Multiplying monomials

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Posted on Jan 02, 2017

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SOURCE: Solving matrix

Go to your matrix button and enter a "3x4" matrix.

Then enter it as follows:

-3 4 5 7

4 3 2 9

-5 5 3 -10

Then exit out and go to "2nd->matrix->math->rref(". Then press enter.

Your screen should look like this:

rref(

Then go to matrix and select your 3x4 matrix, press enter and close it with a parathesis. Your screen should look like this:

rref([A])

Press enter and the screen should say this:

1 0 0 2

0 1 0 -3

0 0 1 5

So,

x=3

y=-3

z=5

Hope this cleared up the confusion!

SJ_Sharks

Posted on Mar 14, 2009

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SOURCE: x-3y=5

2y - x = 3

x = 3y - 5

Add the two equations side by side,

2y - x + x = 3 + 3y - 5

2y = 3y - 2

y = 2

Plug this in the second equation to get x,

x = 3(2) - 5

x = 1

So the solution is **x = 1**, **y = 2**

or in ordered pair notation **(1, 2)**

Posted on Jun 29, 2009

SOURCE: How do you use a fx-9750GII calculator and inverse

One not way would be to define a 4x4 matrix Mat A to hold the coefficients of the linear system. Then define a 4x1 column vector Mat V to hold the constants on the right.

Define a third 4x4 matrix Mat B you may leave filled with 0.

On command line, in Run Mat screen enter (Mat A) ^(-1) and store it in the zero-filled matrix Mat B. this is the inverse of Mat A.

If the inverse of Mat A exists, and it does in this case, the solution of the system is obtained as the column vector, resulting from the multiplication of Mat B by column vector Mat V

You can even shorten the procedure by just calculating ((Mat A)^-1)X (Mat V) [EXE]

To summarize

- Create 4x4 Mat A and type in the coefficients of the linear system.
- Create a 4x1 column vector Mat V for the right-hand sides
- Obtain you solution vector as ((Mat A)^-1)X (Mat V) [EXE]

- use catalog or
- in RunMat screen, press [OPTN] followed by [F2:Mat], then [F1:Mat].
- At this point the identifier is on command line, and you have to press [ALPHA] [X,Theta, T] to enter letter A.
- You use a similar key sequence to enter Mat V

Multiplication operator is the regular [times] key.

Posted on Nov 16, 2010

Raise each component to the same power

(ax^2y^3z^5)^6=(a^6)(x^12)(y^18)(z^30)

(ax^2y^3z^5)^6=(a^6)(x^12)(y^18)(z^30)

Aug 28, 2014 | Office Equipment & Supplies

Write the equality in the form y=(5X+3)/(4X-5). Insert parentheses to ensure a correct result.

- Multiply both sides of the equality by (4X-5). This gives (4X-5)y=(5X+3).
- Open the parentheses as 4Xy-5y=5X+3
- Subtract 5X from both sides 4Xy-5y-5X=5X-5X+3
- Add 5y to both sides 4Xy-5X-5y+5y=5y+3 or 4Xy-5X=5y+3
- Factor the X on the left side X(4y-5)=5y+3
- If 4y-5 does not vanish, you can isolate X by dividing both members of the equality by (4y-5).
- You get X=(5y+3)/(4y-5)=f(y)

Jun 24, 2012 | Mathsoft StudyWorks! Mathematics Deluxe...

Todd is 4 years old.

Let Tom's age = X

Let Todd's age = Y

From the given facts:

X = 5Y

and

X + 7 + 6 = 3 ( Y + 7 )

X+13 = 3Y + 21

X = 3Y + 8

So if:

X=5Y

and

X=3Y+8

Then:

5Y=3Y+8

2Y=8

Y=4

I hope that helps.

Joe.

Let Tom's age = X

Let Todd's age = Y

From the given facts:

X = 5Y

and

X + 7 + 6 = 3 ( Y + 7 )

X+13 = 3Y + 21

X = 3Y + 8

So if:

X=5Y

and

X=3Y+8

Then:

5Y=3Y+8

2Y=8

Y=4

I hope that helps.

Joe.

Sep 14, 2011 | Office Equipment & Supplies

Price of bag of chips s X and price of box of pretzels is Y.

Now you can write following equations:

2X+3Y=413

X+2Y=239

From second equation you know that X is 239-2Y, and you put that in first equation:

2*(239-2Y)+3Y=413

or

-4Y+3Y=413-2*239

Finally we have for Y: Y=65 and X=239-2Y=109.

Now you can write following equations:

2X+3Y=413

X+2Y=239

From second equation you know that X is 239-2Y, and you put that in first equation:

2*(239-2Y)+3Y=413

or

-4Y+3Y=413-2*239

Finally we have for Y: Y=65 and X=239-2Y=109.

Aug 13, 2011 | Office Equipment & Supplies

This should start wit X=something and Y=something, sorry I'm not an human algebra calculator....

Jul 29, 2011 | Computers & Internet

In Algebra

Likewise when you see

Special Binomial Products So when you multiply binomials you get ... Binomial Products

And we are going to look at

1. Multiplying a Binomial by Itself What happens when you square a binomial (in other words, multiply it by itself) .. ?

(a+b)2 = (a+b)(a+b) = ... ?

