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

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

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Answer 1 · 2015-01-14T04:51:36

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

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

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

$6 x + 4 y = 2$ $6x+4y = 2$
$2 x - 4 y = 6$ $2x-4y = 6$
______+
$8 x = 8$ $8x = 8$
$x = 1$ $x =1$

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

$6 \cdot 1 + 4 y = 2$ $6*1+4y = 2$
$4 y = 2 - 6$ $4y = 2-6$
$y = - 1$ $y=-1$

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

Substitution Method
From Eqn 2 we have

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

Substitute in Eqn 1
$6 \cdot (3 + 2 y) + 4 y = 2$ $6*(3+2y)+4y = 2$
$18 + 12 y + 4 y = 2$ $18+12y+4y=2$
$16 y = 2 - 18$ $16y=2-18$
$16 y = - 16$ $16y = -16$
$y = - 1$ $y=-1$
Use in eqn 3
$x = 3 + 2 . - 1$ $x = 3+2.-1$
$x = 1$ $x=1$
So we get $x = 1, y = - 1$ $x=1,y=-1$ .

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

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