[if they got 0] You were 0 for 5-- which is understandable. After all, you haven’t gone through the lesson yet. Go through it now. By the time you're done, you'll be able to get ALL these questions right. Drill B Use graph paper to find the answers to the following questions. 1. What is the solution to x + 3 = y and x + y = 3? (A) (3, 0) (B) (3, 3) (C) (0, 4) (D) (0, 3) [A] Nope. Graph the lines carefully, then look at the point where they intersect, and don’t allow yourself to mix up the x and y-coordinates. [B] Sorry. A slip of the pencil, perhaps. Try again, graphing the lines carefully, and marking the point where they intersect. [C] Oops. A small miscalculation. Take another look at your intersection, and count on the graph paper carefully. [D] Right. This is exactly the point that satisfies both equations; it’s where they intersect, after all. 2. What is the solution to 2x + 5 = y, and x - 1 = y? (A) (4, 3) (B) (-3, 4) (C) (-6, -7) (D) (10, 9) [A] Nope. You’ve gotten messed up in the actual drawing of your lines, their intersection is in a different quadrant altogether. Give it another try. [B] Sorry. Try it again, graphing first one line and then the other. In what quadrant do they intersect? [C] Exactly. You graphed the lines and noted the point of intersection. Nicely done. [D] You’re all over the place, kiddo. Graph these equations again. Then, look at where they intersect. 3. What is the solution to x = y and 2y = x + 3? (A) (3, 3) (B) (-3, 0) (C) (-3, -3) (D) (3, -3) [A] Right! It satisfies both equations, and you can see this when you graph the lines of both equations. [B] Oops. This wouldn’t satisfy the first equation, would it? So how could it be where the two lines intersect? Give it another try. [C] Sorry, kid. You’ve had some trouble with the line from the second equation. Give it another try. [D] Nope. This doesn’t satisfy the first equation, does it? Give it another try. 4. What is the solution to -2x + 1 = y and 3x + 2 = 2y? (A) (0, 7) (B) (0, 1) (C) (8, -7) (D) (2, 4) [A] Nope. You’ve had some problem graphing the line for the first equation. Try it again, and remember, y = mx + b. [B] Exactly. This point is on the lines of both equations. Well done. [C] Oh well. This isn’t on either of the lines for these equations; perhaps you’re mixing up your x and y-coordinates? Try again. [D] Sorry. You’ve run into some problem graphing the first equation. Try it again, more slowly. 5. What is the solution to x - 2y = 0, and 3x + y = 7? (A) (1, 7) (B) (1, 2) (C) (-2, -1) (D) (2, 1) [A] Nix. This sure doesn’t fall on the line of the first equation, and to be a solution it must. Try graphing the lines again. [B] Sorry, you seem to have mixed up your x and y-coordinates, or misgraphed your lines. Try again. [C] Oops. You seem to have misgraphed one of the lines; your coordinates for the intersection are in the wrong quadrant. Try it again. [D] Yes! This solution is true of both equations; you can see it where they intersect on your graph paper. Nice going. Drill C 1. To find the solution for the system of open sentences x + 3 = y and 2x - 1 = 3y, you would best substitute which of the following? (A) x + 3 for x in the second equation (B) x + 3 for y in the second equation (C) 2x for y in the first equation (D) 3y for x in the first equation [A] Nope. If you do this, you still end up with an equation with two variables, which is no help at all. Try again. [B] Great idea. This way, you can solve for x in the second equation. [C] Sorry. If 2x were equal to y, and you knew this from the first equation, maybe, but that’s not what the second equation says, so this won’t provide you with useful information. Try it again. [D] Oops. A valiant attempt, but 3y is equal to 2x - 1, not x itself, so this wouldn’t help you towards a solution. Try again. 2. Once you’ve substituted x + 3 for y in the second equation in the system of open sentences x + 3 = y and 2x - 1 = 3y, what does x equal? (A) -10 (B) 4 (C) -4 (D) 10 [A] Right. You simply put in 2x -1 = 3(x + 3), and solved for x. We’re suitably impressed. [B] Sorry. You seem to have substituted x + 3 for 3y, instead of substituting for y and then looking at the coefficient. Try again. [C] Yikes. First of all, you seem to have substituted for 3y rather than for y itself. Second of all, you seem to be getting muddled in you arithmetic. Slow down and try again, carefully. [D] Ouch. You were doing great, but you got tangled up in negatives and positives. Solve for x again, looking at the signs of everything very carefully. 3. Once you’ve found that x = -10 in the system of open sentences x + 3 = y and 2x - 1 = 3y, what is the solution for both x and y? (A) (10, 13) (B) (-10, -7) (C) (-10, 7) (D) (7, -10) [A] Hey now, the question itself says that x is -10. This solution gives you x as positive 10. No go. Try again. [B] Wonderful! Once you know that x is -10, substitute it into x + 3 = y and solve for y. Beautiful. [C] Oops. You’ve got the right value for x all right, but you had some trouble when substituting it into x + 3 = y. Look more carefully at your arithmetic. [D] Ouch. The question says that x is -10, correct? And which coordinate is the x coordinate in a pair? Try again. 