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Recent problems solved by 'Math_Teacher_37'

Math_Teacher_37 answered: 4 problems
Linear-systems/449742: y=2x^2,y=x+3 solve using system of equations algebraically.
1 solutions

Answer 309374 by Math_Teacher_37(4) About Me  on 2011-05-15 19:58:53 (Show Source):
You can put this solution on YOUR website!
+y+=+2x%5E2 and y+=+x+%2B+3
Since +y+=+2x%5E2 and y+=+x+%2B+3, this means that 2x%5E2+=+x+%2B+3.
Now, you have to solve for x, using the quadratic formula. Begin by moving the terms on the right side of the equation to the left side:
2x%5E2+-+x+-+3+=+0
Now, in the quadratic form, ax%5E2+%2B+bx+%2B+c, a = 2, b = -1, and c = -3. Substitute these values into the quadratic formula, as shown below.
x+=+%28-b+%2B-sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29

x+=+%28-%28-1%29+%2B-sqrt%28%28-1%29%5E2+-+4%2A2%2A-3%29%29%2F%282%2A2%29

x+=+%281+%2B-sqrt%281+%2B+24%29%29%2F%284%29

x+=+%281+%2B-sqrt%2825%29%29%2F%284%29

x+=+%281+%2B-+5%29%2F4

x+=+%281%2B5%29%2F4 OR x+=+%281-5%29%2F4

x+=+6%2F4 OR x+=+-4%2F4

x+=+3%2F2 OR x+=+-1

The solutions are x = -1 OR x = 3/2.


Linear-systems/449380: I need help with this question please:
Solve the system of equations
x + y = 5
x - y = -9
1 solutions

Answer 309125 by Math_Teacher_37(4) About Me  on 2011-05-14 19:13:01 (Show Source):
You can put this solution on YOUR website!
We are trying to solve the following system of equations:
x + y = 5
x - y = -9
Notice that the "y" terms are opposite in sign so we can simply add the equations together, like this:
2x + 0y = -4
2x = -4
Divide both sides by 2: x = -4/2 = -2
Now, x = -2 so substitute into either equation. Since the first equation involves addition and it looks easier, let's go with this one. We have:
x + y = 5
-2 + y = 5
Add 2 to both sides:
-2 + 2 + y = 5 + 2
y = 7
The solution is: (-2, 7)


Quadratic_Equations/449377: X^2=-14x-40 Help pleasee what do i do when the x squared is in the other side?????ASAAPP!
1 solutions

Answer 309123 by Math_Teacher_37(4) About Me  on 2011-05-14 19:08:28 (Show Source):
You can put this solution on YOUR website!
We are trying to solve: x%5E2+=+-14x+-+40
First, bring the terms on the right side of the equation to the left side by doing the opposite (add 14x and add 40 to both sides):
x%5E2+%2B+14x+%2B+40+=+0
Now, try to factor the above expression. Since 40 = 10 * 4 and 10 + 4 = 14 (the middle term), the expression above factors into the following:
(x + 10)(x + 4) = 0
Next, we have x + 10 = 0 OR x + 4 = 0, so x = -10 or x = -4.
The solutions are -4 and -10.


Quadratic-relations-and-conic-sections/449330: how do you put 5x^2-y^2-30x-12y+9=0 into standard graphing form and graph it?
1 solutions

Answer 309108 by Math_Teacher_37(4) About Me  on 2011-05-14 17:08:20 (Show Source):
You can put this solution on YOUR website!
5x%5E2-y%5E2-30x-12y%2B9=0
Begin by moving all the "x" and "y" terms together, like this:
5x%5E2-30x-y%5E2-12y%2B9=0
Now move the "9" to the other side and factor by grouping:
5x%5E2-30x + -y%5E2-12y = -9
For 5x%5E2-30x, factor out a "5":
5%28x%5E2-6x%29
And for -y%5E2-12y, factor out a "-1":

(to make the number in front of the y%5E2 term a positive number):
-1%28y%5E2%2B12y%29
Next, complete the square for 5%28x%5E2-6x%29. Remember that you need to take 1/2 of the middle term (1/2*-6 = -3) and square that answer (-3*-3 = 9). Then add this number to the expression to complete the square:
5%28x%5E2-6x+%2B+9%29
Note that if you distribute the above expression out, you will get the following:
5x%5E2-30x+%2B+45 (This means you added 45 to the original expression, something you will need to do to the other side as well.)
Do the same thing for y%5E2%2B12y:
1/2 * 12 = 6
Square 6: 6 * 6 = 36
So, complete the square and get the following:
-1%28y%5E2%2B12y%2B36%29
Distribute and get: -1y%5E2-12y+-+36
Of course, you will need to add -36 to the other side as well.
In summary, we have added 45 and -36 to both sides of the original equation (and need to add these numbers to the right side as well):
%285x%5E2+-30x+%2B+45%29 + %28-1y%5E2+-+12y+-+36%29+=+-9+%2B+45+-+36
Simplify:

%285x%5E2+-30x+%2B+45%29 + %28-1y%5E2+-+12y+-+36%29+=+0
Factor a "5" out of 5x%5E2+-+30x+%2B+45:
5%28x%5E2-6x+%2B+9%29 = 5%28x+-+3%29%28x+-+3%29
Factor a "-1" out of -1y%5E2+-+12y+-+36:
-1%28y%5E2+%2B+12y+%2B+36%29 = -1%28y+-+6%29%28y+-+6%29
SO:
5%28x+-+3%29%5E2+-+1%28y+-+6%29%5E2+=+0
Since the right side of the equation equals 0, this conic section is not a hyperbola and cannot be graphed.