Answer: Correct option is C (9, 6) x+y=15 —– (1) x−y=3 —– (2) First use formula for cross multiplication method: b 1 c 2 −b 2 c 1 x = c 1 a 2 −a 1 c 2 y = a 2 b 2 −a 2 b 1 −1 So, from equation (1) and (2) we can write the value of a, b and c. 1×3−(−1)×15 x = 15×1−1×3 y = 1×(−1)−1×1 −1 3+15 x = 15−3 y = −1−1 −1 18 x = 12 y = −2 −1 18 x = −2 −1 x= −2 −18 =9 12 y = −2 −1 y= −2 −12 y=6 The solution is (9, 6) Reply
solution=(x+y)(x-y) using identities (a+b)(a-b)=a^2-b^2 x^2–y^2 this is ans (Aliter) (x+y)(x-y) x(x-y)+y(x-y) x^2 -xy+xy-y^2 (-xy and +xy is cancelled due to both sign is + and – ,then this sign is put the same value so then ot is cancelled) x^2–y^2 (-xy+xy is cancelled then ans will get ) i hope it will help you Reply
Answer:
Correct option is
C
(9, 6)
x+y=15 —– (1)
x−y=3 —– (2)
First use formula for cross multiplication method:
b
1
c
2
−b
2
c
1
x
=
c
1
a
2
−a
1
c
2
y
=
a
2
b
2
−a
2
b
1
−1
So, from equation (1) and (2) we can write the value of a, b and c.
1×3−(−1)×15
x
=
15×1−1×3
y
=
1×(−1)−1×1
−1
3+15
x
=
15−3
y
=
−1−1
−1
18
x
=
12
y
=
−2
−1
18
x
=
−2
−1
x=
−2
−18
=9
12
y
=
−2
−1
y=
−2
−12
y=6
The solution is (9, 6)
solution=(x+y)(x-y)
using identities (a+b)(a-b)=a^2-b^2
x^2–y^2
this is ans
(Aliter)
(x+y)(x-y)
x(x-y)+y(x-y)
x^2 -xy+xy-y^2 (-xy and +xy is cancelled due to both sign is + and – ,then this sign is put the same value so then ot is cancelled)
x^2–y^2 (-xy+xy is cancelled then ans will get )
i hope it will help you