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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changchengliang's profile pic

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

Posted on

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.

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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)

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