Are there more than one way to solve systems of equations by elimination?

1 Answer
Manikandan S.
Jan 14, 2015

There are more than one way to solve the system of equations

The most utilised methods are elimination and substitution methods. I prefer substitution than elimination.

Other methods like Cramer's rule and other matrix methods such as Gauss elimination, Gauss - Jacobi are available. These are pretty advanced and can solve any number of linear equations.

A comparison of substitution and elimination methods is given below.

Example

6x+4y=26x+4y=2---------->Eqn 1
x-2y=3x−2y=3---------->Eqn 2

Elimination method
Multiply Eqn 2 by '2' an add with Eqn 1.

6x+4y = 26x+4y=2
2x-4y = 62x−4y=6
______+
8x = 88x=8
x =1x=1

Substitute in one of the equations. Using Eqn 1 we have

6*1+4y = 26⋅1+4y=2
4y = 2-64y=2−6
y=-1y=−1

Hence the solution is x=1,y=-1x=1,y=−1

Substitution Method
From Eqn 2 we have

x=3+2yx=3+2y --> Eqn 3

Substitute in Eqn 1
6*(3+2y)+4y = 26⋅(3+2y)+4y=2
18+12y+4y=218+12y+4y=2
16y=2-1816y=2−18
16y = -1616y=−16
y=-1y=−1
Use in eqn 3
x = 3+2.-1x=3+2.−1
x=1x=1
So we get x=1,y=-1x=1,y=−1.

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