Question 825739
{{{(-3)/(y-4)=(1)/(y+4)}}}
Solving equations with fractions is harder than solving equations without fractions. So we will make the problem easier if we eliminate the fractions as soon as possible.<br>
Fractions in an equation can be eliminated by...<ol><li>Find the lowest common denominator (LCD) of all the fractions (on both sides of the equation).</li><li>Multiply both sides of the equation by the LCD.</li></ol>Since the two denominators, (y-4) and (y+4), have no common factors the LCD is simply the product of the two denominators: (y-4)(y+4). So we will multiply both sides by (y-4)(y+4):
{{{(y-4)(y+4)((-3)/(y-4))=(y-4)(y+4)((1)/(y+4))}}}
As we multiply, each denominator will cancel with some part of (y-4)(y+4):
{{{cross((y-4))(y+4)((-3)/cross((y-4)))=(y-4)cross((y+4))((1)/cross((y+4)))}}}
leaving:
{{{(y+4)(-3)=(y-4)(1)}}}
which simplifies to:
{{{-3y-12=y-4}}}
Now we can solve for y. Adding 3y to each side:
{{{-12=4y-4}}}
Adding 4 to each side:
{{{-8=4y}}}
Dividing both sides by 4:
{{{-2=y}}}<br>
Last we check. This is <u>not</u> optional! When both sides of an equation are multiplied by something that might be zero, like (y-4)(y+4), then a check is required. One must make sure that the solution does not make a factor of (y-4)(y+4) equal to zero. A quick visual check should tell us that if y = -2 (our solution) then neither (y-4) nor (y+4) will be a zero. So our solution checks!<br>
I'll leave the second problem up to you to finish. I will just say that the denominators, (w+2) and (w-12), have no common factors between them. So, like the first problem, the LCD is simply their product: (w+2)(w-12)<br>
P.S. This equation has rational expressions. It is not a rational function. It should be posted in the "polynomials, rational expressions ..." category.