McGraw Hill Glencoe Algebra 1, 2017
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McGraw Hill Glencoe Algebra 1, 2017 View details
4. Elimination Using Multiplication
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Exercise 1 Page 363

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.

(3,2)

Practice makes perfect
To solve a system of 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. 2 x- y=4 & (I) 7 x+3 y=27 & (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 3, the y-terms will have opposite coefficients. 3(2 x- y)=3(4) 7 x+3 y=27 ⇒ 6 x- 3y=12 7 x+ 3y=27We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).
6x-3y=12 7x+3y=27
6x-3y=12 7x+3y+( 6x-3y)=27+ 12
â–Ľ
(II):Solve for x
6x-3y=12 7x+3y+6x-3y=27+12
6x-3y=12 13x=39
6x-3y=12 x=3
Now we can now solve for y by substituting the value of x into either equation and simplifying. Let's substitute 3 for x in Equation (I) and solve for y.
6x-3y=12 x=3
6( 3)-3y=12 x=3
â–Ľ
(I):Solve for y
18-3y=12 x=3
- 3y=- 6 x=3
y=2 x=3
The solution, or intersection point, of the system of equations is (3,2).