Exam 1: Tuesday Sept 14 Exam will cover three weeks of class including today. Covers written homework, online homework, and labs. Practice exams are on the website. Thursday-Friday: Recitation in your lab rooms Covers s2.4&s2.4 homework. Bring a review question on a notecard Monday: Homework due. Written homework due at 6pm. Online homework due at 11:59pm. Monday and Tuesday: Exam review in your recitation rooms. Notecard questions will be addressed here. Tuesday: Exam: 7:15pm-8:15pm. Room assignments on the course website. http://www.math.ksu.edu/math100studio/

Solving a system of equations Finding all values of all variables that make all the equations true. Ex: 6a+3b=12 4a-15=12b+2 I want to find all pairs (a,b) that make both equations true at the same time.

Solving a system of equations This is exactly the same as solving one equation! You can use the same tools. Your tools are: The = sign Doing the same thing to both sides. (Usually adding).

Substitution Method Fancy way of saying: using the meaning of =. Ex: 5x-3y=12 y=4x+1 This means that anywhere I see a ‘y’ I can write “(4x+1)” The parentheses are important!

Solving with substitution 5x-3y=12 y=4x+1 5x-3(4x+1)=12 5x-12x-3=12 -7x-3=12 +3=+3 -7x=15 x=-15/7 anywhere I see an ‘x’ I can write “(-15/7)” y=4(-15/7)+1=(-60/7)+(7/7)=-53/7 (x,y)=(-15/7,-53/7)

Elimination Method Fancy way of saying “adding the same thing to both sides.” Ex: Ex: -7x-3=12 -7x-3y=12 +3=+3 2x+3y=3 --------------- ----------------- -7x =15 -5x =15

Solving with elimination -7x-3y=12 2x+3y=3 ------------- -5x =15 x=-3  anywhere I see an x, I can write “(-3)” 2(-3)+3y=3 -6+3y=3 +6 =+6 ------------ 3y=9 y=3 (x,y)=(-3,3)

What if I can’t use either method? Ex: 6a+3b=12 4a-15=12b+2 Solution: Use your equation solving tools (doing the same thing to both sides) to put the equations in a shape where one method will work. Use whichever method you like best. You only ever need one.

Setting up for substitution 6a+3b=12 4a-15=12b+2 6a+3b=12  pick one equation to reshape -3 = -3b ---------------- 6a =12-3b * (1/6) = (1/6) --------------- a=(1/6)(12-3b) a=2-b/2  now it’s in a shape where I can use substitution. 4a-15=12b+2  finish solving using substitution from here. a=2-b/2  anywhere I see an “a” I can write “(2-b/2)” 4(2-b/2)-15=12b+2 Etc… (a,b)=(65/28, -9/14)  final answer

Setting up for elimination 6a+3b=12 4a-15=12b+2 6a+3b=12  pick an equation to reshape. * -(2/3) = -(2/3) ----------------- -4a-2b=-8 4a-15 =12b+2  now I can add both equations together. -15-2b=12b-6 Etc…. (a,b)=(65/28, -9/14)  final answer

You can use any method you want If you can get your equations in the right shape, substitution will always work. If you can get your equations in the right shape, elimination will always work. Pick the one you are most confident with and stick with it.

A jewelry maker has total revenue for necklaces given by R(x)=90 A jewelry maker has total revenue for necklaces given by R(x)=90.75x, and incurs a total cost of C(x)=24.50x+4770, where x is the number of necklaces a and sold. How many necklaces must be produced and sold in order to break even? 70 necklaces 71 necklaces 72 necklaces 73 necklaces Don’t know

solution R(x)=90.75x C(x)=24.50x+4770 Break even when R(x)=C(x) 90.75x=24.50x+4770 hidden substitution -24.50x=-24.50x ----------------------------- 66.25x=4770 x=72. (C:72).

Cannot be determined from the information given Suppose demand is given by p+2q=200, and supply is given by p-5q=60. Find the equilibrium point (in other words, find the solution to the system). p=$60, q=20 units p=$120, q=60 units p=$160, q=20 units p=$20, q=160 units Cannot be determined from the information given

solution p+2q=200 -p+5q=-60 ----------------  I used elimination because it’s faster here. 7q=140 q=20  use a substitution. p+2(20)=200 p+40=200 p=160 (p,q)=(160,20) (C)

Inequalities A statement that some number is larger than some other number. Ex: x+3>12 Means “the number x+3 is bigger than the number 12” That leaves open a lot of possibilities for the value of “x+3”!

x+3>12 X+3 could be: 13 14 2000 12.5 12.00000000000000000000000000001 Etc.

5.25x >1150 5.25x+1150>0 5.25x+1150=0 Both A & B A plumbing store’s monthly profit from the sale of PVC pipe can be described by P(x)=5.25x-1150 dollars, where x is the number of feet of PVC pipe sold. Set up an inequality that would help determine what level of monthly sales is necessary to incur positive profit. 5.25x >1150 5.25x+1150>0 5.25x+1150=0 Both A & B All of the above

Solution P(x)=5.25x-1150 The profit is represented by P(x). So a positive profit is when P(x)>0 5.25x-1150>0 If I add the same thing to both sides, I get 5.25x>1150 (A) Or think of it this way. 5.25x is how much I earn when I sell pipe. 1150 is how much it costs me to make pipe. I make a profit when what I earn is more than my cost.

Inequalities How does doing the same thing to both sides work for an inequality? Let’s use our example of x+3>12

In order for x+3>12 to be true what has to be true about x?

Not always safe Beware multiplying or dividing by a negative! Ex: -4x+3>1

-4x+3>1

-4x>-2

-x>-2/4

X<1/2 NOT x>1/2 !

How to deal with negative signs in inequalities Option 1: Avoid them. -4x+3>1 +4x=+4x ----------------------------- 3>1+4x Option 2: Flip the inequality -x>-1/2 x<1/2  drawing a diagram helps!

Last thought Option 1 and Option 2 are the same thing: -x>-1/2 +x = +x ---------- 0 >x-1/2 +1/2=+1/2 1/2>x x<1/2