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help solving solutions

 
1.  Simplify    x-y    +    x+y
                    x+y          x-y
 
 
I have zero since x-y crosses out x-y and x+y crosses out x+y
 
 
2.  solve 12x^2 + 7x = 10
    
I am trying to use (-3x + 4) (-4x + 3)
 
3.  simplify    x^2 -5x + 4
                      x -1
 
  my answer is x - 4.  Is this correct?
 
4.  simplify  9x^2 - 3x
                     3x
 
 my answer is 3x-1.  Is this correct?
 
 
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4 Answers

1. In this case, you are supposed to treat each combination on top and bottom as a single unit, like this:
(x-y)/(x+y) + (x+y)/(x-y)
Since you are adding, you find a common denominator. The common denominator in this case is (x+y)(x-y). We need to multiply each fraction to get this as a denominator. Remember to do the same thing to the numerator that you do to the denominator. When you do this, you get:
(x-y)(x-y)/(x+y)(x-y) + (x+y)(x+y)/(x-y)(x+y)
You now foil the numerators, and you get this:
(x^2-xy-xy+y^2)/(x+y)(x-y) + (x^2+xy+xy+y^2)/(x+y)(x-y)
combine like terms on top, and FOIL the bottom. When you do, you get this:
(2x^2+2y^2)/(x^2-y^2)
From here you should be able to figure it out.

2. Melissa was correct in pointing out that you were supposed to subtract 10 from both sides, not add. The result is a trinomial that looks like this:
12x^2 + 7x - 10 = 0
This is expressed algebraically by substituting constants for the coefficients, like this:
ax^2 + bx + c = 0
If you have a leading coefficient (the 'a' in the equation) that is NOT 1, here is what you do:
Multiply the 'a' and 'c' coefficients together. In this case, the answer is -120.
Now you find the factors of 120 that, when added, equal our 'b' coefficient, which is +7, in our case. Here is the process, quickly:
"What times what equals 120?"
1 120
2 60
3 40
4 30
5 24
6 20
8 15
10 12
These are all the factors of 120. We know that one has to be a negative number, since our 'c' coefficient is negative. Looking at our list, only 8, 15 provides this for us. If we used -8 and +15, when we add them we get +7, and when we multiply them we get -120. This is what we are looking for.
Stay with me now because it gets a bit tricky.
We use the two numbers we just found (-8, +15) and they replace our 'b' term. We attach each one to x. We combine one of the new numbers with the first term ('a' and variable), and the other with the third term ('c'). It doesn't matter which one we use with which. Here is how I did it:
(12x^2 - 8x) (+ 15x - 10)
We factor any common terms. Now the terms in the parenthesis should be the same. If they aren't, you did something wrong:
4x(3x - 2) 5(3x - 2)
Since the two sets in parenthesis match, as they should, we just use that set as one binomial:
(3x - 2)
And now we combine the two terms outside the parenthesis to make a new binomial:
(4x + 5)
Our new equation looks like this:
(3x - 2)(4x + 5) = 0
If the set within either parenthesis equals zero, it will multiply with the other set to make our equation 0=0. All we need to do is find out what value of x will result in a zero answer for one of our parenthesis sets. In order to do this, we take each binomial and equal it to zero, then solve for x. I will work the first one:
3x - 2 = 0
3x = 2
x = 2/3
If we plug this into the first binomial, then the binomial will equal zero, and when we multiply it by WHATEVER is in the other binomial, we will get zero, as well. Therefore, 2/3 is one of the solutions. There is one more. I will leave it to you to find.

3. That's what I got

4. That's what I got

Hopefully this helps! Please feel free to contact me if you have any questions. Thanks!

Comments

This "(12x^2 - 8x) (+ 15x - 10)" implies multiplication of two numbers, which is wrong.
It should be "(12x^2 - 8x) + (15x - 10)", the addition of two numbers.
Maybe technically, but since the binomials are multiplied later to equal zero, I felt this would not be useful for the student to understand the problem. In fact, I felt it would distract from learning, in this case. Different techniques. Hopefully she gets it, one way or the other.
In this case, yes, if the whole "technically they are added" doesn't confuse you. Sometimes the first term in the second parenthesis is a negative, and it impacts the problem dramatically, which is why I skipped over that. After all, you needed help on this specific problem, not theoretical problems that may come your way.
Congrats for figuring out the other solution! It sounds like you are getting it. 
For problem #1
 
x-y/x +y   +  x+y/x-y,  I got 2+2=4   (Is this correct?)
The final answer is 2(x2 + y2)/ (x2 - y2)
Sorry for any confusion!
1. Simplify (x-y)/(x+y) + (x+y)/(x-y)

