Example of Posing Problems

There are expressions in ordinary language that we use very frequently and that refer to a fraction or ratio, which is very important that we know how to identify. I mean terms such as: speed that refers to the fraction of kilometers, meters, etc. and that we mention as kilometers per hour, meters per second, etc. giving the appearance of a product.

Unit price: which refers to pesos, cents, etc. and that we read as pesos for an article, cents for an article, etc., or also pesos per kilo, pesos per liter, etc. For the treatment of problems where some type of reason intervenes, we can use the following proposition as a formula:

A quantity is equal to the ratio of the base taken C = R X B

a) Number of kilometers = ratio in kilometers per hour x hours
(distance) (speed) (time)
b) Amount of money = ratio in pesos per unit x units
(Cost) (unit price) (units)
c) Amount of work done = ratio of work done each day
x days worked.

In solving the problems we are going to consider the following steps:
1. Correctly interpret the meaning of the spoken or written expression, assigning the last letters of the alphabet (x, y, z) to the variables or unknowns.

2. Write the algebraic expression or expressions trying to refer all the variables to a single one that could be called x This restriction is temporary as long as we learn to solve expressions with more than one variable).
3. Relate the information already symbolized to establish an equation or an inequality.
4. Solve the equation or inequality.
5. Interpret the algebraic solution in terms of ordinary language, checking that it satisfies the stipulated conditions.

EXAMPLES PROBLEMS OF SETUP:

1. Find the dimensions of a rectangular piece of land with a perimeter of 540 meters, if we know that the length is 30 meters more than the width. This is example 2 of the Problem Setting topic, only now we must symbolize using only one variable).

Length measures 30 meters more than width length = x width = x - 30
and the perimeter is 540 meters
perimeter = 2 times the length + 2 times the width 2x + 2 (x - 30) = 540

Equation: 2x + 2 (x - 30) 540
Solution: 2x + 2x - 60 = 540
4x = 600
x = 150

Interpretation:
length = 150 meters width = 120 meters

Verification:
Perimeter = 2 (150) + 2 (120) = 300 + 240 = 540 meters
2, If the sum of two numbers is 21 and one number is triple the other. What are those two numbers?
Two numbers whose sum is 2.1 x, 21 - x
one is triple the other (21 - x) = 3x
Equation: 21 -x = 3x
Solution: 21 = 4x
x = 21/4

Interpretation: one number = 21/4 and the other = (3) 21/4 = 63/4

Verification:
21/4+63/4=84/4=21