The result:

(a+b)2 = a2 + 2ab + b2

You can easily see why it works, in this diagram:

2. Subtract Times Subtract
And what happens if you square a binomial with a **minus** inside?

(a-b)2 = (a-b)(a-b) = ... ?

The result:

(a-b)2 = a2 - 2ab + b2

3. Add Times Subtract
And then there is one more special case... what if you multiply (a+b) by (a-b) ?

(a+b)(a-b) = ... ?

The result:

(a+b)(a-b) = a2 - b2

That was interesting! It ended up very simple.

And it is called the "**difference of two squares**" (the two squares are **a2** and **b2**).

This illustration may help you see why it works:

a2 - b2 is equal to (a+b)(a-b)
Note: it does not matter if (a-b) comes first:

(a-b)(a+b) = a2 - b2

The Three Cases
Here are the three results we just got:

(a+b)2
= a2 + 2ab + b2
} (the "perfect square trinomials")
(a-b)2
= a2 - 2ab + b2
(a+b)(a-b)
= a2 - b2
(the "difference of squares")
Remember those patterns, they will save you time and help you solve many algebra puzzles.

Using Them
So far we have just used "a" and "b", but they could be anything.

Example: (y+1)2

We can use the (a+b)2 case where "a" is y, and "b" is 1:

(y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1

Example: (3x-4)2

We can use the (a-b)2 case where "a" is 3x, and "b" is 4:

(3x-4)2 = (3x)2 - 2(3x)(4) + (4)2 = 9x2 - 24x + 16

Example: (4y+2)(4y-2)

We know that the result will be the difference of two squares, because:

(a+b)(a-b) = a2 - b2

so:

(4y+2)(4y-2) = (4y)2 - (2)2 = 16y2 - 4

Sometimes you can recognize the pattern of the answer:

Example: can you work out which binomials to multiply to get 4x2 - 9

Hmmm... is that the difference of two squares?

Yes! **4x2** is **(2x)2**, and **9** is **(3)2**, so we have:

4x2 - 9 = (2x)2 - (3)2

And that can be produced by the difference of squares formula:

(a+b)(a-b) = a2 - b2

Like this ("a" is 2x, and "b" is 3):

(2x+3)(2x-3) = (2x)2 - (3)2 = 4x2 - 9

So the answer is that you can multiply **(2x+3)** and **(2x-3)** to get **4x2 - 9**

Jul 26, 2011 | Computers & Internet

3X-5y z=6. 2x 4y z=1.x 2y 9z=-2

Feb 20, 2011 | Office Equipment & Supplies

(2x - 3y)(3x + 2y) = 2x*3x + 2x*2y - 3y*3x - 3y*2y

= 6x^2 + 4xy - 9xy - 6y^2

= 6x^2 - 5xy - 6y^2 You multiply each element in the first set of brackets by each element in the second set of brackets and then consolidate like terms and arrange them in sequence of powers, first x and then y. So:

2x by 3x = 6x squared

2x by 2y = 4xy

-3y by 3x = -9xy

-3y by 2y = 6y squared

6x squared -5xy + 6y squared

= 6x^2 + 4xy - 9xy - 6y^2

= 6x^2 - 5xy - 6y^2 You multiply each element in the first set of brackets by each element in the second set of brackets and then consolidate like terms and arrange them in sequence of powers, first x and then y. So:

2x by 3x = 6x squared

2x by 2y = 4xy

-3y by 3x = -9xy

-3y by 2y = 6y squared

6x squared -5xy + 6y squared

Dec 04, 2010 | Yahoo Computers & Internet

You can solve it with following method.

5x+3y=6 2x-4y=5

So 5x=6-3y so 2[(6-3y)/5]-4y=5

So x=(6-3y)/5 so 12-6y-20y=25

so -26y=25-12

so -26y=13

so y= -(1/2)

2x-4y=5

so 2x=5+4y

so 2x=5+4(-1/2)

so 2x=(10-4)/2

so 2x=6/4

so x =3/2

The value of x=3/2 and value of y= -1/2

Let me know if you need further assistance.

Thanks for using FixYa.

5x+3y=6 2x-4y=5

So 5x=6-3y so 2[(6-3y)/5]-4y=5

So x=(6-3y)/5 so 12-6y-20y=25

so -26y=25-12

so -26y=13

so y= -(1/2)

2x-4y=5

so 2x=5+4y

so 2x=5+4(-1/2)

so 2x=(10-4)/2

so 2x=6/4

so x =3/2

The value of x=3/2 and value of y= -1/2

Let me know if you need further assistance.

Thanks for using FixYa.

Mar 03, 2010 | Office Equipment & Supplies

4x+8y=20 || /4

x+2y=5

x=5-2y

-4x+2y=-30 || /2

-2x+y=-15

-2(5-2y)+y=-15

-10+4y+y=-15

5y=-5

y=-1

x=5-2y || y=-1

x=5-2(-1)

x=5+2

x=7

x+2y=5

x=5-2y

-4x+2y=-30 || /2

-2x+y=-15

-2(5-2y)+y=-15

-10+4y+y=-15

5y=-5

y=-1

x=5-2y || y=-1

x=5-2(-1)

x=5+2

x=7

Mar 10, 2009 | Bagatrix Algebra Solved! 2005 (105101) for...

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