To solve linear equations with variables in both numerator and denominator, we have to use inverse operations.
To remove the variable at the denominator, multiply the same on both sides. Then using inverse operations, we can isolate the variable and solve.
Solve the
equation:
Problem 1 :
(3x + 5) / (2x + 7) = 5
Solution :
(3x + 5) / (2x + 7) = 5
Multiply by (2x + 7) on both sides.
3x + 5 = 5(2x + 7)
3x + 5 = 10x + 35
10x - 3x = -35 + 5
7x = -30
x = -30/7
Problem 2 :
(2y + 5) / (y + 4) = 1
Solution :
(2y + 5) / (y + 4) = 1
Multiply by (y + 4) on both sides.
2y + 5 = y + 4
2y - y = 4 - 5
y = -1
Problem 3 :
(5z - 3)/2z = 8/9
Solution :
(5z - 3) / 2z = 8/9
9(5z - 3) = 8(2z)
45z - 27 = 16z
45z - 16z = 27
29z = 27
z = 27/29
Problem 4 :
(1 - 9y)/(19 - 3y) = 5/8
Solution :
(1 - 9y) / (19 - 3y) = 5/8
8(1 - 9y) = 5(19 - 3y)
8 - 72y = 95 - 15y
-72y + 15y = 95 - 8
-57y = 87
y = - 87/57
y = - 29/19
Problem 5 :
(0.4z - 3) / (1.5z + 9) = -7/5
Solution :
(0.4z - 3) / (1.5z + 9) = -7/5
5(0.4z - 3) = -7(1.5z + 9)
2z - 15 = -10.5z - 63
2z + 10.5z = - 63 + 15
12.5z = - 48
z = -48/12.5
z = -3.84
Problem 6 :
(5y - 3) / (2y + 1) = 2/5
Solution :
(5y - 3) / (2y + 1) = 2/5
5(5y - 3) = 2(2y + 1)
25y - 15 = 4y + 2
25y - 4y = 2 + 15
21y = 17
y = 17/21
Problem 7 :
2x/(3x + 1) = -3
Solution :
2x/(3x + 1) = -3
2x = -3(3x + 1)
2x = -9x - 3
2x + 9x = -3
11x = -3
x = -3/11
Problem 8 :
[17(2 - x) - 5(x + 12)]/(1 - 7x) = 8
Solution :
17(2 - x) - 5(x + 12)/1 - 7x = 8
17(2 - x) - 5(x + 12) = 8(1 - 7x)
34 - 17x - 5x - 60 = 8 - 56x
-17x - 5x + 56x = -34 + 60 + 8
34x = 34
x = 34/34
x = 1
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM