ax + by = c , bx + ay =1 + c
solve for x and y for the following pair of equation.
Answers
Answer:
Please check out the attachment given above ⬆
Step-by-step explanation:
Given :-
ax + by = c ,
bx + ay =1 + c
To find :-
Solve for x and y for the following pair of equation?
Solution :-
Given pair of linear equations in two variables are
ax + by = c ---------(1)
and
bx + ay =1 + c--------(2)
On adding (1)&(2) then
ax+by = c
bx+ay = 1+c
(+)
________________
(a+b)x +(a+b)y = 1+2c
________________
=> (a+b)x +(a+b)y = 1+2c
=> (a+b)(x+y) = 1+2c
=>x+y = (1+2c)/(a+b) ---------(3)
On Subtracting (2) from (1) then
ax+by = c
bx+ay = 1+c
(-)
________________
(a-b)x +(b-a)y = (c-1-c)
________________
=>(a-b)x +(b-a)y = (c-1-c)
=> (a-b)x -(a-b)y = -1
=> (a-b)(x-y) = -1
=> x-y = -1/(a-b) ----------(4)
On adding (3)&(4) then
x+y = (1+2c)/(a+b)
x-y = -1/(a-b)
(+)
______________________
2x+0 = [(1+2c)/(a+b)]+[-1/(a-b)]
_______________________
=> 2x = [(1+2c)/(a+b)]+[-1/(a-b)]
=> 2x = [(1+2c)(a-b)+(-1)(a+b)]/(a+b)(a-b)
=> 2x = [(a-b+2ac-2bc -a-b)/(a²-b²)]
=> 2x = (2ac-2bc-2b)/(a²-b²)
=> 2x = 2(ac-bc-b)/(a²-b²)
=> x = (ac-bc-b)/(a²-b²) -----------(5)
On Substituting the value of x in (4)
x-y = -1/(a-b)
=>[(ac-bc-b)/(a²-b²)]- y= -1/(a-b)
=> y = [(ac-bc-b)/(a²-b²)]+[1/(a-b)]
=>y = [(ac-bc-b+1(a+b)]/(a²-b²)
=> y = (ac-bc-b+a+b)/(a²-b²)
=> y = (ac-bc+a)/(a²-b²) ---------(6)
Answer:-
The value of x = (ac-bc-b)/(a²-b²)
The value of y = (ac-bc+a)/(a²-b²)
Check :-
We have
x = (ac-bc-b)/(a²-b²)
y = (ac-bc+a)/(a²-b²)
LHS of (1)
=> ax+by
=>a((ac-bc-b)/(a²-b²))+b((ac-bc+a)/(a²-b²)
=> [a(ac-bc-b)+b(ac-bc+a)]/(a²-b²)
=> (a²c-abc-ab+abc-b²c+ab)/(a²-b²)
=> (a²c-b²c)/(a²-b²)
=> c(a²-b²)/(a²-b²)
=> c
=> RHS
LHS = RHS
LHS of (2)
bx + ay
=> b((ac-bc-b)/(a²-b²))+a((ac-bc+a)/(a²-b²)
=> [b(ac-bc-b)+a(ac-bc+a)]/(a²-b²)
=> (abc-b²c-b²+a²c-abc+a²)/(a²-b²)
=> (a²c+a²-b²c-b²)/(a²-b²)
=>[a²(c+1)-b²(c+1)]/(a²-b²)
=> (c+1)(a²-b²)/(a²-b²)
=> c+1
=> RHS
=> LHS = RHS
Verified the given relations in the given problem