Big Ideas Math Algebra 2, 2014
BI
Big Ideas Math Algebra 2, 2014 View details
3. Modeling with Linear Functions
Continue to next subchapter

Exercise 37 Page 28

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.

( 1617, 1517)

Practice makes perfect
Since none of the variables are isolated and the y-terms have opposite signs, we will use the Elimination Method. To solve a system of linear equations this way, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. 12 x+4 y=4 & (I) 2 x- y=1 & (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 (II) by 4, the y-terms will have opposite coefficients. 12 x+4 y=4 & (I) 4(2 x- y)=4*1 & (II) ⇒ 12 x+ 4y=4 & (I) 8 x- 4y=4 & (II) We can see that the y-terms will eliminate each other if we add (I) to (II).
12x+4y=4 8x-4y=4
12x+4y=4 8x-4y+( 12x+4y)=4+ 4
â–Ľ
Solve for x
12x+4y=4 8 12x=8
12x+4y=4 16+12x=8
12x+4y=4 172x=8
12x+4y=4 17x=16
12x+4y=4 x= 1617
Now we can solve for y by substituting the value of x into either equation and simplifying.
12x+4y=4 x= 1617
12( 1617)+4y=4 x= 1617
â–Ľ
(I):Solve for y
1634+4y=4 x= 1617
817+4y=4 x= 1617
4y=4- 817 x= 1617
4y= 6817- 817 x= 1617
4y= 6017 x= 1617
y= 1517 x= 1617
The solution, or intersection point, of the system of equations is ( 1617, 1517).