College Algebra Week 4Chapter 7.1-7.2

The University of Phoenix

Inst. John Ensworth

Short section – 7.1 to 7.2

• In these two sections, we work a bit more with situations where we have two equations.

• We want to find if and where they meet!

• We want to know what (x,y) they share.

• This is LIKE our overlapping shading, but we’re just looking for one point!

Section 7.1 Graphing and Substituting to find that special

(x,y)

• We’ll take two equations, graph them and see where they cross…

Example 1 A system with only one solution pg452

• y=x+2 done… m=1 and (0,2)

• x+y=4 y=-x+4 m=-1 and (0,4)

• Plot it!

• It looks like

• (1,3)

• Let’s check it!

Checking Example 1

• (1,3) Plug into the original equations

• y=x+2 x+y=4

• 3=1+2 AND 1+3=4

• 3=3 AND 4=4

• Both at the same time and both right!

• GREAT!

**Examples 7-14**

Example 2 – Infinite Solutions pg 453

2(y+2)=x x-2y=4

Solve for y (y=mx+b form)

2y+4=x x-2y=4

2y=x-4 -2y=-x+4

y=1/2 x –2 y=1/2 x –2

The same slope (1/2) and y-intercept (0,-2)!

Example 2

• Plotting:

**Examples 15-16**

Example 3 another graphing pg 453

• 2x-3y=6 and 3y-2x=3

• Get that y=mx+b form!

• -3y=-2x+6 and 3y=2x+3

• y=2/3 x –2 and x= 2/3 y +1

• Giving us graph fodder…

• =2/3 and (0,-2) and m=2/3 and (0,1)

Example 3 the graph

• Oh oh!

• No answer

• Null Set

**Examples 17-20**

Definitions!• So we say we have a system of equations that are

independent, inconsistent or dependent based on how many solutions there are.

Solving an independent system by substitution

Example 4 page 455y=2x-3

y=x+5 plug this x+5 in for y in the 1st equation

2x-3=x+5

x-3=5

x=8 plug into one of the above equations

y=8+5=13 plugging into the other y=2(8)-3=13 Check!

**Examples 35-46**

A quicker method

• The graphing method we’ve used is clunky. You have to make a graph every time. What if the answer is 8.394349 ? Can you read the graph THAT closely?

• So we’ll try another method… we’ll use the “Substitution Method”

• (There must be some substituting of some sort going on here!)

Example 5 Independent system solved by substitution pg455

• 2x+3y=8

• y+2x=6 solve for y y=-2x+6

• Notice that the second equation tells us what y equals. Why not stick it INTO the y in the first equation?

• Are we crazy enough to do THAT?

Ex 5 Substituting

• 2x+3y=8• y=-2x+6• 2x+3(-2x+6)=8• 2x-6x+18=8• -4x+18=8• -4x=-10• x= 5/2 we have half of the answer, what is y?

Ex 5 finding y

• Since x= 5/2 we can use our second equation

y= -2x+6

y= -2(5/2) +6 = -5+6 = 1 So we have (5/2,1)

• Check it!

• 2(5/2) + 3(1) = 8 and 1=-2(5/2)+6

• 5+3 = 8 and 1=-6+6 = 1

• Check and Check!

**Examples 47-52**

Ex 6 A dependent system solved by substitution

(all numbers = answers) pg 456

• 2x+3y=5+x+4y• y=x-5• We have y ready to use so plug it into the first

equation• 2x + 3(x-5) = 5+x+4(x-5)• 2x+3x-15=5+x+4x-20• 5x-15=5x-15 Oh, oh, they are the SAME! All

answers work! It’s dependent. • **Examples 53-54**

Ex 7 How about substituting and finding an inconsistent solution

(no answer)? Pg 456

• x-2y=3 • 2x-4y=7• It’s easiest to solve for x in the first equation• x= 2y+3• Now stick that into the second equation• 2(2y+3)-4y=7• 4y+6-4y = 7• 6=7 BUZZ! It crashed and burned, it’s inconsistent.

