Rational Equations (fractions)
There are two types of equations with rational expressions or fractions. The 1st type have constants or real numbers as denominators and the 2nd type have variables in the denominators.
Equations which have constants or real numbers as denominators are solved easily if we
multiply through by the lowest common multiple (lcm) of the denominator constants.
Once you've done this, the equation becomes a linear or quadratic equation which can be solved using one of the techniques described in those lessons. (6.1 and 6.2)
Rational equations which have variables in their denominators are solved in a similar manner, however we must check that none of our solutions make any denominator equal zero since division by zero is undefined. If one or more of our answers make some denominator equal zero, we must eliminate those values from the solution set.
examples with constant denominators
a)  4(x + 5) + 12(2) = 3(x + 16) 4x + 20 + 24 = 3x + 48 4x – 3x = 48 – 20 – 24 x = 4 |
a) The lcm of 3 and 4 is 12. Multiply through by 12. Remove brackets Transpose, collect like terms and solve. |
b)  2(2z – 4) + (z – 5) = 2 4z – 8 + z – 5 = 2 5z = 15 z = 3 |
b) Multiply through by 6 the lcm. Remove brackets. Transpose, collect like terms and solve. |
c)  4(x2 – x) – 3(x2 – x) – 2 = 0 4x2 – 4x – 3x2 + 3x – 2 = 0 x2 – x – 2 = 0 (x – 2)(x + 1) = 0 x = 2 or x = –1 |
c) The lcm of 3, 4, and 6 is 12. Multiply through by 12, Remove brackets, collect like terms. Solve by factoring the trinomial. |
examples with variable denominators
a)  4(m – 2) = 3(m – 1) 4m – 8 = 3m – 3 m = 5 |
a) The lcm is (m – 2)(m – 1). Multiply, remove brackets, transpose, solve. Check that m = 5 doesn't make any denominator = zero. |
b)  (3s + 5)(2s – 1) = (2s + 3)(3s – 2) 6s2 + 7s – 5 = 6s2 + 5s – 6 2s = –1 s = – ½ |
b) The lcm is (2s + 3)(3s + 5). Multiply through by the lcm, remove brackets, collect like terms and solve. |
c)   (x + 2) = x 0 = – 2 no solutions |
c) We factor the denominators the lcm to be x(x – 2)(x + 2). Multiply by the lcm and get a statement which cannot be true. This equation has no solutions since 0 cannot equal –2. |
d)   m + 1 = 2 m = 1 If m = 1, then (m – 1) = 0 and (m2 – 1) = 0 therefore there are no solutions. |
d) We factor the denominators the lcm is (m – 1)(m + 1). Multiply by the lcm, transpose and solve. Check if m = 1 makes any denominators = 0. It makes both denominators = 0, therefore there are no solutions. |
practice:
Solve the following equations. Check your answers.
1)
2) 
3)
4)
5) What number must be added to both numerator and denominator of the fraction 
to change its value to 
?
6) Find 3 consecutive even integers such that one-quarter of the 1st, plus one-third of the 2nd, plus one-fifth of the 3rd equals 14. (hint: to get from one even integer to the next, add 2)
solutions:
Solve the following equations. Check your answers.
1)  4(2x – 3) = 3(x + 4) – 12(2) 8x – 12 = 3x + 12 – 24 5x = 0 x = 0 |
2)  8(x + 3) = 4(3x – 5) 8x + 24 = 12x – 20 –4x = –44 x = 11 |
3)   x + 2 + 7x2 – 14x – 26 = 0 7x2 – 13x – 24 = 0 (7x + 8)(x – 3) = 0 becomes x = – 8/7 or x = 3 |
4) 8a – 32 – a2 + 3a = a2 – 9a + 18 2a2 – 20a + 50 = 0 is a2 – 10a + 25 = 0 (a – 5)2 = 0 becomes a = 5 |
5) What number must be added to both numerator and denominator of the fraction 
to change its value to 
?
Let x = the number to be added to both numerator and denominator
The new fraction with value 
will look like this: 
When we cross multiply we get: 3( 7 + x ) = 2 ( 8 + x )
This then becomes 21 + 3x = 16 + 2x so x = –5.
We must add – 5 to both numerator and denominator to make the fraction = 
6) Find 3 consecutive even integers such that one-quarter of the 1st plus one-third of the 2nd plus one-fifth of the 3rd equals 14.
Let x = the 1st even integer.
therefore x + 2 = the 2nd even integer.
and x + 4 = the 3rd even integer.
15x + 20 (x + 2 ) + 12 ( x + 4 ) = 60 (14) (multiply through by 60 the lcm of 4, 3, & 5)
The 3 consecutive even integers are 16, 18, and 20.
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