Hot X: Algebra Exposed

Transcription

1 Hot X: Algebra Expose Solution Guie for Chapter 6 Here are the solutions for the Doing the Math exercises in Hot X: Algebra Expose! DTM from p.80-1 (assume that all enominators 0). First we ll solve for, in terms of x, which means we ll preten x is just an other number (which it is! we just on t know which number it is ) So to isolate, we ll subtract x from both sies an get: = x. Next we re suppose to evaluate this for the values x = 0 an x =. Plugging in x = 0, we get: = x = (0) = 0 = Plugging in x =, we get: = x = () = 8 = 6 Now for part b; we ll solve for x in terms of. Starting from the original equation, + x =, we can just subtract from both sies, getting x =, an then ivie both sies b, an get x =. Now we re suppose to evaluate x at the values = 1 an = 9. Plugging in = 1, we get: x = x = (1) x = 1 An plugging in = 9, we get: x = x = 9 x = 7

2 Answer: Part a: = x; =, 6 Part b: x = ; x = 1, 7. Let s take the hint an start b multipling both sies b ; let s see what happens. We get: = () = (the s cancel on the left) =. Much better! To finish isolating, we nee to ivie both sies b, an we get: = x =, in other wors: = Now we re suppose to evaluate this for the values x = 0,. Plugging in x = 0, we get: = Next we ll evaluate at x =, an we get: = = 1 = = + 1 = Now for part b; we ll solve for x in terms of. Let s start from the unerline equation above, because x is alrea partiall isolate: =. To finish, we ll just subtract 1 from both sies an get x = 1. That wasn t so ba! Now we re suppose to evaluate this for the values = 1 an 9. When = 1, we get: x = 1 x = (1) 1 x = 1 x = 1 Next we ll plug in = 9, an we get: x = 1 x = (9) 1 x = 18 1 x = 17 Done! Answer: Part a: = ; = 1, Part b: x = 1; x = 1, 17

3 . Let s take the hint an start b multipling both sies b 1; let s see what happens as we work to isolate. We get: x + 1 = x 1(x + ) 1 = 1x (the 1 s cancel on the left) x + = 1x. To finish isolating for, we subtract x from both sies an get: x + x = 1x x = 1x. Now we re suppose to evaluate this for the values x = 0,. Plugging in x = 0, we get: Plugging in x =, we get: = 1x = 1(0) = 0 = 1x = 1() = Now for part b; we ll solve for x in terms of. Let s start from the unerline equation above, because x is alrea partiall isolate: = 1x. To isolate x, we just ivie both sies b 1, an get: x = For = 1, we get: x = For = 9, we get: x = 1 1. We re suppose to evaluate this at the values = 1, 9. 1 x = 1 1. x = 9 1 x =. Done! Answer: Part a: = 1x; = 0, Part b: x = 1 ; x = 1 1, DTM from p.87-8 (assume that all enominators 0). Let s take the hint an factor out x. Then we get: x(a + b) = c. Next, to finish isolating x, we just ivie both sies b (a + b), an we get: x(a + b) (a + b) = c (a + b).

4 Since the (a + b) s cancel on the left, we en up with: x = c. An lookie there, (a + b) we re one! (We can rop the parentheses because the re not reall oing anthing.) Answer: x = c a + b. Let s take the hint an istribute the x an the 7, an we get: x + xz + x = 7. Next, to isolate the x, the we ll collect the stuff with x in it to the right (subtract 7x from both sies), an we get: x + xz x = 1. Next we factor out the x an we get: x( + z ) = 1. Finall, to finall isolate x, we just ivie both sies b the stuff in the parentheses, an we get: x( + z ) ( + z ) = 1 ( + z ) parentheses cancels on the left, we get our answer: x =. Since the stuff in the 1. Again, we can rop ( + z ) the parentheses because we re not oing an more manipulation of the equation an the re not separating anthing from anthing else anmore! Answer: x = 1 + z. First we ll simplif this fraction. We i this on p.8-9; we use a trick to get a common factor on top an bottom, an coax out that hiing copcat! An we learne that (c ) ( c) =. So our new problem becomes: + x = 5. To isolate x, remember to treat like an other number, so we first a to both sies, an get: x = 5 +. Next we ivie both (entire) sies b, an we get: x = 5 +. Since the s cancel on the left, we get our answer: x = 5 +. Remember that the s on t cancel on the right fraction, because it s not a common factor to all terms in the numerator! Answer: x = 5 +

5 5. Let s take the hint an simplif the complex fraction. First we ll combine the little fractions in the numerator, pretening nothing else in the worl exists. We ll multipl the copcat fraction times the first fraction to get a common enominator, an we get: x + x 6 looks like this: x which becomes 6 + x 6 = x 6 = x. Nice! So now the complex fraction x, an since it s in tall an skinn form, it s rea for the means an extremes shortcut from p.5. It becomes: x = 6x 6 = x. M gooness, that entire complex fraction became sweet little x! So now the entire problem has boile own to: x = x 1. To isolate x, let s subtract x from both sies, getting: x = 1. Multipling both sies b 1, we get the answer: x = 1. Done! Answer: x = 1