PRACTICE PROBLEMS ON SOLVING RATIONAL EQUATIONS

What is a rational equation ?

A rational equation is an equation that has one or more rational expressions or more simply put, it's an equation with fractions.

And solving equations with rational expressions can be using two different methods.

Method 1 :

Step 1 :

Simplify both sides of the equation by adding/subtracting fractions to get one fraction on either side. 

Step 2 :

Now, we use the Cross multiplication.

Step 3 :

Solve the transformed equation.

Method 2 :

Step 1 :

Multiply both sides of the equation by the Least common denominator to get rid of the fractions.

Step 2 :

Solve the transformed equation.

Problem 1 :

Solution :

By cross multiplication, we get

3(y + 3) = 2(2y)

3y + 9 = 4y

9 = 4y – 3y

9 = y

Cross – check :

By applying y = 9 in the given equation, we get

(Y + 3)/2y = 2/3

(9 + 3)/2(9) = 2/3

12/18 = 2/3

2/3 = 2/3

So, the solution of y is 9.

Problem 2 :

Solution :

By cross multiplication, we get

9b = 3(2b + 7)

9b = 6b +21

9b – 6b = 21

3b = 21

 b = 21/3

b = 7

Cross – check :

By applying b = 7 in the given equation, we get

9/(2b + 7) = 3/b

9/(2(7) + 7) = 3/7

9/(14 + 7) = 3/7

9/21 = 3/7

3/7 = 3/7

So, the solution of b is 7.

Problem 3 :

Solution :

By cross multiplication, we get

y(4y – 1) = 3(5y)

4y2 – y = 15y

4y2 - y - 15y = 0

4y2 16y = 0

By factoring 4y, we get

4y(y - 4) = 0

4y = 0 and y - 4 = 0

y = 0 and y = 4

Cross – check :

By applying y = 4 in the given equation, we get

(4y – 1)/5y = 3/y

(4(4) – 1)/5(4) = 3/4

(16 – 1)/20 = 3/4

15/20 = 3/4

3/4 = 3/4 (true)

By applying y = 0 in the given equation, we get

(4y – 1)/5y = 3/y

(4(0) – 1)/5(0) = 3/0

(– 1)/0 = 3/0

=  infinity

Since it has an infinity, we can't take the y  =  0.

So, the solution of y is 4.

Problem 4 :

Solution :

By cross multiplication, we get

4(2) = 3w + 7

8 = 3w + 7

8 – 7 = 3w

1 = 3w

1/3 = w

Cross – check :

By applying w = 1/3 in the given equation, we get

4/(3w + 7) = 1/2

4/[3(1/3) + 7] = 1/2

4/[(3/3) + 7] = 1/2

4/(1 + 7) = 1/2

4/8 = 1/2

1/2 = 1/2

So, the solution of w is 1/3.

Problem 5 :

Solution :

By cross multiplication, we get

12(y – 3) = 9y

12y - 36 = 9y

-36 = 9y – 12y

-36 = -3y

12 = y

Cross – check :

By applying y = 12 in the given equation, we get

12/y = 9/(y – 3)

12/12 = 9/(12 – 3)

1 = 9/9

1 = 1

So, the solution of y is 12.

Problem 6 :

Solution :

By cross multiplication, we get

c(c – 5) = c – 5

c2 – 5c = c – 5

c2 – 5c – c + 5 = 0

c2 – 6c + 5 = 0

By factorization, we get

(c – 1) (c – 5) = 0

 c = 1 and c = 5

Cross – check :

Step 1 :

By applying c = 1 in the given equation, we get

(c – 5)/(c – 5) = 1/c

(1 – 5)/(1 – 5) = 1/1

-4/-4 = 1/1

1 = 1 (true)

Step 2 :

By applying c = 5 in the given equation, we get

(c – 5)/(c – 5) = 1/c

(5 – 5)/(5 – 5) = 1/5

0/0 = 1/5 (false)

So, the solution of c is 1.

Problem 7 :

Solution :

By cross multiplication, we get

   6(y + 15) = 15y

 6y + 90  = 15y

90 = 15y – 6y

90 = 9y

10 = y

Cross – check :

By applying y = 10 in the given equation, we get

1/15 + 1/y = 1/6

1/15 + 1/10 = 1/6

By taking the least common multiple, we get

2/30 + 3/30 = 1/6

5/30 = 1/6

1/6 = 1/6

So, the solution of y is 10.

Problem 8 :

Solution :

By cross multiplication, we get

(2p – 1)(p + 3) = (p – 1)(2p + 5)

2p2 + 6p – p – 3  = 2p2 + 5p – 2p – 5

2p2 + 5p – 3 = 2p2 + 3p – 5

2p2 + 5p – 3 - 2p2 – 3p + 5 = 0

2p + 2 = 0

2p = -2

p = -2/2

p = -1

Cross – check :

By applying p = -1 in the given equation, we get

(2p – 1)/(2p + 5) = (p – 1)/(p + 3)

(2(-1) – 1)/ (2(-1) + 5) = (-1 – 1)/(-1 + 3)

(-2 – 1)/(-2 + 5) = (-2)/(2)

(-3)/3 = -1

-1 = -1

So, the solution of p is -1.

Problem 9 :

Solution :

By cross multiplication, we get

 (3x – 6)(2) = (3)(2 – x)

6x – 12 = 6 – 3x

6x – 12 – 6 + 3x = 0

9x – 18 = 0

9x = 18

x = 2

Cross – check :

By applying x = 2 in the given equation, we get

(3x – 6)/(2 – x) = 3/2

(3(2) – 6)/(2 – 2) = 3/2

(6 – 6)/0 = 3/2

0/0 = 3/2 (false)

So, there is no solution for x.

Problem 10 :

Solution :

By cross multiplication, we get

(y + 3)[y(y + 2)] = (3y + 4)(y + 2)

(y + 3)(y2 + 2y) = 3y2 + 6y + 4y + 8

y3 + 2y2 + 3y2 + 6y =  3y2 + 10y + 8

y3 + 2y2 + 3y2 + 6y - 3y2 - 10y – 8 = 0

y3 + 2y2 - 4y – 8 = 0

By grouping, 

(y3 + 2y2) + (-4y – 8) = 0

By taking the common factor, we get

y2(y + 2) – 4(y + 2) = 0

(y2 – 4)(y + 2) = 0

   (y2 – 22)(y + 2) = 0

(y + 2)(y – 2)(y – 2) = 0

y + 2 = 0 and y – 2 = 0

y = -2 and y = 2

Cross-check :

Step 1 :

By applying y = -2 in the given equation, we get

(y + 3)/(y + 2) = (2/y) + (1/(y + 2))

(-2·· + 3)/(-2 + 2) = (2/-2) + 1/(-2 + 2)

1/0 = 1/-1 + 1/0

= infinity

Since it has an infinity, we can’t take the y = -2.

Step 2 :

By applying y = 2 in the given equation, we get

(y + 3)/(y + 2) = (2/y) + (1/(y + 2))

(2 + 3)/(2 + 2) = (2/2) + 1/(2 + 2)

5/4 = 1/1 + 1/4

By taking the least common multiple, we get

5/4 = 4/4 + 1/4

5/4 = 5/4 (true)

So, the solution of y is 2.

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