X is partly constant and varies partly as y.when y=10,x=24and when y=14,x=36find x when y is 5
x = ay + b
so use the data given and you have
10a+b=24
14a+b = 36
subtract and you get
a=3, so b = -6
when y=5,
5*3 - 6 = 9
To find the value of x when y is 5, we need to determine the relationship between x and y.
Given that "X is partly constant and varies partly as y," we can assume that x has both a constant and variable component. Let's break down the information given:
When y = 10, x = 24.
When y = 14, x = 36.
From these values, we can see that the constant part of x is 24 and the variable part is 12 (36 - 24).
Now, we need to determine the relationship between the variable part of x (12) and the variable part of y as y changes from 10 to 14.
The change in y is (14 - 10) = 4, and the change in x is (36 - 24) = 12. Therefore, the rate of change between x and y is 12/4 = 3.
So, for every 1 unit increase in y, x increases by 3 units.
To find x when y is 5, we first need to determine the change in y from the given values:
Change in y = (10 - 5) = 5
Next, we calculate the corresponding change in x by multiplying the change in y by the rate of change:
Change in x = 5 * 3 = 15.
Lastly, we find the value of x when y is 5 by adding the change in x to the initial value of x (when y = 10):
x = 24 + 15 = 39.
Therefore, when y is 5, x is equal to 39.
To solve this problem, we can use the concept of direct variation. Direct variation is an equation of the form y = kx, where y and x are variables, and k is a constant.
In this case, we are given that x is partly constant and varies partly as y. This means that x can be expressed as the sum of a constant term and a variable term that is directly proportional to y. Mathematically, we can write it as:
x = c + ky
Given the values of x and y when y=10 (x=24) and y=14 (x=36), we can find the values of c and k by substituting these values into the equation.
When y = 10, x = 24:
24 = c + (10k) (equation 1)
When y = 14, x = 36:
36 = c + (14k) (equation 2)
Now we have a system of two linear equations. We can solve this system to find the values of c and k:
Subtracting equation 1 from equation 2:
36 - 24 = c + (14k) - (c + 10k)
12 = 4k
Dividing both sides by 4:
k = 3
Substituting the value of k into equation 1 to find c:
24 = c + (10 * 3)
24 = c + 30
c = -6
Now, we have the values of c and k.
Using the equation x = c + ky, we can find x when y is 5:
x = -6 + (3 * 5)
x = -6 + 15
x = 9
Therefore, when y is 5, x is 9.