College Algebra & Trigonometry, 2018a

1.9. RATIONAL EQUATIONS 75
In some situations, we can create a single common denominator for every fraction
in the problem and then clear them all at once.
Example
Solve for x.
2
x +3 − 4
3x − 1 =
x
3x 2 +8x − 3
2
x +3 − 4
3x − 1 =
2
x +3 ∗ 3x − 1
3x − 1 − 4
3x − 1 ∗ x +3
x +3 =
x
3x 2 +8x − 3
x
3x 2 +8x − 3
2(3x − 1) − 4(x +3)
(x + 3)(3x − 1)
=
x
(x + 3)(3x − 1)
2(3x − 1) − 4(x +3)=x
The missing step above is the clearing of both denominators:
(x + 3)(3x − 1) ∗
✘✘✘✘
(x +3) ✘ (3x ✘✘✘✘
− 1) ∗
2(3x − 1) − 4(x +3)
(x + 3)(3x − 1)
2(3x − 1) − 4(x +3)
✘✘✘✘
(x +3) ✘ (3x ✘✘✘✘
− 1)
=
x
∗ (x + 3)(3x − 1)
(x + 3)(3x − 1)
x
=
✘✘✘✘
(x +3) ✘ (3x ✘✘✘✘
− 1) ∗ ✘✘✘✘
(x +3) ✘ (3x ✘✘✘✘
− 1)
2(3x − 1) − 4(x +3)=x