Grade 8

1. Prepared by: Ma. Rolinda J. Nadong
8

2. Recall
Find the LCD of the following.
A.
𝑎
3
,
5
12
B.
𝑛
6
,
𝑛
4
C.
5
4
,
3
𝑥
,
1
2
D.
1
𝑥
,
1
𝑥2 ,
1
4
E.
9
2𝑎+6
,
5
3𝑎+9

3. Check your work…
Find the LCD of the following.
A.
𝑎
3
,
5
12
LCD : 12
B.
𝑛
6
,
𝑛
4
LCD : 12
C.
5
4
,
3
𝑥
,
1
2
LCD : 4x
D.
1
𝑥
,
1
𝑥2 ,
1
4
LCD : 4x2
E.
9
2𝑎+6
,
5
3𝑎+9
LCD: 6(a+3)

4. How to solve rational equation:
1. Find the LCD of all the denominators.
2. Multiply each term of the equation by
the LCD.
3. Simplify and solve for x.
4. Check the solution. If the value makes
the LCD equal to zero, then there is no
solution. Such false solutions are called
extraneous solutions.

5. Solve and check the following rational
equations.
A.
1
2
−
2
7
=
3
2𝑥
B.
𝑥
𝑥−2
− 2 =
𝑥
𝑥−2
C.
2
𝑥−3
=
3
𝑥+1
Examples

6. Solution

1
2
−
2
7
=
3
2𝑥
 14x
1
2
−
2
7
=
3
2𝑥
14x
 7x – 4x =21
 3x = 21

3𝑥
3
=
21
3
 x = 7
Checking

1
2
−
2
7
=
3
2𝑥
Substitute 7 for x

1
2
−
2
7
=
3
2 𝟕

7−4
14
=
3
14

𝟑
𝟏𝟒
=
𝟑
𝟏𝟒
true
Example 1

7. Solution

𝑥
𝑥−2
− 2 =
𝑥
𝑥−2
 x-2(
𝑥
𝑥−2
− 2 =
𝑥
𝑥−2
)x-2
 x – 2(x-2) = x
 x-2x + 4 =x
 -x+4=x
 -x+4-x-4 = x-x-4
 -2x=-4

−2𝑥
−2𝑥
=
−4
−2
 x = 2
Checking

𝑥
𝑥−2
− 2 =
𝑥
𝑥−2
substitute 2 to x

𝟐
𝟐−2
− 2 =
2
𝟐−2

2
0
− 2 =
2
0
 0-2=0
 -2≠0
extraneous solution,it does not
solve the equation
Example 2

8. 
2
𝑥−3
=
3
𝑥+1
substitute 11 to x

2
11−3
=
3
11+1

2
8
=
3
12

1
4
=
1
4
true
Solution

2
𝑥−3
=
3
𝑥+1
 LCD: (x-3)(x+1)
 (x-3)(x+1)(
2
𝑥−3
=
3
𝑥+1
)(x-3)(x+1)
 2(x+1) = 3(x-3)
 2x+2 =3x – 9
 2x+2-3x-2=3x-9-3x-2
 -x = -11

−𝑥
−1
=
−11
−1
 x = 11
Checking
Example 3

9. Solve and check the following rational
equations.
A.
𝑥
5
+
1
4
=
𝑥
2
B.
𝑥
𝑥+2
=
3
5
C.
𝑦+3
𝑦−1
=
4
𝑦−1
Try this…

10. 
𝑥
5
+
1
4
=
𝑥
2
substitute
5
6
to x

5
6
5
+
1
4
=
5
6
2

5
6
𝑥
1
5
+
1
4
=
5
6
x
1
2

1
6
+
1
4
=
5
12

2+3
12
=
5
12

5
12
=
5
12
True
Solution

𝑥
5
+
1
4
=
𝑥
2
 20(
𝑥
5
+
1
4
=
𝑥
2
)20
 4x +5 = 10x
 4x+5-10x-5=10x-10x-5
 -6x = -5

−6𝑥
−6
=
−5
−6
 x =
5
6
Checking
Check your work…

11. 
𝑥
𝑥+2
=
3
5
substitute 3 to x

3
3+2
=
3
5

3
5
=
3
5
True
Solution

𝑥
𝑥+2
=
3
5
 5(x+2)(
𝑥
𝑥+2
=
3
5
)5(x+2)
 5x = 3x +6
 5x-3x = 3x +6 – 3x
 2x = 6

2𝑥
2
=
6
2
 x=3
Checking
Check your work…

12. 
𝑦+3
𝑦−1
=
4
𝑦−1
Substitute 1 to y

1+3
1−1
=
4
1−1

4
0
=
4
0
 0=0
True
Solution

𝑦+3
𝑦−1
=
4
𝑦−1
 y-1(
𝑦+3
𝑦−1
=
4
𝑦−1
)y-1
 y+3 = 4
 y+3-3 = 4-3
 y=1
Checking
Check your work…