) Case 2: Q has a repeated linear factors If Q has factors (x – a)m , then that portion of P(x)/Q(x) is x + 4 (x + 3)(x -1)2 Case 3: Q has a nonrepeated irreducible quadratic factor If Q has factor ax2 + bx + c, that portion of P(x)/Q(x) is 3x2 -8x + 1 (x - 4)(x2 +1) 2x4 –x3 + 13x2 -2x +13 (x - 1)(x2 + 4)2

15. Partial Fraction Decomposition (Case - 1) Case 1:Q has only nonrepeated linear factors If Q can be factored as Q(x) = (x – a1)(x – a2)…(x – an) A1…An are constants 3x + 26 (x – 3)(x + 4) Method 2 (x – 3) (x + 4) Method 1 3x + 26 = A(x + 4) + B(x – 3) 3x + 26 = Ax + 4A + Bx -3B 3x + 26 = x(A + B) + (4A – 3B) So… 4A – 3B = 26 A + B = 3 So, A = 3 – B 4(3 – B) – 3B = 26 12 -4B -3B = 26 12 -7B = 26 -7B = 14 B = -2 Substitute -2 in for B => A = 5 • 3x + 26 = A + B _ • (x – 3)(x + 4) (x – 3) (x + 4) • 3x + 26 = A(x + 4) + B(x – 3) • Let x=3 (it zeros out an unknown) • 3(3) + 26 = A(3 + 4) + B(3 – 3) • + 26 = 7A + 0 • = 7A • A = 5 • 3x + 26 = 5(x + 4) + B(x – 3) • Let x=0(or any number-B is last 1 to find) • 3(0) + 26 = 5(0 + 4) + B(0 – 3) • 26 = 20 -3B => 6 = -3B => B = -2 OR

16. Partial Fraction Decomposition (Case - 2) Case 2: Q has a repeated linear factors If Q has factors (x – a)m , then that portion of P(x)/Q(x) is x + 4 (x + 3)(x -1)2 (x + 3) (x - 1)2 x + 4 = A + B + C __ (x + 3)(x - 1)2 (x + 3) (x – 1) (x-1)2 x + 4 = A(x-1)2 + B(x + 3)(x – 1) + C(x + 3) Let x = -3 -3 + 4 = A(-3 – 1)2 + B(-3 + 3)(-3 – 1) + C (-3 + 3) 1 = 16A + 0 + 0 A = 1/16 x + 4 = (1/16)(x-1)2 + B(x + 3)(x – 1) + C(x + 3) Let x=1 1 + 4 = (1/16)(1-1)2 + B(1 + 3)(1 – 1) + C(1 + 3) 5 = 0 + 0 +4C C = 5/4

17. Partial Fraction Decomposition (Case - 3) Case 3: Q has a nonrepeated irreducible quadratic factor If Q has factor ax2 + bx + c, that portion of P(x)/Q(x) is 3x2 -8x + 1 (x - 4)(x2 +1) (x – 4) (x2 + 1) 3x2 – 8x + 1 = A + Bx + C (x – 4)(x2 + 1) (x – 4) (x2 + 1) 3x2 – 8x + 1 = A(x2 + 1) + (Bx + C)(x – 4) Let x=4 3(4)2 – 8(4) + 1 = A(42 + 1) + (B(4) + C)(4 – 4) 17 = 17A + 0 A = 1 3x2 – 8x + 1 = (x2 + 1) + (Bx + C)(x – 4) Let x=0 3(0)2 – 8(0) + 1 = (02 + 1) + (B(0) + C)(0 – 4) 1 = 1 -4C 0 = -4C C = 0 3x2 – 8x + 1 = (x2 + 1) + (Bx)(x – 4) Let x = 1 3(1)2 – 8(1) + 1 = ((1)2 + 1) + (B(1))(1 – 4) -4 = 2 -3B -6 = -3B B = 2

18. Partial Fraction Decomposition (Case - 4) 2x4 –x3 + 13x2 -2x +13 (x - 1)(x2 + 4)2 (x - 1) (x2 + 4)2 • 2x4 + -x3 +13x2 -2x + 13 = A + Bx + C + Dx + E • (x - 1)(x2 + 4)2 x - 1 x2 + 4 (x2 + 4)2 • 2x4 + -x3 +13x2 -2x + 13 = A(x2 + 4)2 + (Bx + C)(x – 1)(x2 + 4) + (Dx + E)(x – 1) • Let x = 1 • 2 – 1 + 13 – 2 + 13 = A(5)2 + (B + C)(0)(5) + (D = E)(0) • 25 = 25A => A = 1 • 2x4 + -x3 +13x2 -2x + 13 = (x2 + 4)2 + (Bx + C)(x – 1)(x2 + 4) + (Dx + E)(x – 1) • = x4 + 8x2 + 16 + (Bx2 –Bx +Cx –C)(x2 + 4) +Dx2 –Dx +Ex – E • = x4+8x2+16+Bx4–Bx3+Cx3–Cx2 + 4Bx2 – 4Bx + 4Cx – 4C +Dx2 –Dx +Ex –E • = x4(1 + B) + x3 (-B + C)+ x2(8 – C + 4B +D) + x(-4B+4C– D+E)+(16-4C–E) • B + 1 = 2 B = 1 • B = C = -1 C = 0 • 8 – C + 4B + D = 13 D = 1 • -4B +4C – D + E = -2 E = 3 • 16 – 4C – E = 13 • 1 + x + x + 3 • x – 1 x2 + 4 (x2 + 4)2