Solve : 2x + 3y + 5 = 0 3x – 2y – 12 =0 (Use either substitution, elimination or cross-multiplication method)
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""" ❤️ Answer ❤️ """
By cross multiplication method
x=
(a
1
b
2
−a
2
b
1
)
(b
1
c
2
−b
2
c
1
)
&
y=
(a
1
b
2
−a
2
b
1
)
(c
1
a
2
−c
2
a
1
)
Where
a
1
b
1
c
1
0
&
a
2
b
2
c
2
0
are 2 lines
⇒ x=
2(−2)−(3×3)
3(−6)−(−2)(−17)
=
−13
−52
= 4
y=
2(−2)−(3×3)
(−17)3−(−6)2
=
−13
−39
= 3
⇒
x=4&y=3 .
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