# 7.5: Solving Rational Equations

```Solving Rational
Equations
Section 7.5 beginning on page 392
Solving by Cross Multiplying
When you have two equal ratios, their cross products are equal.
Example 1: Solve
3
9
=
π₯ + 1 4π₯ + 5
3(4π₯ + 5) = 9(π₯ + 1)
12π₯ + 15 = 9π₯ + 9
3π₯ + 15 = 9
3π₯ = −6
π₯ = −2
CHECK:
3
9
=
(−2) + 1 4(−2) + 5
3
9
=
−1 −8 + 5
−3 =
9
−3
−3 = −3
Solving by Using the LCD
Example 3a : Solve
4π₯ 5
7 4π₯
9 4π₯
β
+
=− β
β
1 π₯
4 1
π₯ 1
** Multiply each term by the LCD.
πΏπΆπ· = 4π₯
** Check
20π₯ 28π₯
36π₯
+
=−
π₯
4
π₯
20 + 7π₯ = −36
7π₯ = −56
π₯ = −8
** Simplify
** Solve
5
7 2
9
+ β =−
(−8) 4 2
(−8)
5 14 9
− +
=
8 8
8
−5 + 14 9
=
8
8
9 9
=
8 8
Solving by Using the LCD
Example 3b: Solve
1−
8
3
=
π₯−5 π₯
π₯(π₯ − 5) β 1 −
πΏπΆπ· = π₯(π₯ − 5)
8
π₯(π₯ − 5) 3 π₯(π₯ − 5)
β
= β
π₯−5
1
π₯
1
π₯(π₯ − 5) −
8π₯(π₯ − 5) 3π₯(π₯ − 5)
=
π₯−5
π₯
π₯ 2 − 5π₯ − 8π₯ = 3(π₯ − 5)
π₯ 2 − 13π₯ = 3π₯ − 15
π₯ 2 − 16π₯ + 15 = 0
π₯ − 15 π₯ − 1 = 0
π₯ = 15, π₯ = 1
Solving an Equation With an
Extraneous Solution
Example 4: Solve
6
π₯−3
6
8π₯ 2
4π₯
=
−
π₯ − 3 π₯2 − 9 π₯ + 3
8π₯ 2
4π₯
π₯ + 3 (π₯ − 3) =
π₯ + 3 (π₯ − 3) −
π₯ + 3 (π₯ − 3)
π₯+3
6 π₯ + 3 = 8π₯ 2 − 4π₯(π₯ − 3)
6π₯ + 18 = 8π₯ 2 − 4π₯ 2 + 12π₯
6π₯ + 18 = 4π₯ 2 + 12π₯
0 = 4π₯ 2 + 6π₯ − 18
0 = 2π₯ 2 + 3π₯ − 9
0 = (2π₯ − 3)(π₯ + 3)
3
π₯= ,
2
π₯ = −3
π₯ + 3 (π₯ − 3)
Solving a Real – Life Problem
50π + 1000
Example 6 : The function π =
represents the average cost c (in
π
dollars) of making m models using a 3-D printer. Find how many models must be
printed for the average cost per model to fall to \$90.
π β 90 =
50π + 1000
βπ
π
90π = 50π + 1000
40π = 1000
π = 25
The average cost per model falls to
\$90 after 25 have been printed.
Monitoring Progress
Solve each equation and check your solution(s).
1)
3
2
=
5π₯ π₯ − 7
3)
1
π₯
=
2π₯ + 5 11π₯ + 8
5)
3π₯
5
3
−
=
π₯ + 1 2π₯ 2π₯
2
π₯ = − ,π₯ = 2
3
π₯ = −3
π₯ = −1,
π₯=4
1
3
2)
−4
5
=
π₯+3 π₯−3
π₯=−
4)
15 4 7
+ =
π₯ 5 π₯
π₯ = −10
6)
4π₯ + 1
12
= 2
+3
π₯+1
π₯ −1
π₯ = −2, π₯ = 5
Monitoring Progress
Solve each equation and check your solution(s).
7)
9
6π₯
9π₯ 2
+
=
π₯ − 2 π₯ + 2 π₯2 − 4
8)
7
6
−5= 2
π₯−1
π₯ −1
π₯ = −3
3
π₯ = − ,π₯ = 2
5
1
**9) Consider the function π π₯ = − 2 . Determine whether the inverse of π is a
π₯
function then find the inverse.
π¦ππ , π‘βπ πππ£ππ π ππ  π ππ’πππ‘πππ.
π π₯ =
1
π₯+2
```