Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
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Exercise 9 Page 411

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.

(-44/19,29/19)

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. 3 x+2 y=-10 & (I) 2 x-5 y=3 & (II) In its current state, this will not happen. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply Equation (I) by 2 and multiply Equation (II) by - 3, the x-terms will have opposite coefficients. 2(3 x+2 y)=2(-10) - 3(2 x-5 y)=- 3(3) ⇒ 6 x+4 y=-20 - 6 x+15 y=- 9 With this, we can see that the x-terms will eliminate each other if we add Equation (I) to Equation (II).
6x+4y=-20 & (I) - 6x+15y=- 9 & (II)
6x+4y=-20 - 6x+15y+ 6x+4y=- 9+( - 20)
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(II): Solve for y
6x+4y=-20 - 6x+15y+6x+4y=- 9- 20
6x+4y=-20 19y=- 29
6x+4y=-20 y= - 2919
6x+4y=-20 y=- 2919
With this, we can now solve for x by substituting the value of y into either equation and simplifying.
6x+4y=-20 y=- 2919
6x+4( - 2919)=-20 y=- 2919
6x-4( 2919)=-20 y=- 2919
6x- 11619=-20 y=- 2919
6x=-20+ 11619 y=- 2919
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Evaluate right-hand side
6x=- 19 * 2019+ 11619 y=- 2919
6x=- 38019+ 11619 y=- 2919
6x=- 26419 y=- 2919
x=- 26419* 16 y=- 2919
x=- 44*619* 16 y=- 2919
x=- 4419 y=- 2919
The solution, or intersection point, of the system of equations is (- 4419, 2919).