= (x-y)^2/((x-y)(x+y)) + (x+y)^2/((x+y)(x-y))

= ((x-y)^2 + (x+y)^2)/((x+y)(x-y))

= (x^2-2xy+y^2 + x^2+2xy+y^2)/(x^2-y^2)

= 2(x^2+y^2)/(x^2-y^2)

NOTICE MINUS SIGN IN DENOMINATOR

2. solve 12x^2 + 7x = 10

12x^2 + 7x – 10 = 0

(ax+b)(cx+d)=(ac)x^2+(ad+bc)x+(bd)
ad and bc are factors of (ac)(bd)

12(-10)=-120

1(-120)
2(-60)
3(-40)
4(-30)
5(-24)
6(-20)
8(-15)
-8(15) <= use these

12x^2 – 8x + 15x – 10 = 0

Factor by Grouping:

(12x^2 – 8x) + (15x – 10) = 0

Take out Greatest Common Factors:

(12x^2 – 8x) + (15x – 10) = 0

4x(3x - 2) + 5(3x - 2) = 0

Both terms have common factor of 3x-2:

(3x - 2)(4x + 5) = 0

Use Zero Product Property:

3x-2 = 0 or 4x+5 = 0

x = 2/3 or x = –5/4


3. Simplify f(x) = (x^2 - 5x + 4)/(x - 1)

Can’t divide by zero, so x ≠ 1.

f(x) = (x - 4) (x - 1) / (x - 1)

Notice that (x-1)/(x-1) = 1 if x ≠ 1,
but (x-1)/(x-1) is UNDEFINED if x = 1.
So the simplified f(x) MUST have the
constraint x ≠ 1 as part of its definition.

f(x) = x - 4, x ≠ 1


4. simplify f(x) = (9x^2 - 3x)/(3x)

Can’t divide by zero, so x ≠ 0.

f(x) = (3x)^2/(3x) - (3x)/(3x), x ≠ 0

Again, x/x = {1 if x ≠ 0, UNDEFINED if x = 0}

f(x) = 3x - 1, x ≠ 0.
1.) It looks like the problem is addition of algebraic fractions.  If so, just like normal fractions (2/3 +1/4) you must find a common denominator.  (It looks like you are treating it as if the problem is multiplication, and you are canceling.)
 
2.) You do solve by factoring, but you don't quite have it.  Remember, if the problem is 12x^2 + 7x =10, you must first move the 10 over to the left side so that the equation is equal to 0.  Now, factor.  You must think of all the factors of 12 (1x12, 2x6, 3x4) and all the factors of 10 (1x10, 2x5) and how these combine.  It is a bit tricky, but keep trying different combinations.  You'll get it!
 
3.) That is what I got!
 
4.) I got the same answer, too!
 No, absolutely Not
 
  You can not cross out(X-y) 's when there is + sign there.
  
    This business of crossing out is a big epidemic symptom among many student. 
 
 
      Think :  
 
           4   + 1       the result is 2 1/4  = 9/2
           2      4                          
 
           If you cross out the 4 's , end up with 1 ,which is dead wrong.
                                                                    2
 
           Crossing out comes from simplifying  a fraction.
 
               16/12 = ( 2. 2. 2.2 ) / ( 2 .2. 3 ) = 4/3
 
               Here we are crossing out two factors of 2 from numerator and denominator by dividing both of
               them with the same common factor to simplifying our fractions to 4/3
 
             Now if we have
 
             16/ 5  * 7/ 12  Since multiplication of fractions implies multiplying numerator and denominators of
                                    both fractions to get the numerator and denominator of the product , and 
               multiplication is Commutative( factors can change their position) , we can cross out factors of 4
               from 16 and 12.
 
               16/ 5 * 7/12 = 4/5 * 7/3 = 28/ 15
 
               Keep in mind that Crossing out can only be done when there is common factor exists in
                 between numerator of one fraction and denominator of other, multiplying 2 or
                 more fractions together.  
 
          2.
               to solve the quadratic
 
                 12X^2 + 7x = 10
 
                   12 X^2 + 7X - 10 =0
 
                     Move everything to one side , let it equal 0 , and solve it either by factoring or using
                   quadratic formula.