Tim Said:

How do you solve this kind of algebra problem?

We Answered:

I've been teaching algebra 1 and 2 for 21 years.
study this way :]

15a=3a+84
-3a -3a<------------------minus 3a from each side
--------------------
+12a=84<------------ divide each side by 12
-------- ------
12 12 As you can see a is equal to 8.


E N J O Y E N J O Y E N J O Y

Sarah Said:

What are the basic formulas to use when solving algebra problems?

We Answered:

450*.15= 67.53

450-67.53=382.5



382.5*.20=76.5

382.5-76.5=306



On this one I feel like I have it figured out, please tell me if I'm correct...

If Leah is 6 years older than Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue?

L=6+S
J=5+L

S+L+J=41


L=6+8=14
J=5+14=19
S+14+19=41
S+33=41
-33 -31

S=8

Sue is 8 years old?
Well done!!

Solve the following equation for A : 2A/3 = 8 + 4A


2A=24+12A

-10A=24
A=-12/5
A=-12/5

Manuel Said:

Help with an algebra problem dealing with Rational Expressions?

We Answered:

r^2 - 5r + 6 = (r-3)(r-2)

r^2 - 4 = (r+2)(r-2)

The (r-2)s cancel leaving answer A

Danielle Said:

Algebra Word Problems. Where's a good free online resource of them?

We Answered:

whats the answer for
5a+4b=4
4a+5b=31/8

Tim Said:

What are the best Math (algebra) tutorials online? I'm preparing myself for a math placement exam.?

We Answered:

Well I'm not so sure what kind of algebra it is, but I'm assuming it's algebra I or algebra II

Go to this site and click the book you want to learn about. then after that click where it's bolded Homework Video Tutors. then select the lesson.

Good luck. :]

Rick Said:

Need help with Algebra 2 problems for test tomorrow.?

We Answered:

Hi,

1. Find the equation in standard form with vertices of (3, -6) and (3, 2) and foci of (3, -8) and (3, 4).

With foci further apart than the vertices, this is a hyperbola, which opens up and down since all points start with 3. The center is at (3,-2), the midpoint of the 2 foci or of the 2 vertices.

"a" = 4, the distance from the center to a vertex and "c" = 6, the distance from the center to a focus.
So a² = 16, c² = 36, and b² = 20 (from a² + b² = c²)

The equation of this hyperbola is:

y + 2. . x - 3
-------- - ------- = 1 <==ANSWER
. 16 . . . 20

2. x² - 4y² - 40y + 6x + 10 = 0 state center, vertices, foci.

x² - 4y² - 40y + 6x + 10 = 0

x² + 6x + __ - 4y² - 40y + __ = -10

x² + 6x + 9 - 4(y² - 10y + 25) = -10 + 9 - 100

(x + 3)² - 4(y - 5)² = -101

(y - 5)². . (x + 3)²
---------- - ----------- = 1
25.25 . . . 101

center: (-3,5)

vertices: (-3,5 + ?(101)/2), (-3,5 - ?(101)/2), (-3 - ?(101),5), (-3 + ?(101),5)

foci: (-3, 5 + ?(303)/2), (-3, 5 - ?(303)/2)



3. Find the ellipse in standard form: 3x² + 4y² + 12x + 24y - 12 = 0.

3x² + 4y² + 12x + 24y - 12 = 0

3x² + 12x + __ + 4y² + 24y + __ = 12

3(x² + 4x + __) + 4(y² + 6y + __) = 12

3(x² + 4x + 4) + 4(y² + 6y + 9) = 12

3(x + 2)² + 4(y + 3)² = 12 + 3(4) + 4(9)

3(x + 2)² + 4(y + 3)² = 60

(x + 2)² . (y + 3)²
---------- + ---------- = 1 <==ANSWER
. 20 . . . . . 15

4) y = a(x - p)(x - q) is the equation of a parabola with x intercepts at p and q.

y = a(x + 4)(x - 2)

-2 = a(-3 + 4)(-3 - 2)

-2 = a(1)(-5)

a = 2/5

y = 2/5(x + 4)(x - 2) <==ANSWER

I hope that helps!! :-)