How do you do these problems ?

Question

Solve .

y varies directly as x^2 .

If y = 112 when x = 4 , find y when x = 6 .






Probem 2




Simplify the expression.

(a^2 - ab + 4a - 4b ) / (a + 4 )

Answer 1

(1) y proportional to x^2:

112 / (4^2) = y / (6^2)
112 / 16 = y / 36
16 y = 36 * 112
y = 36 * 112 / 16
y = 252

(2) simplify

(a^2 - ab + 4a - 4b ) / (a + 4 ) =
(a + 4)(a - b) / (a + 4) =

(a - b)

The key is to factor the numerator and then cancel the factor that matches the denominator.

Answer 2

y varies directly as x^2 means that y = c*x^2, where c is a constant. So you need to find that constant. So if y = 112 and x = 4, then 112 = c * (4)^2 --> 112 / 16 = c or c = 7.
Now since you have c, then our equation becomes y =7x^2. When x = 6, y = 7(6)^2 = 252

(a^2 - ab + 4a - 4b ) / (a + 4 )
look at the numerator and regroup:
a^2 + 4a - ab - 4b =
a(a + 4) - b(a +4) =
(a - b)(a+4)
so
(a - b)(a + 4) / (a + 4) = a - b

Answer 3

Problem one:

There is a constant multiplied to the x^2. Since 4^2 = 16, and 112/16 is 7, the constant is 7. So when x = 6, we take 6^2 which is 36 and multiply it by 7 and we get 252.

Problem two:

We need to factor the top half of the fraction first. and we get:

(a + 4)(a - b)
-----------------
(a + 4)


Since (a+4)/(a+4) = 1, we're left with an answer of (a-b)

Answer 4

y = C(x^2)

when x = 4, x^2 = 16.
C must be 112/16 = 7

y = 7(x^2)

when x = 6, y = 7*64.

(a^2 - ab + 4a - 4b) = a(a-b) + 4(a-b) = (a+4)(a-b)

cancel the (a+4) on top and bottom:
= a-b.

Answer 5

y variesdirectly with x^2
=> y=k*x^2
put the given values in abv eq
112=k*4^2
112=16k
k=112/16
k=7
now if k=7 and x=6
then y=7*6^2
y=7*36
y=252
answer to sec problem
(a^2 - ab + 4a - 4b ) / (a + 4 )=[a(a-b)+4(a-b)]/(a+4)
=(a-4)(a-b)/(a-4)
=a-b