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Posted on Jan 02, 2017

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SOURCE: Factoring polynomials "Algebra"

Mode>choose "5"

choose equation format "4"

input the coefficient for a which is "1" and hit "="

input the coefficient for b which is "4" and hit "="

input the coefficient for c which is "3" and hit "="

input the coefficient for d which is "12" and hit "="

Hit "=" for X1

Hit "=" again for X2

Hit "=" again for X3

Posted on Oct 20, 2008

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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

SOURCE: 4x+3y=6 how do you solve for y

Hello,

I am not lecturing you but I would rather you understand how to do the manipulations involved in isolating a variable.

You want to isolate y, Ok

Start stripping it of all that is not y.

3y + 4x =6. (Addition is commutative, I can change the order of addition)

The term with y is** added **to 4x. If I want the term in y alone on one side I perform the** inverse operation of addition**, a **substraction**. I subtract 4x from both sides.

3y+4x-4x=6-4x. But 4x-4x=0, and we are left with

**3y= - 4x+6**

This operation is sometimes summarized as make one term change side while changing its sign

It would do no harm to put the right side of the foregoing equation in parentheses. I do that to avoid errors)

3y= (-4x+6).

Now y is **multiplied** by the number 3. To isolate y I have to perform the **inverse operation** of the multiplication, namely the division by 3

3y/3 =(-4x+6)/3. The left hand side is just y**y= (-4x+6)/3.**

While result is correct, I can also open the parentheses

y= -4x/3 +6/3**y= -(4/3)*x +2.**

Hope it helps.

Posted on Oct 30, 2009

SOURCE: 4x-y-2z=-7 -4x+8y+5z=46 8x-3y+z=13

Type solve(exp1 and exp2 and exp3, {x,y,z})

See cap images

Posted on Dec 13, 2011

then x = 3y-12, then 4(3y-12) +6y=-12, then (12y-48)+6y=-12, then 12y-48+6y=-12, then 18y=36, then y=2. x=3y-12, x=3(2)-12, x=6-12, x=-6.

Mar 05, 2015 | Office Equipment & Supplies

This is best written as two separate equations:

8x+3y = -23 and 34x+ 5y = -23

Solving the first one for x:

8x = -23-3y

x = -23/8 - 3/8y

Substituting this value for x into the second equation:

34(-23/8 - 3/8y) + 5y = -23

-97.75 - (34)(.375)y + 5y = -23

-97.75 - 12.75y + 5y = -23

-97.75 -7.75y = -23

-7.75y = 97.75-23=74.75

**y **= -74.75/7.75 =** -9.645161**

Substitution back into the equation for x:

x = -23/8 - 3/8(-9.645161)

x = -2.875 + 3.616935

**x** **=.741935**

8x+3y = -23 and 34x+ 5y = -23

Solving the first one for x:

8x = -23-3y

x = -23/8 - 3/8y

Substituting this value for x into the second equation:

34(-23/8 - 3/8y) + 5y = -23

-97.75 - (34)(.375)y + 5y = -23

-97.75 - 12.75y + 5y = -23

-97.75 -7.75y = -23

-7.75y = 97.75-23=74.75

Substitution back into the equation for x:

x = -23/8 - 3/8(-9.645161)

x = -2.875 + 3.616935

Dec 12, 2014 | Bagatrix Algebra Solved! 2005 (105101) for...

Let us backtrack so as to better jump.

An algebraic expression may contain one or several**algebraic terms**, separated by a plus sign or a or a minus and a sign.

Each algebraic term is the product of a constant coefficient and a power of some variable, or the powers of several variables.

Example of an algebraic term 3(x^2)(y^6)....

**If the exponents of the various powers are positive integers, the term is called a monomial. **In short no square roots, or fractionary powers of the variables appear in monomials. Thus 2/x, 3SQRT(x), or 1/x^5 are not monomials.

Finally, a polynomial is an**algebraic expression** made up of one or more monomials.

Example P(X)=(1/3)X^7-(SQRT(5)*X^4+ 16X-25 is a polynomial of degree 7 in the indeterminate/variable X

Q(X,Y)= 3(X^3)*Y^2 + 4X-5Y+10 is a polynomial of degree 5 in the variables X and Y.

