View Full Version : can you check my probs?

momina4468

09-29-2008, 11:51 PM

So I have to do these 10 problems for school and I was wondering if someone can please check to see if you will get the same answers as me. The directions simply say to simplify each problem. I attached the problems because it was easier for me to type them in word.

Here are my answers:

1. ??

2. no solution

3. 1/2

4. {1/3, .2}

5. 2

6. { 5/2, -5/2

7. -2/3 plus and minus square root of 7

8. plus and minus 3

9. x+5/ x+2

10. 4

mmm4444bot

09-30-2008, 03:06 AM

... I was wondering if someone can please check to see if you will get the same answers as me. The directions simply say to simplify...

Your answers are consistent with an instruction to "solve the equation." (The instruction "to simplify" has a different traditional meaning.) If you are sure that you only need to do some type of simplification on Exercise 1, then let us know; the following is guidance to solve the equation for x.

1. ??

Exercise 1 contains two algebraic ratios set equal to one another.

Do you know that the symbol 'a' represents some unimportant constant? In other words, the solution will contain the symbol 'a'; we will neither know nor assign any value to the symbol 'a' in this exercise.

When two ratios are set equal to one another, we can cross-multiply to eliminate the fractions from the equation. (We can do this because the result is exactly the same as multiplying both sides by both denominators and simplifying.)

(x + a) * (3x - 6) = 3x * (x - 1)

Now continue by multiplying the two binomials on the lefthand side using FOIL.

On the righthand side, use the Distributive Property to multiply the binomial (x - 1) by the quantity 3x.

Here's a hint. After you do these multiplications on each side, notice that there is one term that appears the same on both sides. Do you see it? Whenever you see the same term on both sides, you can get rid of them both by subtraction.

Next, get all of the terms that contain the variable x to the left side, and get all of the terms that do not contain an x to the right side.

Factor out the x from the resulting expression on the lefthand side.

You should now have on the lefthand side the variable x times an expression inside of parentheses containing numbers and 'a' which can be simplified)

Dividing both sides of the equation by this expression in parentheses solves for x.

Simply the algebraic ratio on the righthand side.

2. no solution ? Exercise 2 has one solution.

3. 1/2 ? Exercise 3 has two solutions; x = 1/2 is one of them.

4. {1/3, .2} ? 0.2 is not a solution (if the 'period' key on the keyboard were near the actual character that should appear before the digit 2, then I might possibly think that .2 is a typographical error). x = 1/3 is correct.

5. 2 ? x = 2 is one of the three solutions.

6. { 5/2, -5/2 } x = ±5/2 are two of the three solutions.

7. -2/3 plus minus square root of 7 ? This is not stated correctly; I'll correct it for you.

-2/3 plus or minus 1/3 times square root of 7

OR

(-2 plus or minus square root of 7)/3

OR

-2/3 plus or minus (square root of 7)/3

8. plus and minus 3 ? These are correct.

9. x+5/ x+2 ? This is not a solution; it is an algebraic ratio in x. Exercise 9 has one solution. (x = ?)

10. 4 ? x = 4 is correct.

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