solve the following pairs of equations by reducing them to a pair of linear equations: 4/x+3y=14 3/x-4y=24
Answers
Step-by-step explanation:
Given:-
4/x+3y=14 ;
3/x-4y=24
To find:-
Solve the following pairs of equations by reducing them to a pair of linear equations:
4/x+3y=14 and 3/x-4y=24?
Solution:-
Given equations are
4/x+3y=14 ---------(1)
It can be written as
4(1/x) +3y = 14
Put 1/x = a then
4a + 3y = 14
On multiplying with 3 both sides
12 a + 9 y = 42--------(2)
and 3/x-4y=24-----(3)
It can be written as
3(1/x) - 4y = 24
Put 1/x = a then
3a -4y = 24
On multiplying with 4 both sides
12 a - 16 y = 96 --------(4)
On Subtracting equation (2) from equation (4)
12 a - 16 y = 96
12 a + 9 y = 42
(-)
____________
0 + -25 y = 54
____________
=> -25 y = 54
=> y = 54/-25
The value of y = -54/25
On Substituting the value of y in (2)
12 a + 9 y = 42
=> 12 a +9(-54/25) = 42
=> 12 a + (-486/25) = 42
=> 12 a = 42+(486/25)
=> 12 a = (1050+486)/25
=> 12 a = 1536/25
=> a = 1536/(12×25)
=> a = 128/15
The value of a = 128/25
we have
a = 1/x
=> x = 1/a
=> x = 1/(128/25)
=> x = 25/128
The values of x and y are 25/128 and -54/25 respectively.
Answer:-
The solution for the given pair of linear equations in two variables is
( 25/128 , -54/25 )
Check:-
The values of x and y are 25/128 and -54/25 respectively.
On Substituting these values in the equation
4/x+3y
=> 4/(25/128)+3(-54/25)
=>[ (4×128)/25] +[(3×-54)/25]
=> (512/25) -(162/25)
=> (512-162)/25
=> 350/25
=> 14
LHS = RHS is true for x = 25/128 and y = -54/25
and
The values of x and y are 25/128 and -54/25 respectively.
3/x-4y
On Substituting these values in the equation
3/(25/128)-4(-54/25)
=>[ (3×128)/25 ] + [(4×54)/25]
=> (384/25) +(216/25)
=> (384+216)/25
=>600/25
=>24
LHS = RHS is true for x = 25/128 and y = -54/25
Verified the given relations