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.]
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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
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