Assignment 10

course Mth 158

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18:19:58

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

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5y + 6 = -18 - y

5y + 6 - 6 = -18 - y - 6

5y= -24 - y

5y + y= -24 - y + y

6y = -24

6y/6 = -24/6

y = -4

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18:20:28

** 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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18:24:27

query 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

3*[(2x+1)/3 + 16] = 3x * 3

2x + 1 + 48 = 9x

2x + 49 = 9x

2x + 49 - 49 = 9x - 49

2x = 9x - 49

2x - 9x = 9x - 9x - 49

-7x = -49

-7x/-7 = -49/-7

x=7

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18:24:36

** 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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18:27:14

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

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

-x - 6 = 6x + 9

-x = 6x + 15

-7x = 15

x = 15/-7

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18:27:22

** 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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18:32:25

query 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+3)(x-3)[x/(x+3)(x-3)+4/(x+3)]=(x+3)(x-3)[3/(x+3)(x-3)]

x + 4(x-3) = 3

x + 4x - 12 = 3

5x = 15

x = 3

3 causes a denominator to equal 0 so there is no real solution.

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18:32:37

** 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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18:37:08

query 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 -->

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

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

40w^2 + 25w + 56w + 35 = 40w^2 - 28w - 30w + 21

40w^2 + 81w + 35 = 40w^2 - 58w + 21

40w^2 + 81w = 40w^2 - 58w - 14

81w = -58w - 14

139w = -14

w = -14/139

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18:37:19

** 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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18:38:35

query 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 -->

1 - a x = b

-a x = b - 1

x = (b-1)/-a

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18:38:49

** 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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18:41:43

query 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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RESPONSE -->

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

x(x^2 + 6 - 7) = 0

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

-7 + 7 = 0, 1-1 = 0 and x = 0

x = 0

x = -7

x = 1

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18:41:52

** 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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19:09:38

query 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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86 + 80 + 84 + 90 = 340

340 / 4 = Average 85

(85 + 2x) / 3 = final avg.

(85 + 2x) / 3 = 80

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

85 + 2x = 240

2x = 155

x = 77.5

(85 + 2x) / 3 = 90

3*[(85 + 2x) / 3 = 90 * 3

85 + 2x = 270

2x = 185

x = 92.5

Student has to make a 77.5 to get a B(80) or make a 92.5 to get an A(90).

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19:09:49

** 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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19:12:22

query 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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v = g t + v0

v + gt = v0

gt = v0 - v

t = (v0 - v)/g

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19:12:30

** 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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19:12:37

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

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none

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19:12:43

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

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Assignment 10

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