# if y=3 when x=2, find x when y=5

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When y = 3, x = 2, then the ratio of x:y = 2:3.

Therefore, `2/3 = x/5`

Cross multiply to solve.

`2*5=3x`

`10 = 3x` Divide both sides by 3.

`10/3 = (3x)/3`

`x = 10/3`

**The solution for x when y is 5 is x = `10/3.`**

y=3 when x=2, find x when y=5

just set it up as a ratio and cross multiply

`3/2 = 5/x`

`2 xx 5 = 10`

`3xx x = 3x`

`10=3x`

`divide by 3 `

`(10)/3 = (3x)/3`

`(10)/3 = x ` you can divide 10 by 3 to get the decimal answer

so x= `(10)/ 3` or`3.33`

y=3 when x=2, find x when y=5

Assuming you have a linear function; just set up proportions and solve for x.

`(2)/(3) = (x)/(5)`

`10 = 3x`

`x = 10/3`

``First cross multiply. Then solve for x.

if y=3 when x=2, find x when y=5

2/3=x/5

cross multiply

10=3x

x=10/3

The problem states that when x = 2, y = 3 and the value of x has to be determined when y = 5.

Now the variables x and y can be related in any way. For example, if we take the relation y = x^2 - 1, it also gives the value of y as 3 when x is equal to 2. Using this, the value of x when y is 5 would be `+-sqrt 6`

Even if the variables are linearly related, y = 3 when x = 2 does not give a unique value of x for y = 5. To illustrate the same, consider the following graphs:

Given just y=3 when x=2, it is not possible to find a unique value of x for which y = 5.