11

course Mth 158

6/22 10

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

010. `* 10

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

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Your solution:

First, I’ll subtract 6 from both sides.

5y + 6 = -18 – y

5y = -24 – y

Now, I will add y to both sides

6y = -24

Finally, I will divide each side by 6/

Y=-4

confidence rating #$&* 3

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Given Solution:

* * 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 in the above note is 5y + 6 = -18 - y.

The solution to this equation is found by practically the same steps but you end up with y = -4.

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Self-critique (if necessary):OK

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Self-critique rating #$&*

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Question:

1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x

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Your solution:

(2x+1)

--------------- + 16 = 3x

3

Multiply both sides by the LCD,which is 3.

2x+1 + (16 *3=48) = (3*3x)=9x

combine like terms

2x + 49 = 9x

subtract 2x from each side.

49=7x

divide each side by 7

x=7

confidence rating #$&* 3

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Given Solution:

* * 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 to get

49 = 7x

Divide both sides by 7 to get

x = 7.

STUDENT QUESTION

I was wondering at the end since it ended up 49 = 7x and you divide by 7 and say x = 7…would you have to

make it a -7 if you move it to the opposite side of the equation?

INSTRUCTOR RESPONSE

It's not a matter of 'moving things around', but a matter of adding or subtracting the same quantity on both sides, or multiplying or dividing both sides by the same quantity.

In this case both sides are divided by 7, which doesn't involve any negative signs.

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Self-critique (if necessary):OK

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Self-critique rating #$&*

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Question:

* was 1.1.44 \ 36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2

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Your solution:

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

Use the distributive property.

x2 -3x +2x -6 = x2+3x +3x +9

Combine like terms.

x2 –x -6 = x2 +6x +9

subtract the x2 from the left and right.

-x -6= 6x+9

add x to each side.

-6=7x+9

subtract 9 from each side

15=7x

-6-9=-15

divide by7

x=2.14

confidence rating #$&* 3

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Given Solution:

* * 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

x = -15/7

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Self-critique (if necessary):I don’t understand how this number is negative. It may just be late at night but I’m just not putting it together.

see my note above

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Self-critique rating #$&*

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Question:

* 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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Your solution:

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

First, simplify the (x2 -9) which equals (x+3)(x-3)

Second, multiply each side by the LCD, which is (x+3)(x-3)

x + 4(x-3) =3

Distribute.

x + 4x-12 = 3

Combine like terms.

5x-12=3

add 12 to each side.

5x=15

divide by 5

x=3

confidence rating #$&* 3

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Given Solution:

* * 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.

Apply the Distributive Law, rearrange and solve:

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.

STUDENT COMMENT

x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) Since you have like terms (x^2 – 9) on both sides, they cancel each other out

INSTRUCTOR RESPONSE

If something 'cancels' by multiplication or division, it has to 'cancel' from all terms. (x^2 - 9) is not a multiplicative or divisive factor of the term 4 / (x + 3) so that factor does not 'cancel'.

You can multiply or divide both sides by the same quantity, or add and subtract the same quantity from both sides.

Anything called 'cancellation' that doesn't result from these operations is invalid.

Because 'cancellation' errors are so common among students at this level, my solutions never mention anything called 'cancellation'.

If you multiply both sides of the equation

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

by (x^2 - 9), you get

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

x / (x^2-9) * (x^2 - 9) + 4 / (x+3) * (x^2 - 9) = 3 / (x^2-9) * (x^2 - 9). The (x^2 - 9) does then 'cancel' from two of the three terms, but not from the third. You get

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

You're still stuck with an x^2 - 9 factor on one of the terms, and a denominator x - 3.

However this equation does represent progress. If you factor x^2 - 9 into (x-3)(x+3), things quickly simplify.

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Self-critique (if necessary):Ok

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Self-critique rating #$&*

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Question:

* 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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Your solution:

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

First, multiply each side by the LCD, which is (10w-7)(5w+7)

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

distribute.

80w2- 56w+50w – 35+40w2 +56w+25w+35= 40w2 -28w+20w2+28w-30w+21-15w-21

simplify.

