College Algebra Week 4 Chapter 8 1 8

College Algebra Week 4 Chapter 8. 1, 8. 2 The University of Phoenix Inst. John Ensworth

Short section – 8. 1 and 8. 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 8. 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 • • • 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 origional 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!

Example 2 another graphing • • • 2 x-3 y=6 and 3 y-2 x=3 Get that y=mx+b form! -3 y=-2 x+6 and 3 y=2 x+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 2 the graph • Oh oh! • No answer • Null Set

Example 3 Another by graphing • • 2(y+2) = x and x-2 y=4 Get that y=mx+b form! 2 y+4 =x and -2 y=-x+4 2 y=x-4 and y= ½ x – 2 Giving us graph fodder… =1/2 and (0, -2) and m=1/2 and (0, -2) Something strange going on?

Example 3 the graph and solution • They are the SAME! • There an INFINITE number of solutions • {(x, y)|x-2 y=4}

Definitions! • So we say we have a system of equations that are independent, inconsistent or dependent based on how many solutions there are.

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

Ex 4 finding y • x= 5/2 we can use our second equation y= -2 x+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!

Ex 5 How about substituting and finding an inconsistent solution (no answer)? • • x-2 y=3 2 x-4 y=7 It’s easiest to solve for x in the first equation x= 2 y+3 Now stick that into the second equation 2(2 y+3)-4 y=7 4 y+6 -4 y = 7 6=7 BUZZ! It crashed and burned, it’s inconsistent.

Ex 6 A dependent system solved by substitution (all numbers = answers) • 2 x+3 y=5+x+4 y • y=x-5 • We have y ready to use so plug it into the first equation • 2 x + 3(x-5) = 5+x+4(x-5) • 2 x+3 x-15=5+x+4 x-20 • 5 x-15=5 x-15 Oh, oh, they are the SAME! All answers work! It’s dependent.

The cookbook…

Ex 7 An application • Let’s say we have a swimming pool who’s rectangular length is 2 X it’s width. If the perimeter is 120 feet, what is it’s Length (L) and Width (W)

Ex 7 continued • • It told us: L=2 W And 2 L+2 W=120 Here we go… we have L so stick it into the second equation 2(2 W)+2 W=120 4 W+2 W=120 6 W=120 W=20 feet. That means (using the first equation) L=2(W) or 2(20) or 40 L=40 feet.

Ex 8 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 X investment 10% 0. 1 x Second Y investment 12% 0. 12 y

Ex 8 continued • x+y=20000 • 0. 10 x+0. 12 y=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. 12 y = 2160 • 2000 – 0. 1 y +0. 12 y = 2160 • 0. 02 y = 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 8. 1 • • • Here we go again! Definitions Q 1 -6 Solve by graphing Q 7 -16 Match graphs to system of equations Q 17 -20 Solve using the substitution method Q 21 -44 Application problems Q 45 -58

Section 8. 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 3 x-5 y = -9 4 x+5 y=23 Notice that if you added – 5 y to 5 y, they kill each other!

The addition method continued 3 x-5 y = -9 + 4 x+5 y= 23 ________ 7 x + 0 = 14 solve 7 x=14 x=2

Getting y now in the addition method (Ex 1 continued) • We know x=2 so plug it into EITHER original equation! • 3 x-5 y = -9 4 x+5 y=23 • 3(2)-5 y = -9 4(2) +5 y=23 • 6 -5 y = -9 8 +5 y = 23 • -5 y = -15 5 y=15 • y=3 • It worked both ways! So the solution is {(2, 3)}

Example 2: What if the equations don’t help us… make them help us! • 2 x-3 y = -13 • 5 x-12 y=-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 • • • 2 x-3 y = -13 5 x-12 y=-46 So step one is multiply the first one by – 4 (-4)2 x-(-4)3 y=(-4)(-13) -8 x+12 y=52 Now well add them -8 x+12 y=52 + 5 x – 12 y=-46 ________ -3 x + 0 y = 6 x= -2

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

Example 3 Multiplying BOTH to make things nice • -2 x+3 y=6 • 3 x- 5 y = -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(-2 x+3 y) = 3(6) 2(3 x-5 y) = 2(-11) The next step – simplify -6 x+9 y=18 6 x-10 y=-22 Hey! We’re ready!

Example 3 doing the addition -6 x+9 y=18 + 6 x-10 y=-22 ________ 0 x –y = -4 so y=4 Plug that into one of the original equations… -2 x+3(4)=6 -2 x +12=6 -2 x =-6 x=3 • So the (3, 4)

Example 4 Here is one with inconsistent equations -4 y = 5 x + 7 + 4 y = -5 x + 12 It looks EASY! ________ 0 = 0 x + 19 0 = 19 BONK!!

Example 5 – We can do the same thing with fractions and decimals! • ½ x – 2/3 y = 7 • 2/3 x – 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 6 to the first (making 1/x into 3 x) and 12 to the second (making 2/3 x into 8 x)?

Example 5 continued • • • 6[ 1/2 x – 2/3 y] = 6(7) 12[ 2/3 x – 3/4 y] = 12(11) Working both 3 x- 4 y = 42 8 x-9 y = 132 nice, everything became an integer • NOW can we kill something … lets keep hitting on the x’s

Example 5 continued • 3 x- 4 y = 42 • 8 x-9 y = 132 • • • Let’s multiply both equations by the other equation’s coefficient of x -8(3 x-4 y) = -8(42) note I stuck in a – sign for ease 3(8 x-9 y) = 3(132) Multiply them out… -24 x+32 y=-336 24 x – 27 y= 396

Example 5 adding -24 x+32 y=-336 + 24 x– 27 y= 396 __________ 0 x +5 y = 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)

The Addition Method Cookbook

Example 6 Applications • 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? • 4 x+3 y=48 • 3 x+2 y=34

Example 6 continued 4 x+3 y=48 3 x+2 y=34 multiply both by the other equation’s x coefficient (to kill the x’s) -3(4 x+3 y) = -3(48) make it – for ease 4(3 x+2 y) = 4(34) Multiply this out on the next frame…

Example 6 adding! -12 -9 y = -144 + 12 x+8 y = 136 ---------------0 x –y = -8 so y=8 Plug that into 4 x+3(8) = 48 4 x+24=48 4 x=24 x=6 So you have (6, 8) as the solution! Fajita dinners cost $6, Burritos cost $8.

Example 7 • Working with mixing oils, we know the following: %fat Canola Oil Amount of oil X 7 Amount of fat 0. 07 x Corn Oil Y 14 0. 14 y Canola and Corn Oil 280 11 0. 11(280) which is 30. 8 gallons

Example 7 • x+y = 280 (total amounts) • 0. 07 x+0. 14 y = 30. 8 (amounts of fat) What if we multiply the FIRST equation by -0. 07 we get -0. 07 x -0. 07 y = -19. 6 0. 07 x+0. 14 y = 30. 8

Example 7 – to the end -0. 07 x -0. 07 y = -19. 6 + 0. 07 x+0. 14 y = 30. 8 _____________ 0 x + 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

Exercises 8. 2 • Definitions Q 1 – Q 6 • Solve using the addition method Q 7 -36 • Application Q 37 -50