# How do I solve functions with fractions like: y=2/3x+1 or y=3/4x-2.

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I think by y = 2/3x + 1, you mean y = (2/3)*x + 1.

So to find the value of y in this case, multiply x with (2/3) and add 1.

Similarly for y = (3/4)x - 2, multiply x with (3/4) and subtract 2 from the result.

In case you mean y = 2/(3x) +1, you will:

- first have to multiply x with 4
- then divide 2 by the result you get
- and finally add 1

For y = 3/(4x) + 1,

- first multiply x with 4
- divide 3 by the result
- and finally add 1.

In case you mean y =2/(3x+1),

- first multiply x and 3
- add one to the result
- finally divide 2 by what you get after step 2.

Similarly for y =3/(4x - 2),

- first multiply x with 4
- then subtract 2 from it
- and finally divide 3 by the result after step 2.

[It is **essential to use brackets** when you write these kinds of functions so that the **order of operations is clear**.]

It depends on how it is written the denominator.

For instance, if the first function is f(x) = 2/(3x+1), we'll have to impose constraints of existence of the given function.

Why is that?

Because the given function is a fraction and it only exists if the denominator is not zero.

So, we'll determine the value of x that makes the function impossible:

3x + 1 = 0

We'll subtract 1 both sides:

3x = -1

We'll divide by 3:

x = -1/3

So, the function is meaningless for x = -1/3.

Now, let's discuss the other case, when the denominator of the frst function is 3x.

f(x) = (2/3x) + 1

We'll change f(x) by y:

y = (2/3x) + 1

The first step is to multiply the constant 1 by 3x:

y = 2/3x + 3x/3x

y = (2+3x)/3x

Now, we'll multiply by 3x both sides:

3xy = 2 + 3x

We'll subtract 3x both sides:

3xy - 3x = 2

We'll factorize by 3x:

3x(y-1) = 2

We'll divide by 3(y-1):

x = 2/3(y-1)

We'll discuss the same way the second function f(x) = 3/(4x-2) or f(x) = (3/4x) - 2.

y = (2/3)x+1....(1)

y = (3/4)x-2.....(2).

Both are the equations of the straight lines of the form y = mx +c in the slope intercept forms.

So we know that y coordinates at the point of intersection is equal.

So (1)-(2):

y-y = (2/3x)+1-{(3/4)x-2}

0 = (2/3)x-(3/4)x+1+2.

0 = {(2*4-3*3)/12}x+3

Multiply by 12:

0 =(8-9)x+3*12

0 = x+12.

So x = -12.

We put x= -12 in (1): y = (2/3)x+1 = (2/3)(-12)+1 = -8+1 = -7.

So x= -12 , y = -7