Question

x/7=y/3=3x+5y/.......=7x-9y/....... Fill in the blank

Answer 1

Answer:

x/7=y/3=(3x+5y)/36.=(7x-9y)/22

Step-by-step explanation:

Let say

x/7=y/3=(3x+5y)/A.=(7x-9y)/B

X/7=y/3

=> 3x = 7y

=> 3x + 5y  = 7y  + 5y = 12y

=> y/3 = 12y/ A

=> A = 12 * 3

=> A = 36

7x-9y

= 7(7y/3) - 9y

= (49y - 27y)/3

= 22y/3

=> y/3 = 22y/ 3B

=> B = 22

x/7=y/3=(3x+5y)/36.=(7x-9y)/22

Answer 2

Answer:

$\rm \dfrac x7=\dfrac y3=\dfrac{3x+5y}{36}=\dfrac {7x-9y}{22}.$

Step-by-step explanation:

To be filled:

x/7=y/3=3x+5y/.......=7x-9y/.......

Since,

$\rm \dfrac x7 = \dfrac y3\\\therefore x=\dfrac 73 y.$

Using this value of x for the first blank whose value is, let's say, P,

$\rm \dfrac y3=\dfrac{3x+5y}{P}=\dfrac{3\times \dfrac 73 y+5y}{P}=\dfrac{7y+5y}{P}=\dfrac{12y}{P}\\ \dfrac y3=\dfrac{12y}{P}\\\Rightarrow P=\dfrac{12y}{y}\times 3=36.$

Now, using this value of x for the second blank whose value is, let's say, Q,

$\rm \dfrac y3=\dfrac{7x-9y}{Q}=\dfrac{7\times\dfrac {7y}3-9y}{Q}=\dfrac{\dfrac{49y}3-9y}{Q}=\dfrac{\dfrac{49y-27y}3}{Q}=\dfrac{22y}{3Q}\\\dfrac y3=\dfrac{22y}{3Q}\\Q=\dfrac{22y}{3y}\times 3=22.$