5/3 = x/y

First we will cross multiply:

==> 5y = 3x .......(1)

It is given that:

x- y = 5

==> x= y + 5 ........(2)

Now we will substitute with x values in (2)

5y = 3x

==> 5y = 3(y+5)

==> 5y = 3y + 15

==> 2y = 15

==> y= 15/2 = 7/5

==> x= y+5 = 7.5 + 5 = 12.5

(5/3)=(x/y) If x-y=5

Given two equations and two variables we can solve for both:

5/3 = x/y

can be rewritten by cross multiplying.

5y = 3x

x - y = 5

can be written by bringing the y to the other side:

x = 5 + y

Plug in this x value into the other equation:

5y = 3 ( 5 + y )

Now we can solve for y:

5y = 15 + 3y

2y = 15

y = 15/2

Plug y into one of the equations to solve for y:

x = 5 + (15/2) = 12. 5

**We'll use the proportion rule:**

**a/b = c/d <=> (a-b)/b = (x-y)/y**

We'll apply this rule to the first equation:

(5/3)=(x/y) <=> (5-3)/3 = (x-y)/y (3)

But, from the second equation, we have x-y=5.

We'll substitute the difference x-y by 5, in (3).

(5-3)/3 = 5/y

2/3 = 5/y

We'll cross multiply:

2y = 15

We'll divide by 2:

y = 15/2

**y = 7.5**

Now, we'll substitute y by 7.5, into the second equation:

x-y=5

x - 7.5 = 5

We'll add 7.5 both sides:

x = 5 + 7.5

**x = 12.5**

I assume your question is to find the values of x and y for which the conditions (5/3)=(x/y) and x-y=5 are true.

Take x-y=5 first, we get x=5+y

Now replace x in (5/3)=(x/y) by the corresponding value of y.

We get (5/3)=(5+y)/y

=> 5y=3*(5+y)

=>5y= 15+3y

=>5y-3y=15

=>2y=15

=>y=15/2 = 7.5

As x=5+y , it is equal to 7.5+5=12.5.

**Therefore x=12.5 and y=7.5**

(5/3)=(x/y) If x-y=5

To solve for x and y.

The 1st equation is 5/3 = x/y . Multiply by 3y and we get:

5y = 3x. ...(1)

Multiply 2nd equation by 3:

3x-3y = 15......(2)

Eliminate 3x from (1) and (2) by adding:

5y +3x-3y = 3x+15

2y = 15

y = 7.5.

Using (1) 5y = 3x, 5(7.5) = 3x. Or x = 5*7.5/3 = 12.5.

So (x,y) = (12.5 , 7.5)

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