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Big Ideas Math: Modeling Real Life, Grade 8

BI

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

3. Solving Systems of Linear Equations by Elimination

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Exercise 11 Page 214

Are there any variables isolated?

Solution: (-1,-7)
Method: See solution.

Practice makes perfect

Since the y in the first equation is already isolated, the easiest way to solve the system of equations is to use the Substitution Method. Let's substitute y=6x-1 into the second equation.

y=6x-1 & (I) y=3x-4 & (II)

(II):y= 6x-1

y=6x-1 6x-1=3x-4

▼

Solve for x

(II):LHS-3x=RHS-3x

y=6x-1 6x-1-3x=3x-4-3x

(II):Subtract terms

y=6x-1 3x-1=-4

(II):LHS+1=RHS+1

y=6x-1 3x-1+1=-4+1

(II):Add terms

y=6x-1 3x=-3

(II): .LHS /3.=.RHS /3.

y=6x-1 3x/3=-3/3

MovePartNumRight

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

y=6x-1 3/3 * x=-3/3

MoveNegNumToFrac

(II): Put minus sign in front of fraction

y=6x-1 3/3 * x=-3/3

(II): a/a=1

y=6x-1 1 * x=-1

(II): Identity Property of Multiplication

y=6x-1 x=-1

Great! Now, to find the value of y, we need to substitute x=-1 into either one of the equations in the given system. Let's use the first equation.

y=6x-1 x=-1

(I):x= -1

y=6( -1)-1 x=-1

(I): Multiplication Property of -1

y=-6-1 x=-1

(I): Subtract terms

y=-7 x=-1

The solution, or point of intersection, to this system of equations is the point (-1,-7).

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