## Analyzing and Integrating MI

When solving mathematical problems, we often need to borrow from our knowledge of different branches of math, such as measurement, geometry, algebra, number systems, or probability and statistics. Most often, we use the information from these areas in converting relevant information to mathematical equations. Thus, it is imperative that we keep a good understanding of the interrelated branches in math and use them appropriately in constructing solutions for a problem.

Let’s look at a few examples that will require us to use procedures and concepts from different fields of math.

Example 1: A rectangle has a perimeter of 20 cm. The length of the rectangle is 2 units longer than the width. Find the dimensions of the rectangle (geometry sense).

Constructing solution: We will follow the same steps as we did in previous examples.

Define the problem: What are the dimensions (length and width) of a rectangle if its perimeter is 20 cm?

Identify relevant information:

Identify relevant information:

Length = width + 2 cm

Define the variable: We want to find the length and width of the rectangles, so we have two variables here. Let length be equal to l and width be equal to w.

Convert relevant information to mathematical equation: There are two pieces of relevant information that will help us convert information into a mathematical equation.

1. The perimeter of the rectangle is equal to 20 cm.
Note: Perimeter of a rectangle = 2 * (length + width)
Therefore: 20 = 2 * (l + w)
Or, 10 = (l + w)
Or, l + w = 10……..(i)
2. The length of the rectangle is 2 units longer than its width. Therefore:
l = w + 2……..(ii)
Solve mathematical equation:
There are two equations:
l + w = 10…..(i)
l = w + 2……(ii)
Substitute the value of l in equation (i) from equation (ii)
(w + 2) + w = 10
w + 2 + w = 10
2w + 2 = 10
2w = 10 – 2
2w = 8
Dividing both sides by 2:
w = 4
Now substitute the value of w in equation (ii)
l = w + 2 = 4 + 2 = 6

Hence, the dimensions of the rectangle are 6 (length) and 4 (width) centimeters.

Note: If we did not know the appropriate formula for the perimeter of the rectangle, we would not have been able to solve this problem.

Let’s look at an example that uses procedures from number systems for constructing its solution.

Example 2: The sum of three consecutive even integers is 30. Find the three numbers.

Constructing solution:

Define the problem: Find three consecutive even integers such that their sum is equal to 30.

Identify relevant information: The integers are consecutive and even and their sum is equal to 30.

Define the variable: Remember that even numbers are multiples of 2. So we will need to define our variable such that it is a multiple of 2. Therefore, instead of defining our first number as x, we will define it as 2x.

Now, let our first even integer be 2x. Therefore, our next even integer will be 2x + 2, and our third even integer will be 2x + 4. (Note: The difference between two consecutive even integers is always

Convert relevant information to mathematical equation:

The sum of three consecutive even integers is 30” can be written as:

2x + (2x + 2) + (2x + 4) = 30 ……(i)

Solve mathematical equation:

We will solve equation (i), and find x:

2x + (2x + 2) + (2x + 4) = 30

Or, 2x + 2x + 2 + 2x + 4 = 30

Or, (2x + 2x + 2x) + (2 + 4) = 30 (combining like terms)

Or, 6x + 6 = 30

6x = 30 – 6

6x = 24

Dividing both sides by 6:

x = 4

1st even integer = 2 * x = 2 * 4 = 8.

2nd even integer = (2 * x + 2) = (2 * 4 + 2) = 10

3rd even integer = (2 * x + 4) = (2 * 4 + 4) = 12

Thus, the three consecutive even integers such that their sum is 30 are 8, 10, and 12.

#### Applying strategies for finding a solution to a given problem

A problem can be approached in different ways and still lead us to the same solution. Sometimes we use different math procedures to solve a problem, and sometimes we use different techniques. No one technique is better than another, and as long as what we are doing is in the realm of good reasoning and sound math, we are on the right track.

In this section we are going to learn how we can use different strategies/procedures to construct solutions and decide which approach works best.

Example 3:

A class has 30 students. The number of boys is twice the number of girls. Find the number of boys and girls in the class.

Approach 1: Constructing a solution using simultaneous linear equations

Define the problem: Find the number of boys and girls in the class if the total number of students is equal to 30.

Identify relevant information:

The total number of students = 30

The number of boys is twice the number of girls.

Define the variable: Let the number of boys be equal to x, and the number of girls be equal to y.

Convert relevant information to mathematical equation:

The total number of students is equal to 30:

x + y = 30 ….. (i)

The number of boys is twice the number of girls:

x = 2 * y ….. (ii)

Solve mathematical equation:

We will solve equation (i) by substituting value of x from equation (ii):

x + y = 30

(2 * y) + y = 30

3y = 30

Dividing both sides by 3:

y = 10

Substituting y = 10 in either of the equations (i) or (ii) we get:

x = 2 * y = 2 * 10 = 20.

Hence, there are 20 boys and 10 girls in the class.

Approach 2: Construct solutions using ratio properties

The first two steps for constructing the solution, i.e., defining the problem and identifying relevant information, remain the same.

Define the variable: It is given that the number of boys is twice (two times) the number of girls. So, the ratio of boys: girls = 2:1. Hence, for every 1 girl in the class there are 2 boys. Let the total number of girls be x. Thus, the total number of boys in class is 2x.

Convert relevant information to mathematical equation:

The total number of students is equal to 30.

x + 2x = 30 ….. (i)

Solve mathematical equation:

We will solve equation (i) by solving for x in equation (i):

x + 2x = 30

3x = 30

Dividing both sides by 3:

x = 10

Hence, the total number of girls is 10. Because the number of boys is twice the number of girls, the number of boys = 2 * 10 = 20. Therefore, there are 20 boys and 10 girls in the class.

#### Try these questions

1. Find the solution of 2x -3 = 7.
a. 5
b.10
c. 12
d. 7
Solution: 2x – 3 = 7
2x – 3 + 3 = 7 + 3
2x = 10
2x / 2 = 10/ 2 x = 5

2. Solve 2y + 9 = 4
a. -2/5
b. -5/2
c. 2/5
d. 5/2
Solution:
2y + 9 = 4
2y + 9 – 9 = 4 – 9
2y = -5
2y/2 = -5/ 2
y = - 5/ 2

3. Solve: x/3 + 5/2 = -3/2
a. 10
b. 12
c. -12
d. -10
Solution: x/3 + 5/2 = - 3/2
x/3 + 5/2 – 5/2 = -3/2 – 5/2
x/3 = - 8/ 2 = - 4
x/3 * 3 = - 4 * 3
x = -12
4. Solve: 15/ 4 – 7x = 9
a. -4/3
b. -3/4
c. 5/6
d. -3/8