Suppose y varies directly with x. Write a direct variation equation that relates x and y.
y= 7 1/2 when x=3
y=3/2x
y=2/3x*
y=5/2x
y=2/5x
To write a direct variation equation that relates x and y, we need to find the constant of variation.
In this case, we are given that when x = 3, y = 7 1/2.
Since y varies directly with x, we can write the equation as:
y = kx
where k is the constant of variation.
To find the value of k, substitute the given values into the equation:
7 1/2 = k * 3
To solve for k, divide both sides of the equation by 3:
k = (7 1/2) / 3
Now, simplify the right side of the equation:
k = 15/2 / 3
To divide fractions, multiply the first fraction by the reciprocal of the second:
k = 15/2 * 1/3
Multiply the numerators together:
k = 15/6
Simplify the fraction:
k = 5/2
Therefore, the direct variation equation that relates x and y is:
y = (5/2)x
To write a direct variation equation that relates x and y, we need to determine the constant of variation, which is the ratio of y to x.
In this case, we are given that y varies directly with x, so we can use the general form of a direct variation equation: y = kx, where k is the constant of variation.
We are given that y = 7 1/2 when x = 3. Plugging these values into the equation, we get:
7 1/2 = k(3)
To solve for k, we need to convert the mixed number 7 1/2 to an improper fraction. The improper fraction equivalent of 7 1/2 is 15/2. So, the equation becomes:
15/2 = k(3)
To solve for k, we divide both sides of the equation by 3:
(15/2) / 3 = k
Simplifying the equation, we have:
15/6 = k
5/2 = k
Therefore, the direct variation equation that relates x and y is:
y = (5/2)x
Thank you Steve!
Hmmm (2/3)*3 = 2, not 7
y/x = (7 1/2)/3 = (15/2)/3 = 15/6 = 5/2
y = 5/2 x
You need to learn to check your answers. At least do a sanity check. 2/3x < x, so how could y = 2/3 x relate 3 and 7?