Algebra Honors (Period 6 & 7)

Rate-Time- Distance Problems 4-8

D = rt

Uniform Motion

Three types of problems:
Motion in opposite direction
Motion in same direction
Round Trip

Motion in opposite direction
For this we used different students bicycling ... Josh K and Josh P in 6th period and Maddie & Jamie from 7th period.
They start at noon -->60 km apart riding toward each other. They meet at 1:30 PM. If Josh K speed is 4 km/h faster than Josh p ( Maddie is greater by the same from Jamie's rate) What are their speeds?

We set up a chart

Motion in Same Direction

Next we had a fictitious story about ANdrew's Helicopter and David's plane ( or Lucas' helicopter and Kitt's Smiling Plane) taking off from Camarillo Airport flying north. The helicopter flies at a speed of 180 mi/hr. 20 minutes later the plane takes off in the same direction going 330 mi/hr. How long will it take David (or Kitt) to over take Andrew's ( or Lucas') helicopter?

Let t = plane's flying time

Make sure to convert the 20 minutes ---> 1/3 hours.

We set up a chart

When the plane over takes the helicopter they have traveled the exact same distance so set them equal
180(t + 1/3) = 330t

180 t + 60 = 330t
60 = 150t

t = 2/5
which means 2/5 hour. or 24 minutes.

Round Trip

A ski life carries Sara ( or Ryan) up the slope at 6 km/h Sare or Ryan snowboard down 34 km/h. The round trip takes 30 minutes.

Did you see the picture?

Let t = time down

then set up a chart

6(.5 -t) = 34 t
3 - 6t = 34 t
3/40 = t
Now, what's that?

0.75 hr or 4.5 minutes

How far did they snowboard... plug it in
34(0.075) = 2.55 km

Math 6 Honors ( Periods 1, 2, & 3)

Problem Solving: Using Mathematical Expressions 2-6


Seventeen less than a number is fifty six

I suggest lining up and placing the "equal sign" right under the word is
then complete the right side = 56
after that take your time translating the left
Start with a "let statement."
A "let statement" tells your reader what variable you are going to use to represent the number in your equation.
So in this case Let b = the number
it becomes
b - 17 = 56
Now solve as we have been practicing for a couple of weeks.
b - 17 = 56
+17 = +17 using the +prop=
b + 0 = 73
b = 73 by the ID(+)

How could we check?
A FORMAL CHECK involves three steps:
1) Re write the equation ( from the original source)
2) substitute your solution or... "plug it in, plug it in...."
3) DO the MATH!! actually do the math to check!!

so to check the above
b - 17 = 56
substitute 73 and put a "?" above the equal sign...

73 - 17 ?=? 56
Now really do the math!! Use a side bar to DO the MATH!!
That is, what is 73- 17? it is 56
so 56 = 56


Practice some of the class exercises on Page 50, Just practice setting up the equations from the verbal sentences.

Problem Solving: Using Mathematical Expression 2-6

Inequalities Continued

We know about > greater and as well as < less than
so now we look at
≥ which means " greater than or equal to" and
≤ which means " less than or equal to"

This time the boundary point ( or endpoint) is included in the solution set.
The good news is that we still solve these inequalities the same way in which we solved equations-- using the properties of equality.
w/4 ≥ 3
we multiply both sides by 4/1
(4/1)(w/4) ≥ 3(4/1) by the X prop =
1w ≥ 12
w ≥ 12 by the ID(x)

Which means that any number greater than 12 is part of the solution AND 12 is also part of that solution

Take the following:
b - 3 ≤ 150 ( we need to add 3 to both sides of the equation)
+3 = +3 using the + prop =
b + 0 ≤ 153
or b ≤ 153 using the ID (+)

Algebra Honors (Period 6 & 7)

Transforming Formulas 4-7

Formulas are used throughout real life applications-- the book gives an example of the formula for the total piston displacement of an auto engine... we discussed a number of formulas that students recalled such as the following:
A =lw
d = rt
I = Prt
A = ∏r²
A = P(1 + rt)
C = 5/9(F - 32)
y = mx + b
A = bh
C = ∏d
F = Ma
A = ½(b₁b₂h
E= mc2
A² + B² = C²
and even the quadratic formula-- which we will study later this year..



b = ax ; x
just divide both sides by a
b/a = x

Solve P = 2L + 2W for the width, w
P-2L = w
2

C = 5/9(F - 32) Solve for F
We need to multiply both sides by the reciprocal of 5/9
(9/5)C = F - 32
now add 32 to both sides
(9/5)C + 32 = F


Next we tackled

We also discussed the restrictions and found that the denominator could not equal zero.


S = v/r ( solving for r) became one of the homework problems that caused some discussion-- until students realized that they had actually found the reciprocal of r or 1/r instead of solving for r!!
We discussed how to solve that dilemma.

Math 6 Honors ( Periods 1, 2, & 3)

Solving Other Equations & Inequalities 2-5 cont'd
Parent: "What did you do today in Math?"
Student: "Today Mrs. Nelson taught us how to wrap and unwrap a present!!!"
Parent: "Huh..."
Student: "Well, it all has to do with PEMDAS... and undoing equations to solve them!!

Someone was paying attention today... :)


In order to solve two step or multi-step equations you must UNDO in the reverse order of PEMDAS...

UNDO by doing the reverse of PEMDAS

4x + 172 = 248

CHECK:

Step 1: Rewrite the problem
Step 2: Substitute the solutions for the variable
Step 3: DO THE MATH

4x + 172 = 248
We found that x = 19 above-- > in the first problem
so for the check
4x + 172 = 248
4(19) + 172 ?=? 248
76 + 172 ?=? 248
248 = 248 CHECK!!!

For the above, you actually pace a “?” above the equal sign….rather than to each side…