Hot X: Algebra Exposed

Transcription

1 Hot X: Algebra Exposed Solution Guide for Chapter 10 Here are the solutions for the Doing the Math exercises in Hot X: Algebra Exposed! DTM from p To see if the point is on the line, let s plug it in and see if we get a true statement! Plugging in (, 1), we get: x + y = 4 + ( 1) = 4? 4 = 4? Yep! So the point is indeed on the line. For part b, it s already in standard form, so nothing to do there! For part c, we just need to solve for y, so we can do that by subtracting x from both sides, and we get: y = x + 4. For part d, what s the slope? Well, what s the m? It s that negative sign being multiplied times the x; in other words, it s 1. The y-intercept is the b which in this case is 4, and the x-intercept is what x equals when y equals zero. To find that, we plug in 0 for y and solve for x: x + y = 4 x + 0 = 4 x = 4. And that s the x-intercept! For part e, we already know from the two intercepts that the points (0, 4) and (4, 0) are on this line. It looks like another easy point would be (2, 2), so let s plot those three points and draw a line through em. Done! Answer: a. Yes; b. x + y = 4; c. y = x + 4; d. slope = 1; y-intercept = 4;

2 x-intercept = 4; e. (graph below) 3. To see if the point is on the line, let s plug it in and see if we get a true statement! Plugging in 1 2, 1 we get: y = 2x 1 = 2 1 2? = = 1? Yep! So the point is indeed on the line! For part b, to put this in standard form, let s subtract y from both sides and we get: 0 = 2x y; in other words: 2x y = 0. For part c, we just solve for y, which is easier to do starting from the original y = 2x; we just divide both sides by, and we get: y = 2 x. For part d, we can see that the slope is 2 and the y-intercept is 0. What s the x-intercept? Well, what does x equal when y equals 0? Let s see: 0 = 2 x. Multiplying both sides by and dividing both sides by 2, those things disappear on the left and we end up with: 0 = x. So the x-intercept is 0.

3 For part e, we have two points to plot so far: 1 2, 1 and (0, 0). Because the y-intercept and x-intercept are the same (the origin), we need to find a third point. And hm, fractions are tricky to plot accurately, so let s actually find two new points to avoid plotting 1 2, 1. What s an easy point to find for the line y = 2 x? Well, when x =, then the fraction will disappear! Plugging in x =, we get: y = 2 () y = 2, so we have the point (, 2). And let s also use x =, and plugging that in, we ll get y = 2. So that s the point (, 2). Let s plot and draw a line. Done! Answer: a. Yes; b. 2x y = 0; c. y = 2 x ; d. slope = 2 ; y-intercept = 0; x-intercept = 0; e. (graph below) 4. Okay, let s plug this point in and see if it gives a true statement. Plugging in (2, 1), we get: 2y 4x = 7 2(1) 4(2) = 7? 2 8 = 7? 6 = 7? Nope! So the point is not on the line. For part b, it s almost in standard form, but let s write it with the x term before the y term, and we ll also need to multiply both sides by 1, so we get: 4x 2y = 7.

4 For part c, let s start with the original equation and solve for y by first adding 4x to both sides: 2y = 4x + 7, and then dividing both sides by 2, and we get: y = 2x For part d, now we can see that the slope is 2 and the y-intercept is 7. For the x-intercept, 2 let s plug in 0 for y and solve for x: 0 = 2x (subtracting 7 2 from both sides) 7 2 = 2x (dividing both sides by 2) 7 = x. So that s the x-intercept! 4 For part e, we have two points to graph so far; the two intercepts: 0, 7 2 and 7 4, 0. We need one more. Let s plug in x = 1, and we get: y = 2x y = 2(1) y = y = y = 11 2, so we get the point 1, Let s graph! Answer: a. No; b. 4x 2y = 7; c. y = 2x ; d. slope = 2; y-intercept = 7 2 ; x-intercept = 7 4 ; e. (graph below)