4. What is the value of y in this system of open sentences: 3x + 3y = 4, and 3x - y = -4 (A) - (B) 4 (C) 6 (D) 2 [A] Sorry. You seem to have a possible value for x here, not the solution for y. Subtract one equation from another, and solve for y. Go on, we know you can do it. [B] Oops. You were close, but you solved for 2y, not y itself. Give it another go. [C] Sorry. You want to subtract one equation from another, and then solve for y. Try again. [D] Superb. You’ve subtracted one equation from the other, and then solved for y. Nice job. 5. What is the value of x in this system of open sentences: 4y + 2 = 3x and 3y - x = 0 (A) (B) (C) (D) [A] Yup. Just multiply the second equation by 3 so it becomes 9y -3x = 0, revamp it so it becomes 9y = 3x, and subtract one equation from another. From there it’s a snap (sort of) to solve for y, and you can substitute the y back in to find x. Of course, you could also have multiplied the equations by something else to subtract the y’s. Either way, you’ve done an excellent job in finding this answer. [B] Sorry. You’ve found y when the question wanted x. But, since you have y, why not put it into one of the equations and find x that way? [C] Nope. You seem to have gotten confused in your solution; you’ve got the inverse of y here. Try again. [D] Drat. You were very close, but you got muddled when you were solving for x. Try again, being careful of the arithmetic involved. Drill D 1. {**Show a graph of the lines x = 2y and 3y = x + 1 with the points (2, 1) and (4, 2) plotted for the first equation, but the equation itself not present, and the points (2, 1) and (6, 2) for the second equation, though the equation itself still missing} Which of the following pairs of equations could be the equations for the two lines that intersect at point (2, 1)? (A) x = y and x = 2y + 1 (B) 2x = y and 3y = x + 1 (C) x = 2y and 3y = x + 1 (D) x = 2y and x = 2y + 1 [A] Sorry. The first equation wouldn’t satisfy the point of intersection, would it? To review these equations and what their graphs mean, click on Graphing in the list of topics. [B] Nope. You’ve gotten your x and y coordinates mixed up with the first equation. Take a closer look, and click on Graphing in the list of topics, to review. [C] Exactly. Both these equations could include the point (2, 1) that is the intersection of the lines. [D] Sorry. The second equation here couldn’t have (2, 1) be part of its solution set. You might want to click on Graphing in the list of topics to review. 2. To solve the equations x + 3 = 4y and 2x - 1 = y, which of the following would be a reasonable first step to substitute? (A) substitute 4y for 2x in the second equation (B) substitute 2x - 1 for y in the first equation (C) substitute x + 3 for y in the second equation (D) substitute 2x - 1 for x in the first equation [A] Unfortunately, 4y is not equal to 2x, so the substitution wouldn’t make much sense. Give it another try, and click on Substitution in the topics list to review. [B] Great. You know that 2x - 1 is equal to y, thus it is reasonable to substitute, and once you’ve done this, you can solve for x. [C] Sorry, x + 3 isn’t equal to y, it’s equal to 4y, so it wouldn’t make sense to substitute it. Try again, and click on Substitution in the list of topics for some help. [D] Nope. You would still be left with an equation with two variables, and an untrue equation at that. Click on Substitution in the list of topics to review this procedure. 3. Once you substitute 2x - 1 for y in the first equation to solve the equations x + 3 = 4y and 2x - 1 = y, which of the following equations does that yield? (A) x + 3 = 8x - 4 (B) x + 3 = 2x - 1 (C) 2x - 1 = 4y (D) 2x - 1 = 4/3y [A] Right you are! You substituted 2x - 1 for y, and then multiplied it by 4 just as the y it is replacing was multiplied by 4 to make the equation equal. Perfect. [B] Oops. The y that this is replacing was multiplied by 4, wasn’t it? So you must substitute for that y, doing everything to your substitution that was done to y. You should also take a peek at Substitution in the list of topics. [C] Um, er, doesn’t this look like the same equation you started out with? Don’t you want to substitute something here? Try again, and click on Substitution in the list of topics to refresh your memory about it. [D] Sorry, sweetie. You still have an equation with two variables, and that is what substitution is supposed to help. Try it again, first clicking on Substitution in the list of topics to get your facts straight. 4. Once you get x + 3 = 8x - 4, what does x equal? (A) 7 (B) (C) - (D) 1 [A] Sorry, bud. It’s your old pal solving for x, and you probably want to slow down and try it again. [B] Nope. You got a teeny bit sloppy when you were solving for x. Try again, and think about reviewing that solving for x lesson. [C] Oh well. A glitch in your solving-for-x work. Remember, whatever you do, do it to both sides of the equation. Try again. [D] Exactly right. You put the variables on one side, and the number on the other, and you got it. Beautiful. 