# Can't figure out how to begin with solving this one.

#### jaytheseer

##### New member
Dominic spent 1/4 of his money on a pair of socks, 1/5 of it on a magazine and 50 on a snack. If he had 82 left, how much money did he start with?

#### MarkFL

Staff member
Let's let $M$ be the amount of money in units of currency with which Dominic started.

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 $M$?

#### jaytheseer

##### New member
Sorry, but I still can't solve it with this formula. #### MarkFL

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

#### Deveno

##### Well-known member
MHB Math Scholar

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.

#### jaytheseer

##### New member
To be honest, I can't follow your instructions anymore. Sorry to bother you guys but I think I already forgot the basics. 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.

#### MarkFL

Staff member
To be honest, I can't follow your instructions anymore. Sorry to bother you guys but I think I already forgot the basics. 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.
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:

$$\displaystyle 1-\frac{1}{4}-\frac{1}{5}$$ ?

#### Deveno

##### Well-known member
MHB Math Scholar

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

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?

#### jaytheseer

##### New member
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:

$$\displaystyle 1-\frac{1}{4}-\frac{1}{5}$$ ?
Somehow I got the point but I think I need to review factoring for that.

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. #### jaytheseer

##### New member
MarkFL and Devono, 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.

#### MarkFL

Staff member
Somehow I got the point but I think I need to review factoring for that.

Is this correct?

1-1/20
No, but you do have the right common denominator. I would look at it as:

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

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.
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:

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

#### Deveno

##### Well-known member
MHB Math Scholar
My advice is to become acquainted with the Laws of Arithmetic, and how to use them to solve equations in one unknown.

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.