Core Connections Algebra 1, 2013
CC
Core Connections Algebra 1, 2013 View details
2. Section 8.2
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Exercise 62 Page 390

Practice makes perfect
a 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.

6 x-2 y=10 & (I) 3 x- y=2 & (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 2, x- and y-terms will have opposite coefficients. 6 x-2 y=10 2(3 x- y)=2(2) ⇔ 6x - 2y=10 6x - 2y=4 We can see that the x- and y-terms will eliminate each other if we subtract (I) from (II).

6x-2y=10 6x-2y=4
6x-2y=10 6x-2y-( 6x-2y)=4- 10
6x-2y=10 6x-2y-6x+2y=4-10
6x-2y=10 0=- 6

Solving this system of equations resulted in a contradiction; 0 can never be equal to - 6. Therefore, the lines are parallel and do not have a point of intersection.

b In this system of equations at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using substitution, there are three steps.
  1. Isolate a variable in one of the equations.
  2. Substitute the expression for that variable into the other equation and solve.
  3. Substitute this solution into one of the equations and solve for the value of the other variable. For this exercise y is already isolated in one equation, so we can skip straight to solving!

    x-3y=1 & (I) y=16-2x & (II)
    x-3( 16-2x)=1 y=16-2x
    â–¼
    (I): Solve for x
    x-48+6x=1 y=16-2x
    7x-48=1 y=16-2x
    7x=49 y=16-2x
    x=7 y=16-2x

    Great! Now, to find the value of y we need to substitute x=7 into either one of the equations in the given system. Let's use the second equation.

    x=7 y=16-2x
    x=7 y=16-2( 7)
    â–¼
    (II): Solve for y
    x=7 y=16-14
    x=7 y=2

    The solution, or point of intersection, to this system of equations is the point (7,2). To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

    x-3y=1 & (I) y=16-2x & (II)

    (I), (II): x= 7, y= 2

    7-3( 2)? =1 2? =16-2( 7)

    (I), (II): Multiply

    7-6? =1 2? =16-14

    (I), (II): Subtract terms

    1=1 2=2

    Because both equations are true statements, we know that our solution is correct.