5 . Okay, let s plug this point in and see if it gives a true statement. Plugging in (1, 3), we get: y 3 + x = = 2? = 2? 2 = 2? Yep! so the point is indeed on 3 the line. For part b, for standard form, we need to first multiply both sides by 3, and we get: y + 3x = 6; in other words, 3x + y = 6. For part c, we solve for y by subtracting 3x from both sides, and we get: y = 3x + 6. For part d, we can now see that the slope is 3 and the y-intercept is 6. For the x-intercept, we ll plug in 0 for y and solve for x: y = 3x = 3x + 6 3x = 6 x = 2, and that s the x-intercept. For part e, we have three points already; (1, 3) and the two intercepts, (0, 6) and (2, 0). Let s plot em and graph! Answer: a. Yes; b. 3x + y = 6; c. y = 3x + 6; d. slope = 3; y-intercept = 6; x-intercept = 2; e. (graph below)

6 DTM from p We see <, so we know we ll use a dotted line. We ll start by graphing the line y = 3x + 1. Finding points, we see the y-intercept is 1, so we have the point (0, 1). Then plugging in something easy like x = 1, we get: y = 3x + 1 y = 3(1) + 1 y = y = 4. Great! Another point: (1, 4). For a third point, let s choose another easy x-value, like 1, and we get: : y = 3x + 1 y = 3( 1) + 1 y = y = 2. And so our third point is ( 1, 2). We re ready to graph the line! Now, which side of the line gets shaded? Well, we want to shade all the points on the coordinate plane that satisfy the original inequality: y < 3x + 1, and we now know it ll be all the points on one side of this line; we just don t know which side yet. So let s pick a point on one side and test it, to see if we get a true statement! (That would mean the inequality was satisfied by the point.) Looking at the graph below, how about the point (0, 0)? Plugging it in, we get: y < 3x < 3(0) + 1? 0 < 0 + 1? 0 < 1? Yep! So we shade the side that includes the point (0, 0). Done! Answer: (see graph below)

7 3. Since we see, we won t be using a dotted line; just a regular ol line! So first we need to graph the line y = 2x 4. Finding points to plot, first we can plot the y-intercept, which from looking at this slope-intercept form of the line, we know is (0, 4). Next let s pick, oh I don t know, how about x = 1? Then we get: y = 2x 4 y = 2( 1) 4 y = 2 4 y = 2. So we get the point ( 1, 2). I liked neutralizing that first negative term by using a negative x-value, so let s do it again. This time let s use x = 2, and we get: y = 2x 4 y = 2( 2) 4 y = 4 4 y = 0. Hey, it looks like we discovered the x-intercept: ( 2, 0). Now we have our three points and we can graph the line. So which side of the line should we shade? Let s pick a point on one side (not on the line itself, though!). How about (0, 0) just because it s easy, and see if we get a true statement from the original inequality: y 2x 4 0 2(0) 4? 0 0 4? 0 4? Yep! So that s the side we shade; the side that includes the point (0, 0). Done! Answer: (see graph below)

8 4. Since we see, we won t be using a dotted line. First we need to graph the line x =. This is actually a really easy line to draw. No matter what y is, the x-value will always be! So we ll have points like (, 2), (, 0), (, 2), etc. Let s graph it! Now, which side of the line gets shaded? Happily, the easy point (0, 0) is not on the line, so we can use it to test. Of course, we ll only be able to use the x-value from the (0, 0) to plug in, and we get: x 0? Yep! So we shade to the left of the line, where all of the smaller values of x are which makes sense when you look at the expression x, doesn t it? Answer: (see graph below). Since we see >, we ll use a dotted line. First we graph the line 2x + 3y = 6. I always like to rewrite lines in slope-intercept form before graphing them. So, subtracting 2x from both sides and then dividing both sides by 3, we get: y = 2 x + 2. So we can already 3 see one point to graph the y-intercept, (0, 2). For the next point, let s plug in 3 for x, so

9 that the fraction disappears, and we get: y = 2 3 x + 2 y = 2 (3) + 2 y = y = 0. Looks like we discovered the x-intercept, (2, 0). For the third point, let s pick another multiple of 3, so that the fraction goes away again. How about 6? So when x = 6, we get: y = 2 3 x + 2 y = 2 (6) + 2 y = y = 2. And our third point is (6, 2). So, which side of this graph should we shade? Let s pick a point on one side to test in the original inequality (happily we can use (0, 0) since it s not on the line), and we get: 2x + 3y > 6 2(0) + 3(0) > 6? > 6? Nope! So we shade the side that doesn t include the point (0, 0). Done! Answer: (see graph below)