The cookbook…

Ex 8 An application page 457

• Let’s say we have a swimming pool who’s rectangular length is 2X it’s width. If the perimeter is 120 feet, what is it’s Length (L) and Width (W)

Ex 8 continued• It told us:• L=2W• And 2L+2W=120• Here we go… we have L so stick it into the

second equation• 2(2W)+2W=120• 4W+2W=120• 6W=120• W=20 feet. That means (using the first

equation) L=2(W) or 2(20) or 40 L=40 feet. **Examples 79-90**

Ex 9 Application - Investments• Belinda had $20,000 to invest. She invested

part at 10% and the rest at 20%. Her income was $2160. How much did she invest in both?

• Summary

Amount Rate Interest

First investment

X 10% 0.10x

Second investment

Y 12% 0.12y

Ex 9 continued• x+y=20000

• 0.10x+0.12y=2160

• We’ll solve the first equation for x for simplicity x=20000-y then stick it into the second equation…

• 0.10(20000-y) + 0.12y = 2160

• 2000 –0.1y +0.12y = 2160

• 0.02y = 2160

• y=8000 then plug that into the first x+8000=20000 means x=12,000

• So she invested $12,000 at 10% and $8000 at 12%

Exercises 7.1

• Here we go again!

• Definitions Q 1-6

• Solve by graphing Q7-20

• Match graphs to system of equations Q21-24

• Graphing with a calculator Q25-34

• Solve using the substitution method Q35-78

• Application problems Q79-108

Section 7.2 the last of Algebra I

• Now we’ll keep solving systems of linear equations, but using a new trick.

• This one keeps you from having to substitute big nasty expressions

• We eliminate a variable by adding or subtracting the two equations just right.

The Addition Method

• Example 1 page 464

• 3x-5y = -9

• 4x+5y=23

• Notice that if you added –5y to 5y, they kill each other!

The addition method continued

3x-5y = -9

+ 4x+5y= 23________________

7x + 0 = 14

solve 7x=14

x=2

Getting y now in the addition method (Ex 1 continued)

• We know x=2 so plug it into EITHER original equation!

• 3x-5y = -9 4x+5y=23• 3(2)-5y = -9 4(2) +5y=23• 6-5y = -9 8 +5y = 23• -5y = -15 5y=15• y=3 y=3• It worked both ways! So the solution is {(2,3)}**Examples 7-14**

Example 2 pg 465: What if the equations don’t help us… make

them help us!• 2x-3y = -13• 5x-12y=-46• Nothing will vanish if we add or subtract

these two equations. • So let’s MAKE them cancel somehow.• Look at the y’s if the first equation were

multiplied on both sides by –4, what would happen?

Example 2 continued• 2x-3y = -13• 5x-12y=-46• So step one is multiply the first eq. by –4• (-4)2x-(-4)3y=(-4)(-13)• -8x+12y=52• Now well add them -8x+12y=52

+ 5x –12y=-46 ________________

-3x + 0y = 6 x= -2

Example 2 getting y• We know x=-2• Plug it into either of the first equations again and get X• 2x-3y=-13• 2(-2) –3y=-13• -4 –3y = -13• -3y=-9• y=3 So (-2,3) is IT!

• If you want, you can plug it into either or both of the starting equations and check it. We’ll go on though…

**Examples 15-18**

Example 3 Multiplying BOTH to make things nice pg 456

(Use your LCD brain cells)

• -2x+3y=6

• 3x- 5y = -11

• There are no simple (smaller) numbers we can multiply ONE equation by to make it easily add or subtract from the other.

• But what if we multiplied the first by 3 and the second by –2 ???

Example 3 continued

• Doing two at once…

• 3(-2x+3y) = 3(6)

• 2(3x-5y) = 2(-11)

• The next step – simplify

• -6x+9y=18

• 6x-10y=-22 Hey! We’re ready!

Example 3 doing the addition

-6x+9y=18

+ 6x-10y=-22

________________

0x –y = -4 so y=4

Plug that into one of the original equations…

-2x+3(4)=6 -2x +12=6 -2x =-6 x=3

• So the (3,4) **Examples 19-24**

Example 4 Here is one with inconsistent equations page 466

2x-3y=9 copy 2x-3y=9 Multiply by 2

6y=4x-18 reorder -4x+6y=-18

4x-6y= 18

+ -4x+6y=-18 It looks EASY!

