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#### jaytheseer

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- Jan 28, 2014

- 14

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- Jan 28, 2014

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Now, if he spent 1/4 of this on socks and 1/5 of this on a magazine and 50 on a snack, leaving 82, then we may state:

\(\displaystyle M-\frac{1}{4}M-\frac{1}{5}M-50=82\)

This equation reflects the fact that he began with $M$ units of currency, and then 1/4 was subtracted for the socks, another 1/5 was subtracted for the magazine, and then 50 units was subtracted for the snack, leaving a total of 82 units.

Can you solve this equation for $M$?

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- Jan 28, 2014

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Sorry, but I still can't solve it with this formula.

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My first step here would be to add 50 to both sides. This way we have only terms with $M$ as a factor on the left, and a constant on the right. Then, since all the terms on the left have $M$ as a factor, I would factor $M$ out. What do you have at this point?Sorry, but I still can't solve it with this formula.

- Feb 15, 2012

- 1,967

Always do the same thing on both sides of the equals sign. Sometimes stuff on one side will get simpler. This is good, it means progress.

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- Jan 28, 2014

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Here's my idea (and please tell me if i'm getting anywhere with this).

I will start by subtracting 1/4 - 1/5 - 50/1.

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Those terms are not "like terms" as the fractions 1/4 and 1/5 represent portions of Dominic's beginning funds, and 50 is an absolute rather than relative amount.

Here's my idea (and please tell me if i'm getting anywhere with this).

I will start by subtracting 1/4 - 1/5 - 50/1.

To factor as I suggested, consider the expression:

\(\displaystyle xy+xz\)

We see that both terms have $x$ as a factor, and so we may write:

\(\displaystyle xy+xz=x(y+z)\)

So, our equation (after adding 50 to both sides) is:

\(\displaystyle M-\frac{1}{4}M-\frac{1}{5}M=132\)

What you can do now, is factor $M$ out to get:

\(\displaystyle M\left(1-\frac{1}{4}-\frac{1}{5} \right)=132\)

Do you understand what I did there?

Now you may combine the numeric terms within the parentheses. You will need to get a common denominator. What is:

\(\displaystyle 1-\frac{1}{4}-\frac{1}{5}\) ?

- Feb 15, 2012

- 1,967

$M - \dfrac{1}{4}M - \dfrac{1}{5}M - 50 = 82$.

Following his suggestion, we will add 50 to both sides:

$M - \dfrac{1}{4}M - \dfrac{1}{5}M - 50 + 50 = 82 + 50$

$M - \dfrac{1}{4}M - \dfrac{1}{5}M + 0 = 82 + 50$

$M - \dfrac{1}{4}M - \dfrac{1}{5}M = 82 + 50$

$M - \dfrac{1}{4}M - \dfrac{1}{5}M = 132$.

Does this make sense to you?

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- Jan 28, 2014

- 14

Somehow I got the point but I think I need to review factoring for that.Those terms are not "like terms" as the fractions 1/4 and 1/5 represent portions of Dominic's beginning funds, and 50 is an absolute rather than relative amount.

To factor as I suggested, consider the expression:

\(\displaystyle xy+xz\)

We see that both terms have $x$ as a factor, and so we may write:

\(\displaystyle xy+xz=x(y+z)\)

So, our equation (after adding 50 to both sides) is:

\(\displaystyle M-\frac{1}{4}M-\frac{1}{5}M=132\)

What you can do now, is factor $M$ out to get:

\(\displaystyle M\left(1-\frac{1}{4}-\frac{1}{5} \right)=132\)

Do you understand what I did there?

Now you may combine the numeric terms within the parentheses. You will need to get a common denominator. What is:

\(\displaystyle 1-\frac{1}{4}-\frac{1}{5}\) ?

Is this correct?

1-1/20

- - - Updated - - -

$M - \dfrac{1}{4}M - \dfrac{1}{5}M - 50 = 82$.

Following his suggestion, we will add 50 to both sides:

$M - \dfrac{1}{4}M - \dfrac{1}{5}M - 50 + 50 = 82 + 50$

$M - \dfrac{1}{4}M - \dfrac{1}{5}M + 0 = 82 + 50$

$M - \dfrac{1}{4}M - \dfrac{1}{5}M = 82 + 50$

$M - \dfrac{1}{4}M - \dfrac{1}{5}M = 132$.

