Assignment 10

course MTH 158

ٓ‡[ʆʔ\́ìassignment #010

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

010. `query 10

College Algebra

09-23-2007

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19:44:10

1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y

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RESPONSE -->

First, combine variables.

5y + y + 6 = -18

6y + 6 = -18

6y + 6 - 6 = -18 - 6

6y = -24

y = -4

confidence assessment: 3

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19:45:02

** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y

Subtract 6 from both sides, giving us

5y = 12 - y

Add y to both sides,

5y + y = 12 or 6y = 12

divide both sides by 6

y = 2

INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation as I note it is 5y + 6 = -18 - y.

If that's the correct equation the solution is found by practically the same steps but you end up with y = -4. **

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RESPONSE -->

ok

self critique assessment: 3

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19:48:38

1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x

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RESPONSE -->

Start by multiplying each side by 3:

2x + 1 + 16 = 9x

You need to multiply the entire left-hand side by 3. That multiplication distributes over the 16 also. So you should have

2x + 1 + 48 = 9x.

This is indicated in the given solution and should have been noted in your self-critique.

combine like terms

2x + 17 = 9x

2x + 17 - 17 = 9x - 17

2x - 9x = 9x -9x -17

-7x = -17

-7x/-7 = -17/-7

x= 17/ 7 or 2 3/7

confidence assessment: 3

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19:48:59

09-23-2007 19:48:59

** STUDENT SOLUTION:

(2x + 1) / 3 + 16 = 3x

First, multiply both sides of the equation by 3

2x +1 + 48 =9x or 2x + 49 = 9x

subtract 2x from both sides.

49 = 7x

Divide both sides by 7

7 = x

**

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NOTES ------->

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19:54:09

was 1.1.36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2

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RESPONSE -->

First solve the parentheses:

x^2 - 3x + 2x - 6 = x ^2 + 3x + 3x + 9

combine like terms

x^2 - x - 6 = x^2 + 6x + 9

x^2 - x^2 -x -6= x^2 - x^2 +6x+9

-x -6 = 6x + 9

-6 = 7x + 9

-15 = 7x

-15/7 =x or 2 1/7

confidence assessment: 3

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19:54:18

** STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2

First, we use the distributive property to remove the parenthesis and get

x^2 - x - 6 = x^2 + 6x + 9

subtract x^2 from both sides,

-x - 6 = 6x + 9

Subtract 9 from both sides

- x - 6 - 9 = 6x or -x - 15 = 6x

add x to both sides

-15 = 7x

Divide both sides by 7

-15/7 = x

**

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RESPONSE -->

ok

self critique assessment: 3

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19:58:26

1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/

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RESPONSE -->

Multiply each side by x^2-9, which is (x-3)(x+3)

x + 4(x-3) = 3

x + 4x -12 =3

5x -12 +12 +3 +12

5x=15

x=3

confidence assessment: 3

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19:58:37

** Starting with

x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get

x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ).

Multiply both sides by the common denominator ( (x-3)(x+3) ):

( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify:

x + 4(x-3) = 3. Simplify

x + 4x - 12 = 3

5x = 15

x = 3.

If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation.

When you multiplied both sides by x-3, if x = 3 you were multiplying by zero, which invalidated your solution **

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RESPONSE -->

ok

self critique assessment: 2

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20:06:20

1.1.58 (was 1.1.54). Explain, step by step, how you solved the equation (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7)

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RESPONSE -->

First multiply both sides by the denominators (10w-7) (5w+7)

(8w+5)(5w+7) = (4w-3)(10w-7)

40w^2+73w+35=40w^2-58w+21

subtract each side by 40w^2

73w+35 = -58w +21

+58w + 73w +35 = -58w +58w +21

131w + 35 - 35 = 21 -35

131w = -14

w = -14/131

confidence assessment: 2

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20:06:35

** STUDENT SOLUTION:

1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7)

After cancellation the left side reads:

(5w+7)(8w + 5)

After cancellation the right side reads:

(10w - 7)(4w - 3)

multiply the factors on each side using the DISTRIBUTIVE LAW

Left side becomes: (40w^2) + 81w + 35

Right side becomes: (40w^2) - 58w + 21

3) subtract 40w^2 from both sides

add 58w to both sides

subtract 35 from both sides

Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139)

DER**

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RESPONSE -->

ok

self critique assessment: 3

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20:09:51

1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.

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RESPONSE -->

subtract each side by 1

-ax = b - 1

-ax/-a = (b-1)/-a

x=(b-1)/-a

confidence assessment: 3

You don't want a negative sign on your denominator; if you multiply num. and denom. by -1 you get

x=-1 * (b-1)/(-1 * -a) = (1 - b) / a.

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20:10:02

** Start with

1 -ax = b, a <> 0.

Adding -1 to both sides we get

1 - ax - 1 = b - 1, which we simplify to get

-ax = b - 1.

Divide both sides by -a, which gives you

x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get

x = (-b + 1) / a or

x = (1 - b) / a. **

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RESPONSE -->

ok

self critique assessment: 3

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20:11:52

1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = g t + v0 for t.

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RESPONSE -->

start by v0 = 0

v0 is the name of a variable, and can't be arbitrarily set to 0.

v = gt

divide each side by g

v/g = t

confidence assessment: 3

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20:12:00

09-23-2007 20:12:00

** NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0.

Starting with v = g t + v0, add -v0 to both sides to get

v - v0 = gt. Divide both sides by g to get

(v - v0) / g = t

}or

t = (v - v0) / g. **

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NOTES ------->

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20:12:52

Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I understood this lesson fairly well. the problems v=gt+v0 confused me a little.....

confidence assessment: 3

If you can be more specific I can help you address your confusion. Hopefully my note will be useful.

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Your work looks good. See my notes. Let me know if you have any questions. &#