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Hi,

**2x - 1/5 - 3x/7 + 2 = x + 9/2 - x - 3/10**

Or, 2x - 3x/7 = 1/5 - 2 + 9/2 - 3/10

Or, (14x + 3)/7 = (2 - 20 + 45 - 3)/10

Or, 11x/7 = 24/10 = 12/5

Or, 11x = 84/5

Or, __x = 84/55__

**
Good luck and have a nice day.
Thanks for using Fixya.
Rating the solution is highly appreciated
**

Posted on Jul 29, 2010

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Posted on Jan 02, 2017

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First, we will find y in terms of x. We will use the first equation to determine this.

4x+2y=2

We can subtract 4x from both sides:

2y=2-4x

And then divide both sides of the equation by two:

y=1-2x

Since we now have y in terms of x, we can substitute this into our second equation.

-3x-y=-3

-3x-(1-2x)=-3

Then, we can distribute the minus sign

-3x-1+2x=-3

-x-1=-3

Next, we can add 1 to both sides of the equation.

-x=-2

Finally, we divide both sides by negative one to isolate x.

x=2

Now that we have x's value, we can find y's value.

The first thing that we determined is:

y=1-2x

We can substitute in the value of x to this equation.

y=1-2x

y=1-4

y=-3

Therefore, we now have the values of both variables.

x=2

y=-3

4x+2y=2

We can subtract 4x from both sides:

2y=2-4x

And then divide both sides of the equation by two:

y=1-2x

Since we now have y in terms of x, we can substitute this into our second equation.

-3x-y=-3

-3x-(1-2x)=-3

Then, we can distribute the minus sign

-3x-1+2x=-3

-x-1=-3

Next, we can add 1 to both sides of the equation.

-x=-2

Finally, we divide both sides by negative one to isolate x.

x=2

Now that we have x's value, we can find y's value.

The first thing that we determined is:

y=1-2x

We can substitute in the value of x to this equation.

y=1-2x

y=1-4

y=-3

Therefore, we now have the values of both variables.

x=2

y=-3

Jan 13, 2015 | SoftMath Algebrator - Algebra Homework...

This calculator cannot do algebra so it does not handle polynomials. It does not have an EQN (equation) solving mode so you cannot enter an expression set = 0 and solve it for an unknown variable X.

It does have a table Mode where you can enter an expression y= F(X) and generate a table of values. If in the expression F(x) a numerical coefficient is a fraction just enter it as a fraction, possibly enclosed in parentheses. If it is a decimal number just enter it as a decimal number. Converting it to a fraction is not going to generate any more precision in the result. You lose precision when you go from fraction to decimal, but you do not make a number more precise by converting it from a decimal representation to a fraction.

It does have a table Mode where you can enter an expression y= F(X) and generate a table of values. If in the expression F(x) a numerical coefficient is a fraction just enter it as a fraction, possibly enclosed in parentheses. If it is a decimal number just enter it as a decimal number. Converting it to a fraction is not going to generate any more precision in the result. You lose precision when you go from fraction to decimal, but you do not make a number more precise by converting it from a decimal representation to a fraction.

Nov 16, 2013 | Casio fx-300ES Calculator

I am afraid you cannot do that with the TI83/84 Plus SE. The newer OS
allows you to get results expressed as fractions, but not radicals.
Sorry.

Your question seems to imply that you can do algebra with this calculator but that is impossible, unless you stored numerical values in variables a and b.**In that case that is not doing algebra.**

Your question seems to imply that you can do algebra with this calculator but that is impossible, unless you stored numerical values in variables a and b.

Nov 11, 2012 | Texas Instruments TI-84 Plus Calculator

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...

I am sorry to be the messenger with bad news, but this calculator does not do symbolic algebra. It does not have a solve function either.

You do have 5 memory variables in which you can store numerical values but it will not do you any good in this particular case.

Anyway, there is nothing you can do with such an expression (IT IS NOT an equation, just an expression) , except some cosmetic change 2x(2x-3y) or expand it 4x^2-6xy.

You do have 5 memory variables in which you can store numerical values but it will not do you any good in this particular case.

Anyway, there is nothing you can do with such an expression (IT IS NOT an equation, just an expression) , except some cosmetic change 2x(2x-3y) or expand it 4x^2-6xy.

Feb 11, 2010 | Texas Instruments TI-34II Explorer Plus...

You have some numerical value stored in the variable x, which the calculator is plugging into the resultant equation.

You can clear individual variables with 2ND [VAR-LINK]. You can clear all one-character variables a-z with 2ND [F6] 1. You can initialize the machine for a new problem with 2ND [F6] 2.

You can clear individual variables with 2ND [VAR-LINK]. You can clear all one-character variables a-z with 2ND [F6] 1. You can initialize the machine for a new problem with 2ND [F6] 2.

Feb 10, 2010 | Texas Instruments TI-89 Calculator

6x+6=4x+12

Since 4x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 4x from both sides.

6x+6-4x=12

Since 6x and -4x are like terms, add -4x to 6x to get 2x.

2x+6=12

Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.

2x=-6+12

Add 12 to -6 to get 6.

2x=6

Divide each term in the equation by 2.

(2x)/(2)=(6)/(2)

Simplify the left-hand side of the equation by canceling the common factors.

x=(6)/(2)

Simplify the right-hand side of the equation by simplifying each term.

x=3

Good Luck

Since 4x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 4x from both sides.

6x+6-4x=12

Since 6x and -4x are like terms, add -4x to 6x to get 2x.

2x+6=12

Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.

2x=-6+12

Add 12 to -6 to get 6.

2x=6

Divide each term in the equation by 2.

(2x)/(2)=(6)/(2)

Simplify the left-hand side of the equation by canceling the common factors.

x=(6)/(2)

Simplify the right-hand side of the equation by simplifying each term.

x=3

Good Luck

Sep 10, 2009 | Audio Players & Recorders

i don't no

Mar 07, 2009 | Sharp EL-520WBBK Calculator

Hello,

Sorry, but you cannot use this calculator to factor a general polynomial.

1. It does not know symbolic algebra.

2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients**
(no letters) **you can set [MODE] to **Equation **and use the equation solver
to find the real roots of 2nd degree or 3rd degree polynomials.
Assuming that your polynomials have real roots (X1, X2) for the
polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3,
then it is possible to write

**P2(X) =a*(X-X1)(X-X2)**

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

where a is the coefficient of the highest degree monomial aX^2 +...

or aX^3 +....

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

While the [MODE][5:Equation] only handles quadratic and cubic equations, the [SHIFT][SOLVE=] solver finds the roots of arbitarry expressions (not limited to polynomials). In principle you can use it to find the roots of an expression. If it is a polynomial of dgree higher that 3 you can factor it (approximately).

But I have a hunch that this is not what you wanted to hear.

Hope it helps.

Sorry, but you cannot use this calculator to factor a general polynomial.

1. It does not know symbolic algebra.

2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients

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

where a is the coefficient of the highest degree monomial aX^2 +...

or aX^3 +....

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

While the [MODE][5:Equation] only handles quadratic and cubic equations, the [SHIFT][SOLVE=] solver finds the roots of arbitarry expressions (not limited to polynomials). In principle you can use it to find the roots of an expression. If it is a polynomial of dgree higher that 3 you can factor it (approximately).

But I have a hunch that this is not what you wanted to hear.

Hope it helps.

Dec 09, 2008 | Casio fx-300ES Calculator

assuming the question is what is the circle equation?

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

Oct 26, 2008 | Casio FX-115ES Scientific Calculator

Oct 18, 2017 | Computers & Internet

Oct 18, 2017 | Computers & Internet

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