10

course MTH 158

I am going to try to find someone to explain this to me because I dont have a clue. then i will complete the assignment. sorry for the inconvenince, and thank you for your understanding.

??????????z???assignment #010010. `query 10

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College Algebra

02-04-2007

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assignment #010

010. `query 10

College Algebra

02-04-2007

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13:23:40

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 -->

I can not run ass # 9. also, I am nut able to run 1-1 on my CD. it just freezes so i will try my best.

5y + 6 = -18 - y subtract 6

= 5y = -12 - y add Y

= 6Y = -12 devide by 6

= Y = -2

confidence assessment: 1

Good, except that -18 - 6 = -24.

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13:27:11

** 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 -->

I was correct in keeping my - sign yet I still subtracted from 18 not -18

self critique assessment: 2

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13:42:47

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 -->

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

= (x+1) / 3 + 8 = 3x

= 1 / 3 + 8 = 4x

= 5 = x

= x= 5

confidence assessment: 1

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13:45:30

** 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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RESPONSE -->

I am not shure what to start with.

Why the 3?

self critique assessment: 1

The 3 is a denominator. It's a very good idea to multiply both sides by any denominator you find in the equation, which eliminates fractions and usually makes the equation much easier to work with.

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13:54:22

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 -->

(x+2)(x-3) = (x+3)^2

= (x+2)(x-3) = x^2+9

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

=x^2 = x^2 +15

=x = +15

confidence assessment: 1

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13:59:13

** 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 -->

I am going to reaqd over the distributive property agen.

self critique assessment: 2

Please do and if you don't then completely understand why (x+2)(x-3) is not equal to x^2 - 6, and why (x+3)^2 is not equal to x^2 + 9, be sure to let me know so we can work that out.

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15:48:57

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 -->

x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)

= x / (x^2) + 4 / (x+3) = 3 / (x^2)

Those 9's are parts of two different denominators, not terms of the equation. You can't do that unless the expressions are terms, not tied up with other expressions.

= x^3 + 4 / (x+3) = 3 / x^2

= x^3 + 4 / x+ 9 = x^2

= x^3 + 4 + 9 = x^3

= x^3 + 13 = x^3

= 13 = x

confidence assessment: 1

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15:56:40

** 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 -->

sadly I even took the time to reread the chapter.I just

don't get this

self critique assessment: 2

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See my notes, which will hopefully help a bit.

However you're having difficulties here. I need a little more information so I can try to help you over this 'hump'.

Can you give me a synopsis of your work on Problems 6, 10, 12 and 18? Include a short description of each problem, and the steps you took in solving each.

It would be best if you would email this information as well as submitting it through the form. That way I'll be able to respond as quickly as possible.