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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

4. Applications of Linear Systems

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Exercise 15 Page 391

Are there any variables already isolated?

(-3,-2)

Practice makes perfect

Since the y in the second equation is already isolated, the easiest way to solve the system is to use the Substitution Method. Let's substitute y=3x+7 into the second equation.

2x+7y=-20 & (I) y=3x+7 & (II)

(I):y= 3x+7

2x+7( 3x+7)=-20 y=3x+7

(I):Distribute 7

2x+21x+49=-20 y=3x+7

(I):Add terms

23x+49=-20 y=3x+7

(I):LHS-49=RHS-49

23x=-69 y=3x+7

(I):.LHS /23.=.RHS /23.

x=-3 y=3x+7

Now, that we have found a solution for x, we can substitute that into our second equation to solve for y.

x=-3 y=3x+7

(II): x= -3

x=-3 y=3( -3)+7

(II):Multiply

x=-3 y=-9+7

(II):Add terms

x=-3 y=-2

The solution to this system of equations is (-3,-2).

Subchapter links

Lesson Check p.390

Standardized Test Prep p.392

Mixed Review p.392