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. This means that either the

x

-terms or the

y

-terms must cancel each other out.

{3 x - 2 y = 0 4 x + 2 y = 14 (I) (II)

We can see that the

y -

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

{3 x - 2 y = 0 4 x + 2 y = 14

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

{3 x - 2 y = 0 4 x + 2 y + (3 x - 2 y) = 14 + 0

▼

(II):

Solve for

x

$(II):$ Remove parentheses

{3 x - 2 y = 0 4 x + 2 y + 3 x - 2 y = 14 + 0

$(II):$ Add and subtract terms

{3 x - 2 y = 0 7 x = 14

$(II):$ $LHS / 7 = RHS / 7$

{3 x - 2 y = 0 x = 2

We can now solve for

y

by substituting the value of

x

into Equation (I) and simplifying.

{3 x - 2 y = 0 x = 2

$(I):$ $x = 2$

{3 (2) - 2 y = 0 x = 2

▼

(I):

Solve for

y

$(I):$ Multiply

{6 - 2 y = 0 x = 2

$(I):$ $LHS - 6 = RHS - 6$

{- 2 y = - 6 x = 2

$(I):$ $LHS / (- 2) = RHS / (- 2)$

{y = 3 x = 2

The solution, or point of intersection, of the system of equations is

(2, 3) .

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