Posted by
**Teri** on
.

could someone please help me with these problems i am having trouble.

If y=36 when x =6 find x when y = 42

If y=-18 when x=6 find x when y=6

If y =12 when x= 15 find x when y =21

If y=-6.6 when x=9.9 find y when x =6.6

If y=6 when x=(2/3)find x when y=12

those are the problems you don't have to do all just at least show me step by step how you arrived at the answer thank you so much for helping

take the first one:

y1/y2 = x1/x2

y1=36 x1 is 6

y2=42 x2 is ?

then

36/42=6/x solve for x

and

x=42*6/36 =7

check my thinking.

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These are proportion problems

(1) 6:36=x:42 Which is the same as (6*42)/36=x

(2)6:18=x:6 Which is the same as (6*6)/18=x

(3) 15:12=x:21 I think you can do them now

(4)9.9:-6.6=6.6:y Watch where the y is located here.

(5)(2/3):6=x:12 Be sure to handle the fraction correctly.

In general, a proportion is two ratios a:b=c:d

a and d are called the extremes and b and c the means.

The fundamental relation is a*d=b*c The product of the extremes = the product of the means.

You can also think of it in terms of fractions a/b = c/d. in terms of fractions both ratios are different names for the same fraction. One is a multiple of the other. In the first one we have

6:36 or 6/36=1/6

We want to find x such that x/42 = 1/6 so x=42*(1/6)

You can verify that 1/6 = 6/36 = 7/42.

The fraction 1/6 would be the fraction in lowest terms in integers. In some places finding that lowest reducible integer fraction is very important. Other times we can't find such an integer ratio, e.g. if the numbers are irrational.

If y =12 when x= 15 find x when y =21

y=kx 21