________________

0 = 0 BONK!

Example 4b Here is one with inconsistent equations page 466

-4y = 5x + 7

+ 4y = -5x + 12 It is EASY!

________________

0 = 0x + 19

0 = 19 BONK!!

**Examples 25-30**

Example 5 pg 467 – We can do the same thing with fractions and decimals!

• ½ x – 2/3 y = 7• 2/3x – ¾ y = 11• What can we do? We want something to cancel.

We want to turn the fractions in front of either both x’s or both y’s into integers. (We’re using all the early tools!)

• How about multiplying 6 to the first (making 1/x into 3x) and 12 to the second (making 2/3x into 8x)?

Example 5 continued

• 6[ 1/2x – 2/3y] = 6(7)• 12[ 2/3x – 3/4y] = 12(11)• Working both• 3x- 4y = 42• 8x-9y = 132 nice, everything became an

integer• NOW can we kill something … lets keep

hitting on the x’s

Example 5 continued

• 3x- 4y = 42• 8x-9y = 132 Let’s multiply both

equations by the other equation’s coefficient of x

• -8(3x-4y) = -8(42) note I stuck in a – sign for ease• 3(8x-9y) = 3(132)• Multiply them out…• -24x+32y=-336• 24x – 27y= 396

Example 5 adding-24x+32y=-336

+ 24x–27y= 396___________________

0x +5y = 60 y=12

Plug y=12 into one of our original equations to get x ½ x – 2/3(12) = 7 ½ x-8=7 ½ x = 15 x= 30 Our answer is (30,12)

**Examples 31-38**

Example 6 pg 467– now with decimals

0.05x+0.7y=40

x+0.4y=120

So we’ll kill the decimals by multiplying the first by 100 and the second by 10

100[0.05x+0.7y]=[40]100

10[x+0.4y]=[120]10

Multiply out and onto the next slide…

Ex 6 continued

5x+70y=4000

10x+4y=120

If we multiply the first by –2 we’ll kill the x’s

-10x-140y=-8000

10x + 4y= 120

-136y=-6800 y=50 plug into x+0.4y=120 x+0.4(50)=120 x=100

**Examples 39-46** {100,50}

The Addition Method Cookbook

Example 7 Applications page 468

• At a café the price for 4 fajita dinners and 3 burrito dinners is $48 . The total price for 3 fajita dinners and 2 burrito dinners is $34. What is the price for each dinner?

• 4x+3y=48

• 3x+2y=34

Example 7 continued

4x+3y=48

3x+2y=34 multiply both by the other equation’s x

coefficient (to kill the x’s)

-3(4x+3y) = -3(48) make it – for ease

4(3x+2y) = 4(34)

Multiply this out on the next frame…

Example 7 adding!

-12-9y = -144+ 12x+8y = 136

------------------------------ 0x –y = -8 so y=8

Plug that into 4x+3(8) = 48 4x+24=48 4x=24 x=6So you have (6,8) as the solution!Fajita dinners cost $6, Burritos cost $8. **Examples 65-70**

Example 8 page 469 • Working with mixing oils, we know the

following:

Amount of oil

%fat Amount of fat

Canola Oil X 7 0.07x

Corn Oil Y 14 0.14y

Canola and Corn Oil

280 11 0.11(280) which is 30.8 gallons

Example 8

• x+y = 280 (total amounts of oil)

• 0.07x+0.14y = 30.8 (amounts of fat)

What if we multiply the FIRST equation by -0.07 we get

-0.07x -0.07y = -19.6

0.07x+0.14y = 30.8

Example 8 – to the end

-0.07x -0.07y = -19.6

+ 0.07x+0.14y = 30.8_________________________

0x + 0.07 y = 11.2

So y=11.2/0.07 = 160

Using the first equation we put y =160 in to get x

x+160=280 x=120

It takes 120 gal. of canola and 160 gal. of corn oil to make Crisco Canola and Corn Oil **Examples 65-70**

Exercises 7.2

• Definitions Q1 – Q6

• Solve using the addition method Q7-30

• Using fractions and/or decimals Q31-46

• Practice mix bag Q47-64

• Application Q65-87