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Big Ideas Math: Modeling Real Life, Grade 8

BI

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

Chapter Review

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Exercise 23 Page 229

The Elimination Method can be used to solve a system of linear equations if either of the variable terms would cancel out the corresponding variable term in the other equation when added together.

Solution: (- 5,0)
Explanation: See solution.

Practice makes perfect

Let's consider the given system of equations. x+2 y=- 5 & (I) x-2 y=- 5 & (II) Since the y-variable has opposite coefficients in the equations, we will solve the system of linear equations using the Elimination Method.

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

(II): Add (I)

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

▼

(II):Solve for x

(II): Remove parentheses

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

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

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

(II): Add and subtract terms

x+2y=- 5 2x=- 10

(II): .LHS /2.=.RHS /2.

x+2y=- 5 2x2= - 102

CancelCommonFac

(II): Cancel out common factors

x+2y=- 5 2x2= - 102

(II): Simplify quotient

x+2y=- 5 x= - 102

(II): Calculate quotient

x+2y=- 5 x=- 5

Now we can solve for y by substituting the value of x into Equation (I) and simplifying.

x+2y=- 5 x=- 5

(I): x= - 5

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

▼

(I):Solve for y

(I): LHS+5=RHS+5

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

(I): Add terms

2y=0 x=- 5

(I): .LHS /2.=.RHS /2.

2y2= 02 x=- 5

CancelCommonFac

(II): Cancel out common factors

2y2= 02 x=- 5

(II): Simplify quotient

y= 02 x=- 5

(II): 0/a=0

y=0 x=- 5

The solution, or point of intersection, of the system of equations is (-5,0).

Subchapter links

1 Solving Systems of Linear Equations by Graphing p.227

2 Solving Systems of Linear Equations by Substitution p.228

3 Solving Systems of Linear Equations by Elimination p.228

4 Solving Special Systems of Linear Equations p.229