Does this make sense to you?

So those are the sides we're talking about. Thanks for always making things easier for me to digest.

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- Jan 28, 2014

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Based on how I responded to all of your suggestions, what do you think are the topics that I need to review to be able to solve this type of math problem? I really want to learn more in mathematics.

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No, but you do have the right common denominator. I would look at it as:Somehow I got the point but I think I need to review factoring for that.

Is this correct?

1-1/20

\(\displaystyle 1-\frac{1}{4}-\frac{1}{5}=\frac{20}{20}-\frac{5}{20}-\frac{4}{20}=\frac{20-5-4}{20}=\frac{11}{20}\)

Well, I suggest you review factoring. It is a skill you will definitely need to have to move forward in algebra and beyond. I would also recommend reviewing the solving of linear equations, that is, equations where the variable have an implied exponent of 1. For example, can you easily solve:MarkFL and Deveno, I really am so thankful for your patience in helping me out with my basic math problems.

Based on how I responded to all of your suggestions, what do you think are the topics that I need to review to be able to solve this type of math problem? I really want to learn more in mathematics.

\(\displaystyle y=mx+b\) for $x$? Can you solve it for $m$ and $b$ as well?

- Feb 15, 2012

- 1,967

These laws are:

1. The Law of Associativity of Addition:

For any three numbers a,b and c:

(a + b) + c = a + (b + c)

2. The Law of Commutativity of Addition:

For any two numbers a and b:

a + b = b + a

3. The Identity Law for Addition:

For any number a:

a + 0 = 0 + a = a

4. The Law of Additive Inverses:

For every number a, there is another number -a, with:

a + -a = -a + a = 0

(the quantity a + -b is often written a - b).

5. The Law of Associativity of Multiplication:

For any three numbers a,b,c:

(a*b)*c = a*(b*c)

6. The Law of Commutativity of Multiplication:

For any two numbers, a and b:

a*b = b*a

7. The Identity Law for Multiplication:

For any number a:

a*1 = 1*a = a

8. The Law of Multiplicative Inverses:

For any number a EXCEPT 0, there is a number 1/a with:

a*(1/a) = (1/a)*a = 1

(The quantity a*(1/b) is often written a/b)

9. The Distributive Law of Addition over Multiplication:

For any three numbers a,b and c:

a*(b + c) = (a*b) + (a*c)

(a + b)*c = (a*c) + (b*c)

********************

I haven't said "what a number is" because I don't want to overwhelm you. For the time being, suffice to say that the set of all whole numbers and fractions will do as a "starting place" for this, that you may safely assume those are numbers.

********************

Here is how these laws work in action:

Suppose Betty has 3 times as many apples as Bob, plus an additional 5. If Betty has 26 apples, how many does Bob have?

Solution:

Let A be the number of apples Bob has. From the problem, we see that:

3*A + 5 = 26

since both sides of the equation are how many apples Betty has.

(3*A + 5) + -5 = 26 + -5 (Adding -5 to both sides)

(3*A + 5) + -5 = 26 - 5 (re-writing 26 + -5 as 26 - 5)

(3*A + 5) + -5 = 21 (evaluating the difference 26 - 5)

3*A + (5 + -5) = 21 (using Law 1 with a = 3*A, b = 5, and c = -5)

3*A + 0 = 21 (using Law 4)

3*A = 21 (using Law 3)

(1/3)*(3*A) = (1/3)*21 (multiplying both sides by 1/3)

(1/3)*(3*A) = 21*(1/3) (using Law 6 with a = 1/3 and b = 21)

(1/3)*(3*A) = 21/3 (rewriting 21*(1/3) as 21/3)

(1/3)*(3*A) = 7 (evaluating 21/3)

((1/3)*3)*A = 7 (using Law 5, with a = 1/3, b = 3 and c = A)

1*A = 7 (using Law 8)

A = 7 (using Law 7)

Thus Bob has 7 apples.

If any of this confuses you, we will try to help as best we can.