Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 6.2
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Exercise 65 Page 273

Practice makes perfect
a Note that y is already solved for in the first equation, so it is convenient to use the Substitution Method to solve the system of equations.
y= 23x-4 & (I) 2x-3y=10 & (II)
y= 23x-4 2x-3( 23x-4)=10
â–Ľ
Solve for x
y= 23x-4 2x-3* 23x+12=10
y= 23x-4 2x- 63x+12=10
y= 23x-4 2x-2x+12=10
y= 23x-4 12≠ 10
When trying to solve the equation we reach a contradiction. This means the system of equation has no solution, which is only possible if the lines run parallel to each other.
b Let's solve the second equation for y by performing inverse operations until y is isolated.
2x-3y=10
-3y=- 2x + 10
3y=2x - 10
y=2/3x - 10/3
c The conclusion from Part A was that the lines run parallel. For this to be the case, two things must hold true.
  1. The lines must have the same slope.
  2. The lines must have different y-intercepts

Having solved the second line's equation for y, they are now both written in slope-intercept form y= mx+b. In this form, m is the line's slope. Comparing the equations, we see that this value is identical in our equations. y= 23x-4 y= 23x- 103 Therefore, the slope of each line does confirm our statement in Part A.