Question 825741
{{{r+(r^2-5)/(r^2-1)=(r^2+r+2)/(r+1)}}}
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>The first denominator will factor:
{{{r^2-1 = (r+1)(r-1)}}}
The second denominator, r+1, does not factor. Looking at the second denominator and the factored first denominator, we should be able to figure out that the LCD is: (r+1)(r-1). So we will multiply both sides by (r+1)(r-1):
{{{(r+1)(r-1)(r+(r^2-5)/(r^2-1))=(r+1)(r-1)((r^2+r+2)/(r+1))}}}
First we must use the Distributive Property on the left side:
{{{(r+1)(r-1)(r)+(r+1)(r-1)((r^2-5)/(r^2-1))=(r+1)(r-1)((r^2+r+2)/(r+1))}}}
Now as we multiply, each denominator will cancel with some part of (r+1)(r-1):
{{{(r+1)(r-1)(r)+cross((r+1)(r-1))((r^2-5)/cross((r^2-1)))=cross((r+1))(r-1)((r^2+r+2)/cross((r+1)))}}}
Leaving:
{{{(r+1)(r-1)(r)+ 1*(r^2-5)=(r-1)(r^2+r+2)}}}
Now we solve. First we simplify:
{{{(r^2-1)(r)+ r^2-5=r*r^2+r*r+r*2+(-1)*r^2+(-1)*r+(-1)*2)}}}
{{{r^3-r+ r^2-5=r^3+r^2+2r+(-r^2)+(-r)+(-2)}}}
{{{r^3+ r^2-r-5=r^3+r+(-2)}}}
Next we get a zero on one side. Subtracting the entire right side from both sides:
{{{r^2-2r-3=0}}}
Now we factor:
(r-3)(r+1) = 0
From the Zero Product Property:
r-3 = 0 or r+1 = 0
Solving these we get:
r = 3 or r = -1<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 (r+1)(r-1), then a check is required. One must make sure that the solution does not make a factor of (r+1)(r-1) equal to zero.<br>
A quick visual check should tell us that if r = 3 (one solution) then neither (r+1) nor (r-1) will be a zero. So this solution checks!<br>
A quick visual check should tell us that if r = -1 (the other solution) then (r+1) will be a zero! So this solution fails the check and must be rejected!<br>
So there is only one solution to this problem: r = 3<br>
P.S. This equation has rational expressions. It is not a rational function. It should be posted in the "polynomials, rational expressions ..." category.