# How do I solve this? y= 4/x y=3x-1i got as far as making the y's equal to eachother 4/x=3x-1 4/x-3x+1=0 but im not sure what to do from here

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y= 4/x

y=3x-1

==> 4/x = 3x - 1

Let us multiply by x:

==> 4 = 3x^2 - x

Now group all terms on one side:

==> 3x^2 - x - 4 = 0

Now we have a quadratic equation. We will use the formula to solve:

We know that:

x = [ -b +/- sqrt)b^2 - 4ac)]/ 2a

==> x1= (1 + sqrt(1+48) / 6

= (1+ sqrt(49)/6

= (1+ 7)/6 = 8/6 = 4/3

**==> x1= 4/3**

==> y1 = 4/x = 4/ (4/3) = 3

**==> y1= 3**

==> x2= ( 1- 7)/6 = -6/6 = -1

**==> x2= -1**

==> y2 = 4/x2 = 4/-1 = -4

**==> y2 = -4**

**Then we have 2 solutions:**

**( -1,-4) and ( 4/3, 3)**

y = 4/x …….(1)

y = 3x -1 ……..(2)

Substitute (2) into (1)

3x - 1 = 4/x

Multiply x on both sides:

3x^2 - x = 4

Re-arranging:

3x^2 - x -4 = 0

Factorize the expression on the left hand side (LHS):

(3x-4) (x+1) = 0

x = 4/3 or -1

Substituting these values of x into (1),

Case 1: x = 4/3

y = 4/(4/3) = 4 (3/4) = 3

When **x=4/3, y=3**

Case 2: x = -1

y= 4/(-1) = -4

When **x=-1, y=-4**

By the way, to appreciate this question graphically, click on the link provided at the bottom of this submission.

The two solutions are when the two curves intersect each other.