To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the

x -

terms or the

y -

terms must cancel each other out.

{4 x + 8 y = - 6 6 x + 12 y = - 9 (I) (II)

Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (I) by

3

and multiply (II) by

- 2,

the

x -

terms will be additive inverses.

{3 (4 x + 8 y) = 3 (- 6) - 2 (6 x + 12 y) = - 2 (- 9) ⇓ {12 x + 24 y = - 18 - 12 x - 24 y = 18

We can see that the

x -

terms will eliminate each other if we add (I) to (II).

{12 x + 24 y = - 18 - 12 x - 24 y = 18

SysEqnAdd

$(II):$ Add $(I)$

{12 x + 24 y = - 18 - 12 x - 24 y + (12 x + 24 y) = 18 + (- 18)

RemovePar

$(II):$ Remove parentheses

{12 x + 24 y = - 18 - 12 x - 24 y + 12 x + 24 y = 18 - 18

AddSubTerms

$(II):$ Add and subtract terms

{12 x + 24 y = - 18 0 = 0

Solving this system of equations resulted in an identity;

0

is always equal to itself. Therefore, it is a dependent system and the system has an infinite number of solutions, or intersection points. Let's rewrite the first equation in order to express the intersection points in terms of