Use elimination method to find all possible solutions of the following
pair of linear equations : ax+by-a+b=0 and bx-ay-a-b=0
Answers
Answer:
haa use elimination method friend
Answer:
The solution of the equations is x=1 and y=-1.
Step-by-step explanation:
Given : Equations ax+by-a+b=0ax+by−a+b=0 and bx-ay-a-b=0bx−ay−a−b=0
To find : Solve the equation by elimination method?
Solution :
Let ax+by=a-bax+by=a−b ....(1)
and bx-ay=a+bbx−ay=a+b ....(2)
To eliminate y multiply (1) by a and (2) by b,
a^2x+aby=a^2-aba
2
x+aby=a
2
−ab ....(3)
b^2x-aby=ab+b^2b
2
x−aby=ab+b
2
......(4)
Add (3) and (4),
a^2x+aby+b^2x-aby=a^2-ab+ab+b^2a
2
x+aby+b
2
x−aby=a
2
−ab+ab+b
2
x(a^2+b^2)=a^2+b^2x(a
2
+b
2
)=a
2
+b
2
x=\frac{a^2+b^2}{a^2+b^2}x=
a
2
+b
2
a
2
+b
2
x=1x=1
In similar way,
To eliminate x multiply (1) by b and (2) by a,
abx+b^2y=ab-b^2abx+b
2
y=ab−b
2
....(5)
abx-a^2y=a^2+ababx−a
2
y=a
2
+ab ......(6)
Subtract (5) and (6),
abx+b^2y-abx+a^2y=ab-b^2-a^2-ababx+b
2
y−abx+a
2
y=ab−b
2
−a
2
−ab
(b^2+a^2)y=-(b^2+a^2)(b
2
+a
2
)y=−(b
2
+a
2
)
y=-\frac{(b^2+a^2)}{(b^2+a^2)}y=−
(b
2
+a
2
)
(b
2
+a
2
)
y=-1y=−1
Therefore, The solution of the equations