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 34 Page 392

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

(4,2.5)

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. This means that either the x-terms or the y-terms must cancel each other out. 5 x+8 y=40 & (I) 3 x-10 y=-13 & (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 Equation (I) by 5 and multiply Equation (II) by 4, the y-terms will have opposite coefficients. 5(5 x+8 y)=5(40) 4(3 x-10 y)=4(-13) ⇒ 25 x+ 40y=200 12 x- 40y=-52 We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).
25x+40y=200 12x-40y=-52
25x+40y=200 12x-40y+ 25x+40y=-52+ 200
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(II):Solve for x
25x+40y=200 37x=148
25x+40y=200 x=4
Now, we can now solve for y by substituting the value of x into either equation and simplifying.
25x+40y=200 x=4
25( 4)+40y=200 x=4
â–Ľ
(I):Solve for y
100+40y=200 x=4
40y=100 x=4
y=2.5 x=4
The solution to this system of equations is (4,2.5).