Solving Rational Equations
Purplemath
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Solve the following equation:
![10 / (x + 4) = 15 / [ 4 (x + 1) ]](rational/solve07.gif){kind=small}
The rational expressions in this equation have variables in the denominators. So my first step is to check for which x-values are not allowed, because they'd cause division by zero. Setting each denominator equal to zero and solving, I get:
x ≠ −4, −1
Looking at the equation, I notice that this equation is a proportion; that is, the equation is of the form "(one fraction) equals (another fraction)". So all I need to do here is "cross-multiply"; that is, I can use Method 3:
10(4(x + 1)) = 15(x + 4)
40x + 40 = 15x + 60
25x + 40 = 60
25x = 20
Since this solution won't cause any division-by-zero problems, it is a valid solution to the equation, and my answer is:
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Solve the following equation:
There is only one fraction, so the common denominator is the only denominator; namely, x. Also, setting that denominator equal to zero, I see that the solution to this equation cannot be x = 0.
Method 1: To solve, I can convert everything to this common denominator, and then solve the numerators:
x2 + x = 72
x2 + x − 72 = 0
(x + 9)(x − 8) = 0
x = −9 or x = 8
Method 2: To solve, I can start by multiplying through on both sides by x:
x2 + x = 72
x2 + x − 72 = 0
(x + 9)(x − 8) = 0
x = −9 or x = 8
Either way, the solution is the same. Since neither solution causes a division-by-zero problem in the original equation, both solution values are valid.
x = −9, 8
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Solve the following equation:
![10 / [ x (x − 2) ] + 4 / x = 5 / (x − 2)](rational/solve04.gif){kind=small}
First, I'll solve the denominators. Setting each equal to zero to find the disallowed values, I get:
x ≠ 0, 2
The lowest common denominator of these fractions will be x(x − 2).
Method 1: I can solve the equation by converting all of the rational expressions to the common denominator, and then solving the numerators:
![10/[x(x − 2)] + (4x − 8)/[x(x − 2)] = 5x/[x(x − 2)]](rational/solve13.gif)
10 + (4x − 8) = 5x
10 + 4x − 8 = 5x
4x + 2 = 5x
2 = x
Method 2: I can also solve this equation by multiplying through on both sides of the equation by the denominator. (The equation becomes a bit of a mess. Use caution!)
![[10/(x(x − 2)] [x(x − 2)/1] + (4/x) [x(x − 2)/1] = [5/(x − 2)] [x(x − 2)/1]](rational/solve12.gif)
10 + 4(x − 2) = 5(x)
10 + 4x − 8 = 5x
4x + 2 = 5x
2 = x
Using either method, I get the same answer; namely, x = 2. However, checking back to the beginning, where I first noted the disallowed values for the original equation, I see that x ≠ 2. In other words, my only solution value would actually cause division by zero. Since the only possible solution causes division by zero, then this equation really has no solution. My answer then is:
no solution
It doesn't look like I did anything wrong, mathematically, in the previous exercise. So how did I end up with an entirely invalid solution?
When dealing with rational expressions and equations, we're not allowed to divide by zero. When there are variables in denominators, we then will have certain values which can cause division by zero. Whichever method one uses to solve a given rational equation, one will, at some point, get rid of those denominators. In other words, at some point, one will, in effect, make those division-by-zero problems magically disappear. But they aren't actually gone; they've merely been ignored at some stage. In the end, one must go back to the beginning and check one's solution against those original disallowed values. And it is perfectly possible that a given equation will have no solution at all.
Whenever you solve a rational equation, always check your (interim) solution against the denominators (and their disallowed values) from the original equation. It is entirely possible that a problem will have an invalid (that is, an "extraneous") solution. This is especially true on tests. So always check!
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