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Pearson Algebra 1 Common Core, 2015

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Pearson Algebra 1 Common Core, 2015 View details

3. Solving Systems Using Elimination

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Exercise 1 Page 381

If either of the variable terms would cancel out the corresponding variable term in the other equation, you can use the Elimination Method to solve the system.

$(2, 3)$

Practice makes perfect

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) .

Subchapter links

Lesson Check p.381

Practice and Problem-Solving Exercises p.382-384

Standardized Test Prep p.384

Mixed Review p.384