A math club had 4 members in its first year. It had 19 members in its fourth year and 29 members in its sixth year. Assuming a c
onstant rate of increase, how many members will the math club have in its thirteenth year?
1 answer:
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Answer:
With Method I, they should plan to refinish 92 chairs and with Method II they should plan to refinish 70 chairs.
Step-by-step explanation:
This problem can be solved by means of a system of equations, that is to say a system that in this case will contain 2 linear equations with two variables: "x" and "y".
First you must define what your variables x and y are:
- x: Chairs refinished by method I.
- y: Chairs refinished by method II.
On the one hand, you know that between method I and method II they plan to work 151 hours. In method I, each chair takes 0.5 hours of work, this means that to obtain the total amount of time it takes to work in the "x" chairs by this method, 0.5 must be multiplied by the number of chairs. You can apply the same reasoning to calculate the total amount of time it takes to work on the "y" chairs by method II knowing that each chair takes 1.5 hours. Everything said above is represented by the equation:
<em> Equation (A)</em>
In the equation the values 0.5 and 1.5 are represented in the form of a fraction (1/2 and 3/2 respectively) to be able to solve more in a way how the system of equations.
On the other hand, to state the other equation of the system, it must be taken into account that by method I the material costs $ 9 for each chair and by method II the material costs $ 7 for each chair. To determine the value of the material in each method, multiply the value of each chair by the amount of chairs refinished in each case. And the sum of the value of the materials of both methods must be $ 1318. This is represented by the equation:
<em>9*x+7*y=1318 Equation (B)</em>
Having both equations, you can solve the system. There are several methods to solve it, but one of the easiest and most widely used methods is substitution. This consists of isolating one of the variables from one of the equations and replacing it in the other equation.
In this case you can choose to isolate the variable x from equation B, resulting in:
<em>Equation (C)</em>
It is always preferable to work in fractions for convenience to solve the calculations.
Now you replace the expression obtained from x in equation A, obtaining:
Now you have an equation with a variable, "y", which can be solved, that is, you can get the value of "y". So "y"=70
Remembering that the variable "y" is the number of chairs refinished by method II, <u><em>the value of "y" means that 70 chairs by that method were refinished</em></u>.
To calculate the value of "x", you simply replace the value of "y" in either of the two equations (A) or (B) of the system and solve the equation. Or you can replace the value of "y" in equation (C): Either way the result must be the same: "x"=92
Remembering that the variable "x" is the number of chairs refinished by method I, <u><em>the value of "x" means that 92 chairs by that method were refinished.</em></u>