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Big Ideas Math: Modeling Real Life, Grade 8

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Big Ideas Math: Modeling Real Life, Grade 8 View details

3. Solving Systems of Linear Equations by Elimination

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Exercise 12 Page 214

Are there any variables already isolated?

Solution: (1,1)
Method: 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.

3x=y+2 3x+2y=5

(I): LHS-2=RHS-2

3x-2=y+2-2 3x+2y=5

(I): Subtract terms

3x-2=y 3x+2y=5

(I): Rearrange equation

y=3x-2 3x+2y=5

Now that y is isolated, the easiest way to solve the system of equations is to use the Substitution Method. Let's substitute y=3x-2 into the second equation.

y=3x-2 3x+2y=5

(II): y= 3x-2

y=3x-2 3x+2( 3x-2)=5

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(II):Solve for x

(II): Distribute 2

y=3x-2 3x+2(3x)-2(2)=5

(II): Multiply

y=3x-2 3x+6x-4=5

(II): LHS+4=RHS+4

y=3x-2 3x+6x-4+4=5+4

(II): Add terms

y=3x-2 9x=9

(II): .LHS /9.=.RHS /9.

y=3x-2 9x/9=9/9

MovePartNumRight

(II): a* b/c=a/c* b

y=3x-2 9/9 * x=9/9

(II): a/a=1

y=3x-2 1 * x=1

(II): Identity Property of Multiplication

y=3x-2 x=1

Now, we can solve for y by substituting the value of x into either equation and simplifying.

y=3x-2 x=1

(I): x= 1

y=3( 1)-2 x=1

▼

(I):Solve for y

(I): Identity Property of Multiplication

y=3-2 x=1

(I): Subtract terms

y=1 x=1

The solution, or point of intersection, of the system of equations is (1,1).

Subchapter links

Try It p.212-214

Self-Assessment for Concepts & Skills p.214

Self-Assessment for Problem Solving p.215

Review & Refresh p.216

Concepts, Skills, & Problem Solving p.216-218