Joanne Said:

I Don't Know How to do these Algebra 2 equations, Please help? Something about substitution?

We Answered:

well what u do is subsitute what each of those are....eg....since y=5x...u would put 5x in place of the y in (4x + 2y= 7)....to get
4x + 2(5x)= 7 --> 4x + 10x= 7 --> 14x=7 --> x= 2

and u would do that for the rest of them .....!!!

Veronica Said:

How do you solve these equations (algebra 2)?

We Answered:

1.
5/4 = n-3 / n-4
5(n-4) = 4(n-3) <-- cross multiplying
5n - 20 = 4n - 12 <-- opening the brackets
5n - 4n = -12 - (-20) <-- cross subtracting
n = -12 + 20 = 8 <-- simplifying

2.
1 / a-7 = 3
1 = 3(a-7)
1 = 3a - 21
1 - (-21) = 3a
22 = 3a
a = 22/3

3.
1/x - 3/4 = x/4
4x(1/x) - 4x(3/4) = 4x(x/4) <-- multiply through by 4x
4 - 3x = x^2 <-- simplify
0 = x^2 + 3x - 4 <-- cross subtract
0 = (x + 4)(x - 1) <-- factorise
x = 1 or -3

4.
14/x = 9-x
14 = x(9-x)
14 = 9x - x^2
x^2 - 9x + 14 = 0
(x - 7)(x - 2) = 0
x = 2 or 7

Herman Said:

How do you know how to classify equations? Algebra 2?

We Answered:

Here's a system of tests:

Are both variables squared?
No: It's a parabola. -> 5x+y^2=3y+1 and x^2-2y=1
Yes: Go to the next test....

Do the squared terms have opposite signs?
Yes: It's an hyperbola. -> 2x^2-2y^2+8x-15y+6=0
No: Go to the next test....

Are the squared terms multiplied by the same number?
Yes: It's a circle.
No: It's an ellipse -> x^2+2y62+8x-5y+6=0


I'm guessing your last one is
4x^2+4y^2+=8x-16y+6 - so from the tests above, its a circle,

Check out the site below - really good on this stuff

Justin Said:

help with these 2 cosine algebra equations? 10 points best answer!?

We Answered:

12. cos pi t = -1

pi t = pi
t = 1

7. ( 7/9 ) pi

20. I am assuming that it is pi over 5 not s?

so the answer is: t= 5/3

Now, if it is s then: t = s/3.


11. t = .7322

Ana Said:

How do you solve 2 step algebra equations? X/A=B?

We Answered:

Hi!

Well, in this equation we can't solve because there are multiple variables in just one equation. I think you mean solve FOR a variable. When you solve for a variable, all you need to do is get that variable by itself. So in this equation,

X/A = B

It is solved for B because B is by itself on one side. If you want to solve for X, we multiply A on both sides to get:

X = AB

Now if you wanted to solve for A, you would multiply by A:

X = AB

Then divide by B:

X/B = A

Now this is solved for A. So, if the equation is solved for B it looks like: B = X/A. If it is solved for X, it looks like: X = AB. And if it is solved for A, it looks like: A = X/B

I hope this helps!

Eddie Said:

ALGEBRA EQUATIONS 2 very small sections?

We Answered:

These are just solving equations. For solving equations, the goal is to get all the like terms on one side in order to isolate the variable.

I'll give you a few examples

1. add 8 to both sides to get the numbers on one side and the variable on the other.
b-8 +8 = -2 +8
b=6
2. do the same as in 1. x= -21
3. multiply both sides by -7.
(-y/7) ( -7) = 14 (-7)
y= -98
4. you isolate the x by multiplying it by the reciprocal coefficient, which is (5/4). x=35
5. r= -6

1. two steps, move the 8, divide by 3. b= -6
2. x=1
3. c= -28
4. x= -48

Rose Said:

Need help with 2 algebra equations.?

We Answered:

in order for this equation to be true
your going to need 14 kids on van A and 10 on van B
i used this equation ....
a+2=b-2


for the second problem it would be (x+50)

the reason its 50 is because 50*-2 = -100
(which is the last term in x^2+48-100)

also 50-2=-48
(which is your middle term in x^2+48-100)

these are good and interesting questions !

hope this helped