Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
5. Systems With Three Variables
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Exercise 55 Page 173

If either of the variable terms would cancel out the corresponding variable term in the other equation, you can use the Elimination Method to solve the system.

See solution.

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. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. 4 x+8 y=-6 & (I) 6 x+12 y=-9 & (II) Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (I) by 3 and multiply (II) by -2, the x-terms will be additive inverses. 3(4 x+8 y)=3(-6) -2(6 x+12 y)=-2(-9) ⇓ 12x+24 y=-18 - 12x-24 y=18We can see that the x-terms will eliminate each other if we add (I) to (II).
12x+24y=-18 -12x-24y=18
12x+24y=-18 -12x-24y+( 12x+24y)=18+( -18)
12x+24y=-18 -12x-24y+12x+24y=18-18
12x+24y=-18 0=0
Solving this system of equations resulted in an identity; 0 is always equal to itself. Therefore, it is a dependent system and the system has an infinite number of solutions, or intersection points. Let's rewrite the first equation in order to express the intersection points in terms of x.
12x+24y=-18 0=0
24y=-12x-18 0=0
y=- 1224x- 1824 0=0
y=- 12x- 1824 0=0
y=- 12x- 34 0=0
Therefore, the infinite intersection points are all pairs (x,y) satisfying the equation above, or {(x,y)|y=- 12x - 34 }.