120w2 +75w = 60w2 -45w

subtract 60w2 from each side.

60w2 + 75w= -45w

subtract 75w from each side

60w2 = -120w

divide by 60

w2=-2w

confidence rating #$&* 1

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Given Solution:

* * GOOD 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)

STUDENT QUESTION:

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

work what you can

40w^2 + 35 = 40w^2 +21

take away 40w^2 from both sides

didnt understand this one..;

INSTRUCTOR RESPONSE:

It doesn't look like you used the distributive law to multiply those binomials.

(5w+7)(8w+5) = 5w ( 8w + 5) + 7 ( 8w + 5)= 40 w^2 + 25 w + 56 w + 35 = 40 w^2 + 81 w + 35.

(10w-7)(4w-3) = 10 w ( 4 w - 3) - 7 ( 4 w - 3) = 40 w^2 - 30 w - 28 w + 21 = 40 w^2 - 58 w + 21.

* 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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Your solution:

1 - a x = b, a <> 0

Subtract 1 from each side.

-a x = b a -1

divide each side by –a

X= b a -1 / a

confidence rating #$&*

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Given Solution:

* * 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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Self-critique (if necessary):

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Self-critique rating #$&*

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Question:

* extra problem (was 1.1.72). Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring.

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Your solution:

x^3 + 6 x^2 - 7 x = 0

x(x2 +6x -7)=0

x(x+7)(x-1)=0

confidence rating #$&* 2

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Given Solution:

* * Starting with

x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side:

x(x^2 + 6x - 7) = 0. Factor the trinomial:

x ( x+7) ( x - 1) = 0. Then

x = 0 or x + 7 = 0 or x - 1 = 0 so

x = 0 or x = -7 or x = 1. **

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Self-critique (if necessary):

I see that I forgot to complete the problem. I should have listed what x was equal too.

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Self-critique rating #$&*

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Question:

* 1.1.90 (was 1.2.18). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) scores 86, 80, 84, 90, scores to ave B (80) and A (90).

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Your solution:

Exam = 2/3

Test scores: 80, 84, 86 ,90

Average test score= 85

90=final grade

T=test average

E= exam average

90= t +2E / 3

add in your t average.

90= 85 + 2(e) /3

multiply by the lcd to get discard fractions.

90(3)= 85 + 2(e)

270= 85 +2E

simplify=

270-85=185 = 2E

divide by 2

e=92.5 A

Solve for B (80 – 89)

70(3)= 85 + 2(e)

210= 85 +2E

simplify=

210-85=125 = 2E

divide by 2

E=83 (b)

confidence rating #$&*

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Given Solution:

* * This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation.

Let x be the score you make on the exam.

The average of the four tests is easy to find:

• 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85.

The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have

• final average = (1 * test average + 2 * exam grade) / 3.

This gives us the equation

• final ave = (85 + 2 * x) / 3.

If the ave score is to be 80 then we solve

(85 + 2 * x) / 3 = 80.

Multiplying both sides by 3 we get

85 + 2x = 240.

Subtracting 85 from both sides we have

2 x = 240 - 85 = 155

so that

x = 155 / 2 = 77.5.

We can solve

(340 + x) / 5 = 90

in a similar manner. We obtain x = 92.5.

Alternative solution:

If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be

1/3 * 85 + 2/3 * x = final ave.

For final ave = 80 we get

1/3 * 85 + 2/3 * x = 80.

Multiplying both sides by 3 we have

85 + 2 * x = 240.

The rest of the solution goes as before and we end up with

x = 77.5.

Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. **

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Self-critique (if necessary):

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Self-critique rating #$&*

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Question:

* 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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Your solution:

V= -G T + V0

Subtract V0 from each side.

v-v0= -g t

divide each side by –g

v-v0 / -g = T

confidence rating #$&* 3

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Given Solution:

* * 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

so that

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

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Self-critique (if necessary):

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Self-critique rating #$&*

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

"

&#Good work. See my notes and let me know if you have questions. &#

#$&*