Math 8

CHAPTER 2-3: WRITING TWO-STEP EQUATIONS

Algebraic expressions just are the ones that have variables

Numeric expressions have only numbers

Equations must have an = sign while expressions do not

STRATEGY #1:TRANSLATE WORD BY WORD

Always try this first.

Just be careful of less THAN and subtracted FROM because these are switched from the order that you read/say them:

A number less THAN 12 is 12 – n but if you say a number less 12, this would be n – 12

12 subtracted FROM a number is n – 12, but 12 subtract a number would be 12 – n

The only other translation to be careful of is when you multiply a SUM or DIFFERENCE by a number or variable:

12 times the SUM of a number and 5 is 12(n + 5), but the sum of 12 times a number and 5 would  be 12n + 5

12 times the DIFFERENCE of a number and 5 is 12(n – 5), but the difference of 12 times a number and 5 would be 12n – 5

If you have 2 or more unknowns, use different variables:

The difference of a number and ANOTHER number would be x - y

STRATEGY #2: DRAWING A PICTURE

(When in doubt, draw it out! ;)

I have 5 times the number of quarters as I have dimes.

I translate to: 5Q = D

I check: If I assume that I have 20 quarters, then 5(20) = 100 dimes

Does this make sense? That would mean I have a lot more dimes than quarters.

The original problem says I have a lot more quarters!

My algebra is WRONG! I need to switch the variables.

5D = Q

I check: If I assume that I have 20 quarters, then 5D = 20

D = 4

Does this make sense? YES! I have 20 quarters and only 4 dimes.

Sometimes it helps to make a quick picture.

Imagine 2 piles of coins.

The pile of quarters is 5 times as high as the pile of dimes.

You can clearly see that you would need to multiply the number of dimes

to make that pile the same height as the number of quarters!

STRATEGY #3: MAKE A T-CHART (really and x-y table)

Let’s say you know that every bagel you buy costs the same amount of money, $.65. You buy bagels and spend $4.55. Write the algebraic equation and then solve for the number of bagels that you purchased.

Number of Bagels	Price
1	.65
2	1.30
3	1.95
b	.65b

By looking at the pattern from the left column to the right column, you find that the number of bagels TIMES the unit rate of the price/bagel gives you the total purchase price. Now you use a variable like b to come up with the algebraic expression for the purchase of any number of bagels, .65b.

Finally, set up the algebraic equation for the amount of money you spent (given in the problem):

.65b = 4.55

You can also do this for two-step equations. Let’s say you’re joining a gym and there is an initiation fee of $50 and then a monthly fee of $20 a month. Focus on the amount that is happening repeatedly because that is going to be your coefficient of the variable…since that is the amount that will “vary”, as opposed to the $50 one-time fee that will never change.

You can do this with or without a table!

Number of Months	Price
0	50
1	70
2	90
3	110
m	20m + 50

$20m + $50 = your cost

Say the problem asks how many months you’ve been going if you’ve paid $170:

20m + 50 = 170

Solve for the number of months:

20m = 120

m = 6 months

Math 6A ( Periods 1 & 2)

Multiplying Fractions 2-1

A park has a playground that is ¾  of its width  and ⅘ of its length. What fraction of the park is the playground?

In class, fold a piece of paper  horizontally into fourths and shade three of the four sections (yellow) to represent ¾

Fold the paper vertically into fifths and shade ⅘ of the paper blue!

COUNT the total number of squares. This number is the denominator. The numerator is he number number of squares shaded with both colors!

34 4 5=12 20=35  So the playground covers ⅗ of the entire park!

KEY IDEA:  Multiplying Fractions

Words: Multiply the numerators and then multiply the denominators.

Number Example:  

Algebra:  where b and d are both not equal to ZERO!

Find     Notice you multiply the numerators first, then multiply the denominators

When the numerator of one fraction is the SAME as the denominator of another fraction, you can use mental math to multiply For example:  because you can divide out the common factor 5.

Find         

Using what we know we would do the following     

    

But could we simply before we multiply?  YES!!

Divide out Common Factors FIRST

Looking at   What do you notice?  The GCF( 24, 36)  = 12. We will be using the GCF of numbers to simplify our fractions!  

KEY IDEA: Multiplying Mixed Numbers

Write each mixed number as an improper fraction. Then multiply as you would with fractions

Find                       Write as the improper fraction

Now rewrite the problem

Now, I want you to get comfortable with both the improper fraction and the mixed number, so practice giving both forms as your solution!  In Algebra there are many times when you will be leaving your number as an improper fraction!

