McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
1. Graphing Quadratic Functions
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Exercise 89 Page 553

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.

(2,1)

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- or the y-terms must cancel each other out. 2 x+ y=5 & (I) 3 x-2 y=4 & (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 2 the y-terms will have opposite coefficients. 2(2 x+ y)=2(5) 3 x-2 y=4 ⇓ 4 x+ 2y=10 3 x- 2y=4We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).
4x+2y=10 3x-2y=4
4x+2y=10 3x-2y+( 4x+2y)=4+ 10
â–Ľ
(II):Solve for x
4x+2y=10 7x=14
4x+2y=10 x=2
Now we can solve for y by substituting the value of x into either equation and simplifying.
4x+2y=10 x=2
4( 2)+2y=10 x=2
â–Ľ
(I):Solve for y
8+2y=10 x=2
2y=2 x=2
y=1 x=2
The solution of the system of equations is (2,1).