Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 4.1
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Exercise 10 Page 171

Practice makes perfect
a Examining the system, we see that in the second equation, x is already isolated. Therefore, its most easily solved by using the Substitution Method.
2x+6y=10 x=8-3y The second equation is already isolated for the variable x. Therefore, we can substitute the expression for x in the first equation and solve for y.
2x+6y=10 & (I) x=8-3y & (II)
2( 8-3y)+6y=10 x=8-3y
16-6y+6y=10 x=8-3y
16≠10 x=8-3y
We have reached a contradiction as 16 does not equal 10. This means the system has not solutions.
b To graph a linear function, we need at least two points that fall on its graph. Let's rewrite the equations so that they arein slope-intercept form.
2x+6y=10 x=8-3y ⇔ y=- 13x+ 53 y=- 13x+ 83

Next, we will choose two x-values and calculate their corresponding y-values for each function. |c|c|c| [-1em] x & - 1/3x+5/3 & y [0.8em] [-1em] 2 & -1/3( 2)+5/3 & 5 [0.8em] [-1em] 5 & -1/3( 5)+5/3 & 0 [0.8em] |c|c|c| [-1em] x & - 1/3x+8/3 & y [0.8em] [-1em] 5 & -1/3( 5)+8/3 & 1 [0.8em] [-1em] 8 & -1/3( 8)+8/3 & 0 [0.8em] By plotting each set of points in a coordinate plane, we can draw their graphs.

c From the diagram and slope-intercept form, we see that the functions have the same slope. Since they also have different y-intercepts, they must be parallel lines.