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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 34 Page 383

Are there any variables already isolated?

(- 1,4), see solution.

Practice makes perfect

Since no variable term is isolated or has a coefficient equal to 1, it is not convenient to use the Substitution Method. Therefore, we will use the Elimination Method. In order to eliminate the x-terms, let's multiply the first equation by 4 and multiply the second equation by 3.

3x+2y=5 & (I) 4x+5y=16 & (II)

(I):LHS * 4=RHS* 4

12x+8y=20 4x+5y=16

(II):LHS * 3=RHS* 3

12x+8y=20 12x+15y=48

Now we can continue by subtracting Equation (I) from Equation (II).

12x+8y=20 12x+15y=48

(II): Subtract (I)

12x+8y=20 12x+15y-( 12x+8y)=48- 20

(II): Distribute - 1

12x+8y=20 12x+15y-12x-8y=48-20

(II): Subtract term

12x+8y=20 7y=28

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

12x+8y=20 y=4

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

12x+8y=20 & (I) y=4 & (II)

(I): y= 4

12x+8 ( 4)=20 y=4

(I): Multiply

12x+32=20 y=4

(I): LHS-32=RHS-32

12x=- 12 y=4

(I): .LHS /12.=.RHS /12.

x=- 1 y=4

The solution to the system — the point of intersection — is (-1,4).

Subchapter links

Lesson Check p.381

Practice and Problem-Solving Exercises p.382-384

Standardized Test Prep p.384

Mixed Review p.384