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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

3. Solving Systems Using Elimination

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Exercise 33 Page 383

Are there any variables already isolated?

(2,0), see solution.

Practice makes perfect

Since the variable y in the first equation has a coefficient equal to 1, we can isolate it using the Properties of Equality.

2x+y=4 6x+7y=12

(I): LHS-2x=RHS-2x

y=4-2x 6x+7y=12

Now that y is isolated, the easiest way to solve the system is to use the Substitution Method.

y=4-2x 6x+7y=12

(II): y= 4-2x

y=4-2x 6x+7( 4-2x)=12

(II):Distribute 7

y=4-2x 6x+28-14x=12

(II):Subtract term

y=4-2x -8x+28=12

(II):LHS-28=RHS-28

y=4-2x -8x=-16

(II):.LHS /-8.=.RHS /-8.

y=4-2x x=2

Now that we have found a solution for x, we can substitute that into our first equation to solve for y.

y=4-2x x=2

(I):x= 2

y=4-2( 2) x=2

(I):Multiply

y=4-4 x=2

(I):Subtract term

y=0 x=2

The solution to the system — the point of intersection — is (2,0).

Subchapter links

Lesson Check p.381

Practice and Problem-Solving Exercises p.382-384

Standardized Test Prep p.384

Mixed Review p.384