An algebraic expression may contain one or several

Each algebraic term is the product of a constant coefficient and a power of some variable, or the powers of several variables.

Example of an algebraic term 3(x^2)(y^6)....

Finally, a polynomial is an

Example P(X)=(1/3)X^7-(SQRT(5)*X^4+ 16X-25 is a polynomial of degree 7 in the indeterminate/variable X

Q(X,Y)= 3(X^3)*Y^2 + 4X-5Y+10 is a polynomial of degree 5 in the variables X and Y.

Oct 01, 2014 | Office Equipment & Supplies

To find the solution, first find the value of y for each equation.

Then substitue one equation into the other so that you only the x variable left.

Then just solve for x.

Once you have a value for x, then you can easily solve for y.

So for the first equation:

3y - 6x = -3

3y = 6x - 3

**y = 2x - 1**

Now for the second equation:

2y + 8x = 10

2y = -8x + 10

**y = -4x + 5**

Since both equations equal y, they also equal each other, therefore:

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1

6x = 6

**x=1**

Now substitute x=1 into either original equation:

y = 2x - 1

y = 2 (1) - 1

y = 2 - 1

**y = 1**

Therefore the solution is x=1 and y=1

Good luck, I hope that helps.

Joe.

Then substitue one equation into the other so that you only the x variable left.

Then just solve for x.

Once you have a value for x, then you can easily solve for y.

So for the first equation:

3y - 6x = -3

3y = 6x - 3

Now for the second equation:

2y + 8x = 10

2y = -8x + 10

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1

6x = 6

y = 2x - 1

y = 2 (1) - 1

y = 2 - 1

Good luck, I hope that helps.

Joe.

Nov 09, 2011 | Texas Instruments TI-84 Plus Silver...

There are an infinite number of polynomials with those roots. Assuming you want one with the lowest degree, here are two:

x^3 - 2x^2 - 5x + 6

2x^3 - 4x^2 - 10x + 12

Since the roots of the polynomials are -2, 1, and 3, the values (x+2), (x-1), and (x-3) must be zero.

To get these polynomials, simply multiply

k * (x+2) * (x-1) * (x-3)

where k is any nonzero value.

x^3 - 2x^2 - 5x + 6

2x^3 - 4x^2 - 10x + 12

Since the roots of the polynomials are -2, 1, and 3, the values (x+2), (x-1), and (x-3) must be zero.

To get these polynomials, simply multiply

k * (x+2) * (x-1) * (x-3)

where k is any nonzero value.

Oct 03, 2011 | Texas Instruments TI-84 Plus Calculator

We can write this polynomial as:

You can see this polynomial in following picture:

Notice that it intersects x axis for x=-2, 1 and 3 (because these are roots of polynomial).

- (x-(-2))*(x-1)*(x-3)=
- (x+2)(x-1)(x-3)=
- (x+2)[x*(x-3)-1*(x-3)]=
- (x+2)*(x^2-3x-x+3)=
- (x+2)(x^2-4x+3)=
- x*(x^2-4x+3)+2*(x^2-4x+3)=
- x^3-4x^2+3x+2x^2-8x+6=
- x^3-2x^2-5x+6

You can see this polynomial in following picture:

Notice that it intersects x axis for x=-2, 1 and 3 (because these are roots of polynomial).

Oct 03, 2011 | Office Equipment & Supplies

One way is to use the Polynomial Root Finder and Simultaneous Equation Solver app.

Press the APPS key, then select PlySmlt2 and press ENTER. Press ENTER to get past the opening screen, then select "POLY ROOT FINDER" and press ENTER. Select the order of the polynomial and other settings as desired. Press F5 (the GRAPH key) to go to the next screen. Enter the polynomial coefficients, then press F5 to solve. The next screen will show you the roots (unless you selected real roots and the polynomial doesn't have any real roots).

If you don't have the app installed, you can download it from

http://education.ti.com/educationportal/sites/US/productDetail/us_poly_83_84.html

Press the APPS key, then select PlySmlt2 and press ENTER. Press ENTER to get past the opening screen, then select "POLY ROOT FINDER" and press ENTER. Select the order of the polynomial and other settings as desired. Press F5 (the GRAPH key) to go to the next screen. Enter the polynomial coefficients, then press F5 to solve. The next screen will show you the roots (unless you selected real roots and the polynomial doesn't have any real roots).

