7.1: Systems of Linear Equations (2 Variables)y

x

y

x

y

x

y = ½ x –3y = (-2/3)x + 2 One Solution (x, y)

y = ½ x - 3y = ½ x + 2

No Solutions

y = ½ x + 32y = x + 6

Infinite solutions (x,y)

Solving Systems: Graphical Method

Step 1: Graph both equationsStep 2: Find the point of intersection

y

x

x + 2y = 7x – y = 4

x y x y0 3 ½ 0 -41 3 1 -3

Solution: (5, 1)

Solving Systems: Substitution Method

y

x

x + 2y = 7x - y = 4

Step 1: Solve for x or y in 1 equationStep 2: Substitute into the 2nd equationStep 3: Solve algebraically to find 1 variableStep 4: Plug the ‘found variable’ back in and solve for the second variable.

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)

Solving Systems:Addition Method

x + y = 10x – y = 8

2x = 18

2 2

x = 9

x + y = 109 + y = 10 y = 1

Solution (9, 1)

x + 2y = 7x - y = 42

x + 2y = 72x - 2y = 8

3x = 15

3 3

x = 5

x – y = 45 - y = 4 y = 1Solution (5, 1)

2A + 3B = -13A + 5B = -2

5

-3

10A + 15B = -5-9A - 15B = 6

A = 1

2A + 3B = -12(1) + 3B = -12 + 3B = -1 3B = -3 B = -1

ApplicationsWrite 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?

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 unitsP = 80 - .0008x (Demand Equation) P = price in dollars

As the price of a product increases, demand for it decreasesAs 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

7.2 Three Equations & Three Unknowns

x + y + z = 0x + 2y – 3z = 53x + 4y + 2z = -1

Step 1: Eliminate 1 variable (the same variable) from two sets of equationsStep 2: Solve the resulting 2 equations/2 unknowns w/substitution or additionStep 3: Solve for the 3rd unknown using found variables and the 3rd equation.

x + z = 3x + 2y – z = 12x – y + z = 3

Find solution: {5, -3, -2} Find solution: {1, 1, 2}

Application: InvestmentsKelly 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 investmentIn Treasury bills to be $3000 more than her investment in corporate bonds.How much money should Kelly place in each investment?

NonSquare SystemsLinear Systems where the number of equations is unequal to the number of variables.

ExampleX + 5y + 3z = 7 -2x - 10y - 6z = -142x + 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 zThe solution set is {-53z +47, 10z -8, z)

Review: Graphing Systems of Linear Inequalities

Graph: x + y < 4 x y0 41 3

Test Point (0,0) 0 + 0 < 4 --- TRUE

y

x

Step 1: Graph the lineStep 2: Decide: Dashed or SolidStep 3: Choose a test point on 1 sideStep 4: Plug in test point & checkStep 5: Shade the TRUE side.

Graph the system of inequalities2x – y < 2x + 2y > 6

x y x y0 -2 0 31 0 2 2

y

x

Test and ShadeSee in-class example

7.4 Non-Linear Systems of EquationsFor 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 = -22x2 – y = 0 y = 1/2

Using substitution, orY = 2x2

3x - 2x2 = -2 3(2) –y = -22x2 -3x -2 = 0 y = 8( 2x + 1)(x – 2) = 0X = -1/2 or x = 2

Graphing Systems of Non-Linear Inequalities

Try these Systems:

Y ≥ x2 + 1

Y ≤ -x + 13

Y < 4x + 13

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 factorsIf Q can be factored as Q(x) = (x – a1)(x – a2)…(x – an)

A1…An are constants

Case 2: Q has a repeated linear factorsIf Q has factors (x – a)m , then that portion of P(x)/Q(x) is

Case 3: Q has a nonrepeated irreducible quadratic factorIf Q has factor ax2 + bx + c, that portion of P(x)/Q(x) is

1 2

1 2

... ,n

n

P x AA A

Q x x a x a x a

1 2

2... ,n

m

AA A

x a x a x a

2

Ax B

ax bx c

1 1 2 2

22 2 2... m m

m

A x BA x B A x B

ax bx c ax bx c ax bx c

3x + 26 (x – 3)(x + 4)

x + 4 (x + 3)(x -1)2

3x2 -8x + 1 (x - 4)(x2 +1)

2x4 –x3 + 13x2 -2x +13(x - 1)(x2 + 4)2

Case 1: Q has only nonrepeated linear factorsIf Q can be factored as Q(x) = (x – a1)(x – a2)…(x – an)

A1…An are constants

1 2

1 2

... ,n

n

P x AA A

Q x x a x a x a

3x + 26 (x – 3)(x + 4)

Partial Fraction Decomposition (Case - 1)

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)9+ 26 = 7A + 035= 7AA = 53x + 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

(x – 3) (x + 4) Method 1

OR

3x + 26 = A(x + 4) + B(x – 3)3x + 26 = Ax + 4A + Bx -3B3x + 26 = x(A + B) + (4A – 3B)So…4A – 3B = 26 A + B = 3 So, A = 3 – B

4(3 – B) – 3B = 2612 -4B -3B = 2612 -7B = 26-7B = 14B = -2Substitute -2 in for B => A = 5

Method 2

Partial Fraction Decomposition (Case - 2)Case 2: Q has a repeated linear factorsIf Q has factors (x – a)m , then that portion of P(x)/Q(x) is

x + 4 (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=11 + 4 = (1/16)(1-1)2 + B(1 + 3)(1 – 1) + C(1 + 3) 5 = 0 + 0 +4CC = 5/4

(x + 3) (x - 1)2

1 2

2... ,n

m

AA A

x a x a x a

Partial Fraction Decomposition (Case - 3)Case 3: Q has a nonrepeated irreducible quadratic factorIf Q has factor ax2 + bx + c, that portion of P(x)/Q(x) is

2

Ax B

ax bx c

3x2 -8x + 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

(x – 4) (x2 + 1)

3x2 – 8x + 1 = (x2 + 1) + (Bx)(x – 4)Let x = 13(1)2 – 8(1) + 1 = ((1)2 + 1) + (B(1))(1 – 4) -4 = 2 -3B -6 = -3B B = 2

Partial Fraction Decomposition (Case - 4)

1 1 2 2

22 2 2... m m

m

A x BA x B A x B

ax bx c ax bx c ax bx c

2x4 –x3 + 13x2 -2x +13(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 = 12 – 1 + 13 – 2 + 13 = A(5)2 + (B + C)(0)(5) + (D = E)(0) 25 = 25A => A = 12x4 + -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

(x - 1) (x2 + 4)2

1 + x + x + 3x – 1 x2 + 4 (x2 + 4)2