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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

3. Modeling With Quadratic Functions

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Exercise 36 Page 214

If a variable has the same or opposite coefficient in the equations, then use the Elimination Method.

(2,5)

Practice makes perfect

To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. Since here the y-variable has opposite coefficients, we can add the equations to eliminate y. x+ y=7 & (I) 5 x- y=5 & (II)Let's do it!

x+y=7 5x-y=5

(II): Add (I)

x+y=7 5x-y+( x+y)=5+ 7

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(II):Solve for x

(II): Remove parentheses

x+y=7 5x-y+x+y=5+7

(II): Add and subtract terms

x+y=7 6x=12

(II): .LHS /6.=.RHS /6.

x+y=7 x=2

Now we can find the value of y by substituting 2 for x into Equation (I), and simplifying.

x+y=7 x=2

(I): x= 2

2+y=7 x=2

(I): LHS-2=RHS-2

y=5 x=2

The solution to the system, which is the point of intersection of the lines, is (2,5).

Subchapter links

Lesson Check p.212

Practice and Problem-Solving Exercises p.212-214

Standardized Test Prep p.214

Mixed Review p.214