Understanding Algebra 1: Multiplying (AX + B)(C - D) Simplified

[pointintime]

Does this...

(AX + B)(C - D)

when you multiply

the first numbers... AX and C

mean the same thing as

(AX)C -D(AX) +BC -BD

or this

AXC - DAX + BC - BD

which one is it?

 

[pointintime]

Please I need to know

 

[Дьявол]

Does this...

(AX + B)(C - D)

when you multiply

the first numbers... AX and C

mean the same thing as

(AX)C -D(AX) +BC -BD

or this

AXC - DAX + BC - BD

which one is it?

Both are valid. The brackets do not change anything at this point.

Regards.

 

[pointintime]

ok well the reason why I'm asking because in a physics problem it does so...

like which one is it??

like without the brackets really changes it

if you put the brackets on

 

[Дьявол]

Could you possibly explain what A,X,B,C,D are? Are they numbers, matrices or what? Any concrete problem?

Regards.

 

[pointintime]

ya sure...

- (180 s)(a t3)

this was line before

-[Vo t3 + 2^-1 a t3^2] = 180 s (a t3) - a t3^2 + Vo (180 s) - Vo t3 - 1,100 m + Xo

line before that

-[Vo t3 + 2^-1 a t3^2] = (a t3 + Vo) (180 s - t3) - 1,100 m + Xo

so there would be no ( ) around a t3?

like aren't you doing this...

(a t3)(180 s)
?

and here's the line before that line

1,100 m - [Xo + Vo t3 + 2^-1 a t3^2] = (a t3 + Vo) (180 s - t3)

 

[pointintime]

-[Vo t3 + 2^-1 a t3^2] = 180 s (a t3) - a t3^2 + Vo (180 s) - Vo t3 - 1,100 m + Xo

see like in that line

would you get this

(180 s)a + (180 s)t3

or just (180 s)a t3

 

[pointintime]

were s is seconds

and the 3 is just a subscript

 

[pointintime]

please someone!

 

[symbolipoint]

You really need to understand commutative and associative properties of equality for Addition and Multiplication. Also you must understand the distributive property. This is year-1 Algebra stuff which you must learn to know formally and intuitively. Much of your progress relies on these properties and a few others.

 

[pointintime]

ok...
so which one is it?

 

[pointintime]

oh come on

 

[belliott4488]

You're still not getting it ...

A(BC) = ABC

The parentheses make no difference so long as there are no sums in them.

A(B+C), on the other hand, would give you AB + BC.

Can you see where you were wrong above now? In post #7 one option is definitely correct and the other is definitely incorrect.

 

[pointintime]

ok yes just got anohter question...

Ok well I got down to this and don't know what to do please someone help me...

-2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)

how do I solve this for t3

the 3 is a subscript to t
180 s is 180 seconds and is considered one term
Vo is one term
Xo is one term

 

[belliott4488]

Everything in this equation is a constant except t3, correct?

If so, then you've two terms with t3^2, one with t3 to the first power, and one term with no t3 (on the right side), i.e. a constant.

That makes this a quadratic equation. Do you remember how to solve them?

I'm guessing that you're not all that comfortable with algebra, given your original question, but maybe you are (?). If not, though, you need to brush up on it.

 

[pointintime]

I rember how to solve them but not when it's in this format

 

[pointintime]

y = ax^2 + bx + C

I know how to solve for x... I think... it's been a while though please help...

sorry lol

 

[belliott4488]

Okay ... take what I'm about to type and have it tatooed on the back of one of your hands.

if you have ax^2 + bx + c = 0 (no y, BTW - we're dealing with only one variable here),

then the solution for x is given by the quadratic formula:

x = (-b +/-sqrt(b^2 - 4ac))/2a ("sqrt" = square root)

You must learn this now and never again forget it! ;-)

So ... you need to rearrange your equation for t3 so that it's in the form above, which you can do since you have terms with t3^2, t3 and constants, just like above for x.