5. Once you find that x = 1 in the equations x + 3 = 4y and 2x - 1 = y, solve for y as well. (A) y = -1 (B) y = 0 (C) y = 4 (D) y = 1 [A] Ouch. No big deal, but you stumbled a bit. If x = 1, and 2x - 1 is y, then y is... [B] Oh, drat. Just put the old x is equal to 1 information back into one of the equations, and solve for y. [C] Oops. You tried the first equation, right? But you went a bit too fast. Slow down, and try to solve for y again. [D] Bingo! If x is 1, then 2 (1) - 1 = y, which equals 1. 6. If 3x + 2 = 5y, and x - 2 = 3y, what are the values of x and y? (A) x = -2, y = -4 (B) x = -4, y = -2 (C) x = 2, y = 4 (D) x = -4, y = 2 [A] Oops. You got pretty close, but you got a bit tangled up in the solutions. Try it again, and take a peek at the Simultaneous Equations topic in the list at the top; review could be very helpful. [B] Great job! You probably multiplied the second equation by 3, and then subtracted one equation from the other. That way, you find y and then substitute it back in to find x. Great work. [C] Nope. You’ve gotten a bit muddled in your variables, and the signs of those variables as well. To review how to deal with these, click on Simultaneous Equations in the list of topics. [D] Drat and double drat! So close, and yet... You got confused as to the sign of one of your solutions. Click on Simultaneous Equations in the topics list to review, and try solving again. 7. If you combine the number of Halloran’s pores with twice the number of Kaori’s pores, you’d have 40 pores all together. But if you had three times as many pores as Halloran and you took away the exact number of Kaori’s pores, you’d only have 1. Which of the following pairs of equations correctly depicts this situation? (A) H + 2K = 40, and 3H + K = 1 (B) H + 2K = 1, and 3H - K = 40 (C) H + 2K = 40, and 3H - K = 1 (D) 2H + K = 40, and H - 3K = 1 [A] Sorry, kid. The second equation says you “took away” the number of Kaori’s pores. Is that addition? Try again, and take a stroll up to the list of topics and click on Word Problems with 2 Variables to review. [B] Nope. Look at the first equation; what does it add up to again? You should probably click on Word Problems with 2 Variables in the list of topics to review just how to translate these equations. [C] Fabulous! Your knowledge of pores is only exceeded by your excellent math sense. [D] Ouch. The coefficient goes next to the variable that needs to be increased. Try again, after clicking on Word Problems with 2 Variables in the list of topics for a quick review. 8. Once you’ve determined that the equations are H + 2K = 40, and 3H - K = 1, you would do which of the following to solve? (A) multiply the first equation by 3 and add the two equations (B) multiply the second equation by 2 and subtract the two equations (C) multiply the first equation by 2 and add the two equations (D) multiply the second equation by 2 and add the two equations [A] Sorry. If you added them after multiplying by 3, you’d have 6H + 5K = 121. Still two variables, no closer to finding one of them. Try it again, and take a minute to click on Word Problems with 2 Variables in the list of topics to review. [B] Sorry. If you do this, meaning you subtract, you get -5H + 4K = 38. Still two variables. Remember, when you subtract a negative it’s like adding. Try again after you’ve clicked on Word Problems with 2 Variables in the list of topics. [C] Oh well. Once more, you would end up with an equation with two variables, when what you want is to end up with an equation with only one variable. Try it again, after clicking on Word Problems with 2 Variables in the list of topics. [D] Stunning! Electric! Rave reviews all around. Doing this will give you 7H = 42; an equation with only one variable, which is exactly what you want. 9. Once you multiply the second equation by 2 and add the two equations H + 2K = 40, and 3H - K = 1, you find out Halloran has how many pores? (A) 7 (B) 42 (C) 6 (D) [A] Oops. Just because 7H = 42, you can’t skip over the slow and careful solving for H. Try again. [B] Sorry, kid. Take it slow. Once you have 7H = 42, you must solve for H. 42 isn’t H, it’s 7H. Try again. [C] Great! Once you have 7H = 42, you solve for H. [D] Nope. You got muddled up in adding the equations and solving for H. If you need help with that, wander back to Word Problems with 2 Variables before moving ahead to the games. 10. Once you find out Halloran has 6 pores, and the two equations H + 2K = 40, and 3H - K = 1 are still true, how many pores does Kaori have? (A) 34 (B) 16 (C) 17 (D) 68 [A] Ouch. Very close, but you have 2K here, not K herself. Try again. [B] Oops. This is an arithmetic mistake. Read more carefully, and try it again. [C] Wonderful. You find that if H is 6, then 3H - K = 1, or 18 - K = 1. So K must be 17. You could have used the other equation too, of course–whichever makes you happy. [D] Nope. A teeny little stumble on the arithmetic end of things. You multiplied by 2, instead of dividing. Slow down, and give it another try.