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)
4y^{2} – y = 15y
4y^{2 }- y - 15y = 0
4y^{2 }- 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
c^{2} – 5c = c – 5
c^{2} – 5c – c + 5 = 0
c^{2} – 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)
2p^{2} + 6p – p – 3 = 2p^{2} + 5p – 2p – 5
2p^{2} + 5p – 3 = 2p^{2} + 3p – 5
2p^{2} + 5p – 3 - 2p^{2} – 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)(y^{2} + 2y) = 3y^{2} + 6y + 4y + 8
y^{3} + 2y^{2} + 3y^{2} + 6y = 3y^{2} + 10y + 8
y^{3} + 2y^{2} + 3y^{2} + 6y - 3y^{2} - 10y – 8 = 0
y^{3} + 2y^{2} - 4y – 8 = 0
By grouping,
(y^{3} + 2y^{2}) + (-4y – 8) = 0
By taking the common factor, we get
y^{2}(y + 2) – 4(y + 2) = 0
(y^{2} – 4)(y + 2) = 0
(y^{2} – 2^{2})(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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