If you don't have the app installed, you can download it from

http://education.ti.com/educationportal/sites/US/productDetail/us_poly_83_84.html

Jan 20, 2011 | Texas Instruments TI-83 Plus Calculator

The short story is that this calculator does have a computer algebra system or CAS and thus cannot factor polynomials with arbitrary (unknown) coefficients or known coefficients.

However if the coefficients are given you can ,if you are willing to travel that way, factor approximately a polynomial P(x).

Basically, the idea is that any polynomial P(X) of degree n can be written in the factored form (X-x_1)(X-x_2)...(X-x_n), where x_1, x_2, x_3,...x_n are the roots (real or complex) of the equation P(X)=0.

The procedure ( for a 3rd degree polynomial) is as follows:

If you want to factor a cubic polynomial P3(X) = aX^3 bX^2 cX d , you write the corresponding cubic equation as**aX^3 bX^2 cX d =0** , then you divide all terms of the equation by** a** to obtain

**X^3 (b/a)X^2 (c/a)X (d/a)=0.**

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots**X1,X2,and X3.**
Then the polynomial X^3 (b/a)X^2 (c/a)X (d/a) can be cast in the
factored form (X-X1)(X-X2)(X-X3) and the original polynomial P3(X) can
be written as

**P3(X) = a*(X-X1)(X-X2)(X-X3) **

You can handle the quadratic polynomial the same way.

P2(X) =a*(X-X1)(X-X2) where X1, X2 are the two real roots.

However if the coefficients are given you can ,if you are willing to travel that way, factor approximately a polynomial P(x).

Basically, the idea is that any polynomial P(X) of degree n can be written in the factored form (X-x_1)(X-x_2)...(X-x_n), where x_1, x_2, x_3,...x_n are the roots (real or complex) of the equation P(X)=0.

The procedure ( for a 3rd degree polynomial) is as follows:

If you want to factor a cubic polynomial P3(X) = aX^3 bX^2 cX d , you write the corresponding cubic equation as

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots

You can handle the quadratic polynomial the same way.

P2(X) =a*(X-X1)(X-X2) where X1, X2 are the two real roots.

Sep 11, 2010 | Casio FX-9750GPlus Calculator

Hello,

I am not lecturing you but I would rather you understand how to do the manipulations involved in isolating a variable.

You want to isolate y, Ok

Start stripping it of all that is not y.

3y + 4x =6. (Addition is commutative, I can change the order of addition)

The term with y is** added **to 4x. If I want the term in y alone on one side I perform the** inverse operation of addition**, a **substraction**. I subtract 4x from both sides.

3y+4x-4x=6-4x. But 4x-4x=0, and we are left with

**3y= - 4x+6**

This operation is sometimes summarized as make one term change side while changing its sign

It would do no harm to put the right side of the foregoing equation in parentheses. I do that to avoid errors)

3y= (-4x+6).

Now y is**multiplied** by the number 3. To isolate y I have to perform the **inverse operation** of the multiplication, namely the division by 3

3y/3 =(-4x+6)/3. The left hand side is just y

**y= (-4x+6)/3.**

While result is correct, I can also open the parentheses

y= -4x/3 +6/3

**y= -(4/3)*x +2.**

Hope it helps.

I am not lecturing you but I would rather you understand how to do the manipulations involved in isolating a variable.

You want to isolate y, Ok

Start stripping it of all that is not y.

3y + 4x =6. (Addition is commutative, I can change the order of addition)

The term with y is

3y+4x-4x=6-4x. But 4x-4x=0, and we are left with

This operation is sometimes summarized as make one term change side while changing its sign

It would do no harm to put the right side of the foregoing equation in parentheses. I do that to avoid errors)

3y= (-4x+6).

Now y is

3y/3 =(-4x+6)/3. The left hand side is just y

While result is correct, I can also open the parentheses

y= -4x/3 +6/3

Hope it helps.

Oct 22, 2009 | Texas Instruments TI-89 Calculator

B. 2((2-Y)/3)-Y=3

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

Nov 05, 2007 | SoftMath Algebrator - Algebra Homework...

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