Welcome to the Chapter 11 Test Corrections Clinic! Please click through the slides to find the corrections you need to make to your test.

60 hats is the break-even point Question 1 Since the IV (the variable we can change is the number of hats, x= the number of hats, y=the total cost We know just to MAKE the hats, it will cost the band $240 to set up, and $8 per hat. y = 240 + 8x y = 240 + 8x Again, since the number of hats varies, x= the number of hats, y=the total earned y = 12x y = 12x We know that the band is selling hats for $12 per hat, so that’s the money they’ll make. Remember, the break-even point is where you have no profit – the money you’ve earned is the same as your money spent. Now, remember, profit is what you take home after you’ve paid your bills. It’s the money earned – costs. 60 hats is the break-even point 240+8(0)=$240 12(0)=$0 0 - 240= -$240 Here, you could’ve written anything like: I looked to see where my cost = income. I looked to see where my profit = $0. From here, we’re just going to plug in our number of hats for x. 240+8(20)=$400 12(20)=$240 400-240= -$160 240+8(40)=$560 12(40)=$480 480-560= -$80 240+8(50)=$640 12(50)=$600 600-640= -$40 240+8(60)=$720 12(60)=$720 720-720= $0 Remember what we said before, x=number of hats, y=total cost/earned, so your break-even (or intersection point) is 240+8(70)=$800 12(70)=$840 840-800= $40 (60, 720) 240+8(80)=$880 960-880= $80 12(80)=$960

Question 2 In order to graph these, we’ll have to draw our axes. Since you have a negative y-intercept, you know you’ll have to use a 4-quadrant graph. Now that you’ve drawn your axes, you just need to plot your lines. Remember: Begin with your y-intercept. Move with your slope. Because the mx part of this equation is –x, the slope is actually -1. It looks like they intersect at (2, -1), so we’ll say that’s our solution!

Question 3 These lines are perpendicular, because their slopes are negative reciprocals. (Both of these facts needed to be stated for full credit on this question. Remember, in order to determine whether lines are parallel, perpendicular, coincidental, or none of the above (just intersecting), we need to look at their slopes.

Question 4&5 (-3, ) -8 y = 2x - 7 ( ) NO SOLUTIONS! y = 6x+10 y For number 4, both of your y’s are isolated variables, so you could substitute this either way. For the case of this example, we’ll do it like this. y = 6x+10 Now we’ve performed a substitution, so we have the following equation: y = -4x - 20 6x + 10 = -4x – 20 -6x -6x 10 = -10x – 20 +20 +20 30 = -10x -10 -10 -3 = x We still have to solve for y, though. This means we’ll take our new x value of -3, and plug it into either equation to solve for y. For the case of this example, let’s use the first equation. (-3, ) -8 y = 6x + 10 y = 6(-3) + 10 y = -18 + 10 y = -8 y = 2x - 7 For number 5, you have an isolated variable of y, so you can take everything opposite of it, and substitute it into the other equation. Rewriting this equation makes it look something like this: y = 2x - 7 -6x + 3(2x – 7) = 21 distribute your parentheses -6x + 6x – 21 = 21 oh, wait… -6x + 3 ( ) y = 21 Since you have -6x and 6x, you know that those combined = 0, which leaves you with an equation of -21=21. When you see this, you should know this means… NO SOLUTIONS!

Question 6 x + 2y = 4 -2y -2y x = 4 – 2y (4 – 2y) + 4y = 8 Let’s start by getting one of these equations to have an isolated variable! Let’s take a look at that second equation… x + 2y = 4 -2y -2y x = 4 – 2y Now that we have an isolated variable, we can take everything opposite of it… …and substitute it into the other equation. It gives you an equation like this… (4 – 2y) + 4y = 8 4 – 2y + 4y = 8 4 + 2y = 8 -4 -4 2y = 4 y = 2 Right away, we can see that if we have a y coordinate, we’re going to have an x coordinate, which means we have…

Question 7 Remember, the slopes of perpendicular lines are negative reciprocals. Let’s check out the slopes of these linear equations…

Question 8

Question 9 Remember, the slopes of parallel lines are the same. Let’s check out the slopes of these linear equations… Well, that was easy. 