Monthly Archives: January 2013

Learn to Solve Single Variable Linear Equations

In keeping with my new years resolution of making this site more accessible to my nieces and nephews, I’ve added a script that generates random single variable linear equations that ask the user to solve for x. The script also has a “Compute” button that will give the answer, as well as an option to to show the step by step procedure that is used to reach this solution.

In many instances, solving single variable linear equations are a person’s first introduction into variables, or solving for unknown values. This concept, though, remains very important in our daily lives, with questions such as “how many paychecks do I need to save before I can afford to buy a new car?” or “how long will it take me to get home?”

Both of these questions can be represented by a linear equation. For example if each of my paychecks is for $500, and the car I wish to purchase has a listed price of $4500, then the question of “how many paychecks do I need to save before I can afford to buy a new car” can be represented by the following equation, which we would like to solve for x:

500 x = 4500

As a second example, suppose that out of each $500 paycheck $100 must go towards paying my bills. Suppose also that I have already begun saving and have an initial amount of $1300 saved already. Then the question of “how many paychecks do I need to save before I can afford to buy a new car can be represented by the following equation, which we’d also like to solve for x:

500 x + 1300 = 4500 - 100 x

The goal when solving an equation of this form is to isolate the variable, which in the case of these examples, means to get the x on one side of the equals sign ( = ) by itself and some number on the other side of the equals sign. We can do this by remembering a simple rule, Whatever you do to change one side of the equation, you must make the same changes on the other side of the equation. We do this because initially if we have two things being equal, the only way that they can stay equal is if we do the same things to both these things. We decide what to do by looking at what has already been done and performing the inverse operation of that action.

If there was initially addition of some number, we will subtract that same number.
If there was initially subtraction of some number, we will add that same number.
If there was initially multiplication by some number, we will divide by that same number.
If there was initially division by some number, we will multiply by that same number.

While there is no rule as to what the first step should be, it is generally easiest to try to keep reducing the number of terms in the equation, while trying to place all the terms with an x on one side of the equation and all the terms without an x on the other side of the equation. For instance, in the example earlier that states

500 x + 1300 = 4500 - 100 x

We notice that on the left hand side (LHS), x has been multiplied by 500 and the term 1300 has been added. So as a first step, we could either divide both sides by 500 or subtract 1300 from both sides. Likewise, on the right hand side (RHS), x has been multiplied by (-100) and the number 4500 has been added to it. So we could also divide both sides by by (-100) or subtract 4500 from both sides of the equation. We could also add 100 x or 500 x to both sides of the equation.

Initially there are four terms in this equation and division (by 500 or -100) will keep the number of terms in the equation at 4, whereas subtraction of 1300 or 4500 or 500 x or (-100 x) will reduce the number of terms to three. So we choose to do one of these actions. Lets go with subtraction of 1300 from both sides of the equation.

It can now be represented by

500 x + 1300 - 1300 = 4500 - 100 x - 1300

This simplifies to

500 x = 3200 - 100 x

Again we are faced with many options regarding what to do next. However, to get all the terms with an x on one side of the equation, we can add 100 x to both sides of this equation, which then becomes

500 x + 100 x = 3200 - 100 x + 100 x

This simplifies to
600 x = 3200

Now we notice that x is being multiplied by 600. So to solve for x, we only need to divide both sides by 600 which gives us that

x = 3200 / 600

Relating this to fraction arithmetic and Euclid’s Algorithm, we can reduce this fraction by noticing that 3200 = 200 * 16 and 600 = 200 * 3, so 3200 / 600 can be reduced to 16 / 3.

So x = 16 / 3

Since we’re looking for the number of paychecks I would need to save, this needs to be a whole number. 16 / 3 is greater than 5, so 5 paychecks will not be enough. So in order to make this purchase, I would need 6 paychecks.

The script also gives the option to generate a new problem to gain more practice with these types of problems.

Fraction Arithmetic

Fraction Arithmetic

I hope everyone had a good holiday season. I certainly enjoyed mine. Over this season, I had a chance to speak with some youth and their parents. Funny that whenever we discuss that I have a PhD in applied mathematics, the topics of the children struggling in mathematics and the possibility of tutoring their children always seem to come up. I have no problem with tutoring and I actively participate in such sessions in my spare time. However I will say that it is sometimes a difficult task to do this job over such a short time period. Needless to say, I felt bad that I couldn’t have been of more assistance.

So, this being the holiday season and all, I decided to make somewhat of a new years resolution to focus this site more towards some of the things that the youth struggle with to hopefully be able to answer some of their questions.

With that being said, the first area that I decided to look at was fractions. This is one of the first areas where the youth begin to dislike mathematics. I feel like regardless of how much teachers and professors speak of the importance of understanding these processes, many students simply never grasp the procedures involved, partially because they never get used to the rules associated with these matters.

In this first script on fractions, I’ve focused on four types of problems corresponding to the four basic operations of arithmetic: Addition, Subtraction, Multiplication and Division.

To add two fractions of the form

num1
den1
+
num2
den2

We use the formula

num1
den1
+
num2
den2
=
num1
den1
+
num2
den2
=
num1*den2 + num2*den1
den1*den2

Lets take a moment to consider where this formula comes from. In order to be able to add fractions we first need to obtain a common denominator for the two fractions. One way that always works to obtain a common denominator is to multiply the denominators of the two fractions. So in the formula above, the denominator on the right hand side of the equals sign is the product of the two denominators on the left hand side. Once we have a common denominator, we need to rewrite each of the two fractions in terms of this common denominator.

num1
den1
+
num2
den2
=
num1*den2
den1*den2
+
num2*den1
den1*den2

The formula for subtracting fractions is similar, with the notable difference of a subtraction in the place of addition.

num1
den1
-
num2
den2
=
num1*den2 – num2*den1
den1*den2

To multiply two fractions (also known as taking the product of two fractions, the resulting numerator is the product of the two initial numerators, and likewise the resulting denominator is the product of the two initial denominators.

num1
den1
*
num2
den2
=
num1*num2
den1*den2

Finally, remembering that division is the inverse of multiplication, we can derive the formula to divide two fractions by multiplying by the inverse of the fractions:

num1
den1
÷
num2
den2
=
num1
den1
*
den2
num2
=
num1*den2
den1*num2

The next step in each of these operations is to reduce the fraction to lowest terms. One way of doing this is by considering Euclid’s GCD algorithm which is available here.

The script is available to practice your work on fractions at

http://www.learninglover.com/examples.php?id=31