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Posted on Jan 02, 2017

Now is the equation supposed to be y=2x+1? If so,

The graph of y = 2x+1 is a straight line

When x increases, y increases twice as fast, hence 2x

When x is 0, y is already 1. Hence +1 is also needed

So: y = 2x + 1

Here are some example values:

xy = 2x + 1

-1y = 2 × (-1) + 1 = -1

0y = 2 × 0 + 1 = 1

1y = 2 × 1 + 1 = 3

2y = 2 × 2 + 1 = 5

Check for yourself that those points are part of the line above!

The graph of y = 2x+1 is a straight line

When x increases, y increases twice as fast, hence 2x

When x is 0, y is already 1. Hence +1 is also needed

So: y = 2x + 1

Here are some example values:

xy = 2x + 1

-1y = 2 × (-1) + 1 = -1

0y = 2 × 0 + 1 = 1

1y = 2 × 1 + 1 = 3

2y = 2 × 2 + 1 = 5

Check for yourself that those points are part of the line above!

Jan 16, 2015 | SoftMath Algebrator - Algebra Homework...

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...

What you want to do is solve the linear equation 2x=10 to find the value of the unknown x that makes the equality true. Solving an equation usually involves rearranging therms, factors and so on. However your calculator was not designed to handle the solution of equations, no matter how simple they are.

Some Casio scientific calculators can solve some types of equations (polynomials of degree 2 or 3, simultaneous linear equations in up to 3 unknowns). These are the FX-115 ES (Plus) and FX-991 ES (Plus C). Other equivalent models are sold in the world under different names.

Some Casio scientific calculators can solve some types of equations (polynomials of degree 2 or 3, simultaneous linear equations in up to 3 unknowns). These are the FX-115 ES (Plus) and FX-991 ES (Plus C). Other equivalent models are sold in the world under different names.

Oct 16, 2013 | Casio FX350MS Scientific 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...

x-intercept (or zero of the function)

Set y=0 in the equation and solve fo x: 2x-5(0)=20 and x=20/2=10

y-intercept (initial value)

Set x= 0 in the equation and solve for y: 2(0)-5y=20 and y=20/(-5)=-4

If the equation is in the slope-intercept form y=ax+b

1. y-intercept is b

2. To get the x-intercept ( zero of the function) Set y=0=ax+b, and x=-b/a

Set y=0 in the equation and solve fo x: 2x-5(0)=20 and x=20/2=10

y-intercept (initial value)

Set x= 0 in the equation and solve for y: 2(0)-5y=20 and y=20/(-5)=-4

If the equation is in the slope-intercept form y=ax+b

1. y-intercept is b

2. To get the x-intercept ( zero of the function) Set y=0=ax+b, and x=-b/a

Aug 27, 2011 | Texas Instruments TI-84 Plus Calculator

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

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...

Hello,

Sorry this calculator cannot solve equations. It does not have the program. It does not know how to do those things. However it can calculate.**Before you can use the calculator you must prepare all yourself. **When you have solved the problem, you ask the calculator to compute the value of what you found.

Let us clean the equation a bit

2(x-4) -5x=-5

Get rid of the parentheses 2(x-4) becomes 2x-2*4=2x-8 .Put that result where it was in equation

2x-8 -5x=-5

Group together the terms that have x in them 2x-5x=-3x

Then

2x-8-5x=-5, becomes -3x-8=-5

You want to isolate the term with x (have it on one side, and the others on the other side) -3x= -5 -(-8)= -5+8 =3.

The equation becomes

-3x=3 Thus**-x=3/3=1 and x=-1.**

**In general you would have obtained x= (some number/some other number) and that is where the calculator would intervene.**

Hope it helps.

Sorry this calculator cannot solve equations. It does not have the program. It does not know how to do those things. However it can calculate.

Let us clean the equation a bit

2(x-4) -5x=-5

Get rid of the parentheses 2(x-4) becomes 2x-2*4=2x-8 .Put that result where it was in equation

2x-8 -5x=-5

Group together the terms that have x in them 2x-5x=-3x

Then

2x-8-5x=-5, becomes -3x-8=-5

You want to isolate the term with x (have it on one side, and the others on the other side) -3x= -5 -(-8)= -5+8 =3.

The equation becomes

-3x=3 Thus

Hope it helps.

Sep 11, 2009 | Texas Instruments TI-30XA 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

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

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