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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 28 Page 217

Are there any variables already isolated?

Solution: (0,1)
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.

y+5x=1 & (I) 5y-x=5 & (II)

(I): LHS-5x=RHS-5x

y+5x-5x=1-5x 5y-x=5

(I): Subtract terms

y=1-5x 5y-x=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=1-5x into the second equation.

y=1-5x 5y-x=5

(II): y= 1-5x

y=1-5x 5( 1-5x)-x=5

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(II):Solve for x

(II): Distribute 5

y=1-5x 5(1)-5(5x)-x=5

(II): Identity Property of Multiplication

y=1-5x 5-5(5x)-x=5

(II): Multiply

y=1-5x 5-25x-x=5

(II): LHS-5=RHS-5

y=1-5x 5-25x-x-5=5-5

(II): Subtract terms

y=1-5x -26x=0

(II): .LHS /-26.=.RHS /-26.

y=1-5x -26x/-26=0/-26

MovePartNumRight

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

y=1-5x -26/-26 * x=0/-26

(II): - a/- b=a/b

y=1-5x 26/26 * x=0/-26

(II): a/a=1

y=1-5x 1 * x=0/-26

(II): Identity Property of Multiplication

y=1-5x x=0/-26

MoveNegDenomToFrac

(II): Put minus sign in front of fraction

y=1-5x x=-0/26

(II): 0/a=0

y=1-5x x=0

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

y=1-5x x=0

(I): x= 0

y=1-5( 0) x=0

▼

(I):Solve for y

(I): Zero Property of Multiplication

y=1-0 x=0

(I): Subtract terms

y=1 x=0

The solution, or point of intersection, of the system of equations is (0,1).

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