Y is partly constant and partly varies as X and when X=5,Y=8 and when X=6,Y=4.
(A)find the equation connecting Y and X.
(B)find y when X=15.
(C)find X when Y =20.
?
y = kx + c
when X=5,Y=8
8 = 5k + c
whe X=6,Y=4
4 = 6k + c
subtract them
-4 = k
back in 8 = 5k + c
8 = -20 + c
c = 28
so now you have
y = -4x + 28
and finish it off
y = mx+b
The two points give you
5m+b = 8
6m+b = 4
solve for m and b, and then you can easily answer the questions
To find the equation connecting Y and X, we can use the given information and the concept of linear relationships. We know that when X=5, Y=8, and when X=6, Y=4.
(A) The equation connecting Y and X can be written in the form of Y = mX + b, where m is the slope and b is the y-intercept. To find the slope, we can use the formula:
m = (change in Y) / (change in X)
Using the given values, we plug them into the formula:
m = (4-8) / (6-5)
m = -4 / 1
m = -4
Now that we have the slope, we can use the point-slope formula to find the equation:
Y - Y1 = m(X - X1)
Using one of the points (X=5, Y=8), we substitute the values:
Y - 8 = -4(X - 5)
Expanding and simplifying:
Y - 8 = -4X + 20
Rearranging to the desired form:
Y = -4X + 28
Therefore, the equation connecting Y and X is Y = -4X + 28.
(B) To find Y when X=15, we can substitute the value of X into the equation we found in part (A):
Y = -4(15) + 28
Y = -60 + 28
Y = -32
Therefore, when X=15, Y=-32.
(C) To find X when Y=20, we can rearrange the equation we found in part (A):
Y = -4X + 28
Substituting Y=20:
20 = -4X + 28
Rearranging:
-4X = 8
X = -2
Therefore, when Y=20, X=-2.