Find    Write both Mixed Numbers as improper fractions

NOTE: Before you start multiplying the numerators check to see if you can simplify BEFORE you begin!  I see something… do you?

Is this a reasonable answer?  Why?

Math 8

CHAPTER 2-5: Multistep Equations

IDENTITY OR NO POSSIBLE SOLUTION EQUATIONS:

An identity equation is where ANY NUMBER can be substituted for the VARIABLE, the equation will be TRUE. What will happen is that while you’re balancing the equations, you will ultimately end up with the SAME EXACT EXPRESSION ON EACH SIDE of the equation.

You can keep going, but as soon as you have the same thing on both sides, you know you have an IDENTITY EQUATION. If you keep going, the variable will disappear on both sides of the equation, leaving you with a numeric equation (a number of both sides of the equal sign). That numeric equation will be TRUE.

Therefore, ALL REAL NUMBERS will work so we say there are INFINITELY MANY SOLUTIONS.

A no possible solution equation is one where no matter what number you substitute into the equation, the equation will be FALSE.

What will happen is that after you balance the equations, you will ultimately end up without any variable on either side, just like the Identity Equation, but this time, the numbers will be DIFFERENT (which can never be true).

THERE IS NO POSSIBLE SOLUTION OR THE NULL SET ∅

NOTICE SOMETHING ELSE ABOUT

SOLVING EQUATIONS IN GENERAL:

WHENEVER YOU HAVE THE SAME EXACT TERM WITH THE SAME SIGN ON DIFFERENT SIDES OF THE EQUATION, YOU CAN SIMPLY CROSS THEM OUT BECAUSE  WHEN YOU USE THE ADDITIVE INVERSE PROPERTY ON BOTH SIDES TO BALANCE, BOTH TERMS WILL DROP OUT!

If there’s Distributive Property, generally, you will distribute first (there are times when you have the choice to divide first and I’ll show you that in class)

Simplify each side of the equation first...combine like terms on the same side!

Then use the ADDITIVE INVERSE PROPERTY to move variables to the other side of the equation so that all variables are on the same side.

Usually, we try to move the smaller coefficient to the larger because sometimes this avoids negative coefficients,

BUT that is not always the case, and you may move to whatever side you choose!

Here’s a way to remember to use the OPPOSITE SIGN when you move terms to the other side of the equation:

OPPOSITE SIDES

use the

OPPOSITE SIGN

However if like terms are on the same side of the equation, use the sign of the coefficient that you’re given and simply combine them by using integer rules:

SAME SIDE

use the

SAME SIGN

EXAMPLE OF COLLECTING TERMS FIRST:

Sometimes, you will have to COLLECT LIKE TERMS ON THE SAME SIDE OF THE EQUATION before balancing:

8y + 12 – (-2y) = -6

10y + 12 = -6

10y = -18

y = -18/10 =  -9/5

TWO STEPS WITH DISTRIBUTIVE PROPERTY

Usually, you want to do DISTRIBUTE FIRST!

UNLESS THE FACTOR OUTSIDE THE ( ) CAN BE DIVIDED

OUT OF BOTH SIDES PERFECTLY! (which I will show you in class)

EXAMPLE:

5y - 2(2y + 8) = 16

5y - 4y - 16 = 16 [distribute]

y - 16 = 16 [collect like terms]

y = 32 [solve by adding 16 to both sides]

EXAMPLE WHEN YOU DON'T NEED TO DISTRIBUTE FIRST:

-3(4 + 3x) = -9

4 + 3x = 3 [Don't distribute! Divide by -3. The -3 goes into both sides perfectly!)

3x = -1 [Subtract 4 from both sides]

x = -1/3 [Divide both sides by 3]

EXAMPLE WHERE YOU NEED TO BOTH COMBINE ON THE SAME SIDE FIRST, AND THEN MOVE THE VARIABLES TO THE SAME SIDE:

3y - 10 - y = -10y + 12

The 3y and -y are on the SAME SIDE, so just use the SAME SIGNS as in the equation and combine them using integer rules:

2y - 10 = -10y + 12

Now the 2y and the -10y are on OPPOSITE SIDES of the equation so use the OPPOSITE SIGN to move the -10y to the left side:

2y - 10 = -10y + 12

                                     +10   +10y

12y - 10 = 12

NOW DO THE TWO-STEP ;)

12y - 10 = 12

+ 10 +10

12y = 22

12 12

y = 11/6