4x+6y =15, 3x-4/y = 7 in elimination method
Answers
Step-by-step explanation:
Solution
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Given equations are 4x+6y=15 and 3x−4y=7
By cross multiplication method, we know that, for a system of linear equations in x and y, of the form a
1
x+b
1
y+c
1
=0 and a
2
x+b
2
y+c
2
=0, we have:
b
1
c
2
−b
2
c
1
x
=
c
1
a
2
−c
2
a
1
y
=
a
1
b
2
−a
2
b
1
1
∴ x=
a
1
b
2
−a
2
b
1
b
1
c
2
−b
2
c
1
and y=
a
1
b
2
−a
2
b
1
c
1
a
2
−c
2
a
1
On comparing with a
1
x+b
1
y+c
1
=0 and a
2
x+b
2
y+c
2
=0 we have,
a
1
=4,b
1
=6,c
1
=−15 and a
2
=3,b
2
=−4,c
2
=−7
∴ x=
a
1
b
2
−a
2
b
1
b
1
c
2
−b
2
c
1
and y=
a
1
b
2
−a
2
b
1
c
1
a
2
−c
2
a
1
⇒x=
4×(−4)−3×6
6×(−7)−(−4)×(−15)
and y=
4×(−4)−3×6
−15×3−(−7)×4
⇒x=
−16−18
−42−60
and y=
−16−18
−45+28
⇒x=
−34
−102
and y=
−34
−17
⇒x=3 and y=
2
1
This is the required solution of the given system of equations using the cross multiplication method.
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