 

[pointintime]

Ok I graphed into a graphic calculator

y1 = -2 X^2 - 180 x + X^2

y2 = .2^-1 (5.494*180-1100)

and got this

(3.111, -555.4)

I assume I need the 3.111 seconds for my answer
but sense this is physics not algebra how do I rearange for y3?

 

[pointintime]

ok but what does this have to do with the problem

.2^-1 (5.494*180-1100)

the equation on the right side of the equal sign

I just ignore it...?

 

[belliott4488]

Ok I graphed into a graphic calculator

y1 = -2 X^2 - 180 x + X^2

y2 = .2^-1 (5.494*180-1100)

and got this

(3.111, -555.4)

I assume I need the 3.111 seconds for my answer
but sense this is physics not algebra how do I rearange for y3?

Whoa ... I have no idea where you're going with that ... what's y??

 

[pointintime]

x = (-b +/-sqrt(b^2 - 4ac))/2a ("sqrt" = square root)

so is the 2a part...

the a is -2^-1 so does it become part of numerator and I get this?
x = a(-b +/-sqrt(b^2 - 4ac))/2 ("sqrt" = square root)

 

[belliott4488]

ok but what does this have to do with the problem

.2^-1 (5.494*180-1100)

the equation on the right side of the equal sign

I just ignore it...?

Of course not! That's part of the equation. You need to have a zero on the right side to make it look like ax^2 + bx + c = 0, so what to you have to do to get zero on the right side?

 

[pointintime]

i have no idea but the 2a part... read #22 please
also why do I just ignore the other part of the equation that seems rather strange

 

[belliott4488]

x = (-b +/-sqrt(b^2 - 4ac))/2a ("sqrt" = square root)

so is the 2a part...

the a is -2^-1 so does it become part of numerator and I get this?
x = a(-b +/-sqrt(b^2 - 4ac))/2 ("sqrt" = square root)

No, first you have to get the equation into the standard form so that you can clearly identify a, b, and c.

a is not simply -2^-1. Gather all your t3^2 terms together.

 

[belliott4488]

We're leapfrogging posts, here ... I'm going to back off for a couple of minutes until you've read all the posts and replied.

 

[pointintime]

I have no idea just move it all over to the left?

wow in physics you like try to rearange for just like one variable and i don't know this is kinda weird...

 

[pointintime]

I don't know how to get this mess into standard form...

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)

 

[belliott4488]

I have no idea just move it all over to the left?

Well, yeah ... so long as you do it correctly.

wow in physics you like try to rearange for just like one variable and i don't know this is kinda weird...

This is not really about Physics. At this point you're just trying to solve an algebra problem. But yes, you certainly have to use algebra to solve Physics problems, and yes that often requires rearranging equations to get them into a form that you know how to solve.

 

[pointintime]

ok I'll atempt to set it to zero

 

[belliott4488]

I don't know how to get this mess into standard form...

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)

Okay, here's the deal: you can combine terms that have the same power of the variable that you're solving for. The highest power of t3 that you have is 2, so that means you can put all of this on the left side of the equal sign (leaving zero on the right), and then combine it all into no more than 3 terms: one that is t3^2 multiplied by some constant (which we'll identify as a), another that is t3^1 (=t3) multiplied by some constant (which we'll identify as b), and the third which is t3^0 (or just 1) multiplied by some constant (which we'll identify as c).

(Since t3^0 is just 1, you don't write it, so that's why the third term by itself is just c.)

Give that a shot.

 

[pointintime]

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)
(Vo (180 s) - 1100 m + Xo)^-1 a(-2^-1 t3^2 - 180 s (t3) + t3^2) = 0

:O

 

[pointintime]

-2^-1 t3^2 - 180 s (t3) + t3^2

ok how do I combine

-2^-1 t3^2

and

t3^2

:O

 

[pointintime]

sorry I'm kinda bad

 

[pointintime]

Someone told me this was the answer don't know were it comes from

t² - 360t = (360v-2s)/a