Question 473486
your equation is:
{{{(2x+1)/(3x-3) = (5x+1)/(6x-2)}}}
i don't see any common factors, so i think we'll have to do this the hard way.
cross multiply to get:
{{{(2x+1)*(6x-2) = (3x-3)*(5x+1)}}}
multiply those factors out to get:
{{{12x^2 + 2x - 2 = 15x^2 - 12x - 3}}}
subtract 12x^2 and 2x from both sides of the equation, and add 2 to both sides of the equation to get:
{{{0 = 15x^2 - 12x - 3 - 12x^2 - 2x + 2}}}
combine like terms to get:
{{{0 = 3x^2 - 14x - 1}}}
this is the same as:
{{{3x^2 - 14x - 1 = 0}}}
this is a quadratic equation.
you need the quadratic formula to solve this.
the quadratic formula is:
x = {{{-b +- sqrt(b^2-4ac)/(2a)}}}
in your equation:
a = 3
b = -14
c = -1
solving by using the quadratic formula gets you:
x = 4.737034184
x = -.070367517
substitute these values of x into the original equations and you will see that the equations are true.
those are your answer.
this was very messy.
i didn't see an easier way to get the answer.