We are solving natural equations like ln 2x + 1 = 2

The next step is 2x + 1 = e squared

How do I get e squared?

You can use the value of e you know and square it, or you can use the exponential function as follows

[SHIFT][e^x] 2 [=]

Your solution is

x=(e^2 -1)/2= 3.19452804946 =3.19

Posted on Aug 23, 2010

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Posted on Jan 02, 2017

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...

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...

Try casting the equation of the circle into the form (x-h)^2+(y-k)^2=R^2. Here (h,k) would be the center and R the radius.

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

Sep 20, 2011 | SoftMath Algebrator - Algebra Homework...

Apply what you learned, especially that this system is quite simple.

Elimination

1. entails eliminating one variable to find (in this case) a single equation involving the other variable.

2. Solve that new equation, meaning isolate the variable that was not eliminated.

3. Substitute the value found in the last step and replace it in one of the original equation to obtain the other variable.

x+y=10

x-y=-6

Add the left sides to get x+y+x-y=2x

Add the right sides of the system to obtain 10+(-6)=4

Write Sum of left sides =Sum of right sides or 2x=4.

Since this new equation involves only x, you can solve it for x, getting x=4/2=2

Now you know the value of x (=2)

Take one of the original equations (for example x+y=10) and put 2 in place of x.

The equation becomes 2+y=10

Solve it to obtain y=10-2=8

Thus your solution is x=2, y=8

Check: use one equation in which you substitute 2 for x and 8 for y: x-y=-6 becomes 2-8=-6

Verified.

In the general case you will have to multiply by certain values to obtain opposite coefficients for the same variable. Here that was not necessary because the coefficient of y is 1 in the first equation and -1 in the second.

Elimination

1. entails eliminating one variable to find (in this case) a single equation involving the other variable.

2. Solve that new equation, meaning isolate the variable that was not eliminated.

3. Substitute the value found in the last step and replace it in one of the original equation to obtain the other variable.

x+y=10

x-y=-6

Add the left sides to get x+y+x-y=2x

Add the right sides of the system to obtain 10+(-6)=4

Write Sum of left sides =Sum of right sides or 2x=4.

Since this new equation involves only x, you can solve it for x, getting x=4/2=2

Now you know the value of x (=2)

Take one of the original equations (for example x+y=10) and put 2 in place of x.

The equation becomes 2+y=10

Solve it to obtain y=10-2=8

Thus your solution is x=2, y=8

Check: use one equation in which you substitute 2 for x and 8 for y: x-y=-6 becomes 2-8=-6

Verified.

In the general case you will have to multiply by certain values to obtain opposite coefficients for the same variable. Here that was not necessary because the coefficient of y is 1 in the first equation and -1 in the second.

Aug 16, 2011 | Casio Office Equipment & Supplies

Ok so first you need to find what x equals in terms of y. so use the first equation .2x+.7y=1.5 solve for x by the following steps

subract .7y from both sides: .2x= 1.5-.7y --->

divide by .2 on both sides: x= 7.5 - 3.5y. --->

Now plug in x= 7.5 - 3.5y for x in the second equation 0.3 (7.5 - 3.5y) -0.2y=1 --->

Distribute the .3: 2.25 - 1.05y -.2y = 1--->

solve for y by combining like terms (the y's) and subtractive 2.25: -1.25y = -1.25--->

divide by -1.25 and you get y= 1

Now go back to your x= 7.5 - 3.5y you solved for. Plug in 1 for y in the equation and solve for x--->

x=7.5-3.5(1) so x= 4 and y=1 yay your done!

subract .7y from both sides: .2x= 1.5-.7y --->

divide by .2 on both sides: x= 7.5 - 3.5y. --->

Now plug in x= 7.5 - 3.5y for x in the second equation 0.3 (7.5 - 3.5y) -0.2y=1 --->

Distribute the .3: 2.25 - 1.05y -.2y = 1--->

solve for y by combining like terms (the y's) and subtractive 2.25: -1.25y = -1.25--->

divide by -1.25 and you get y= 1

Now go back to your x= 7.5 - 3.5y you solved for. Plug in 1 for y in the equation and solve for x--->

x=7.5-3.5(1) so x= 4 and y=1 yay your done!

May 03, 2011 | Texas Instruments TI-30 XIIS Calculator

I'm assumeing the problem is 2x^2 - 19x+22 = 0, solve for x.

I don't see any easy to factor this so I'll either need to complete the square or use the quadratic equation.

Using the quadratic equation, where A = 2, B = -19, and C = 22

x = [-(-19) +/- squareroot( 19^2 - 4*2*22)] / (2*2) =

(19 +/- squareroot(185))/4 =

8.150367 or 1.349632

I don't see any easy to factor this so I'll either need to complete the square or use the quadratic equation.

Using the quadratic equation, where A = 2, B = -19, and C = 22

x = [-(-19) +/- squareroot( 19^2 - 4*2*22)] / (2*2) =

(19 +/- squareroot(185))/4 =

8.150367 or 1.349632

Feb 01, 2011 | SoftMath Algebrator - Algebra Homework...

The 30XIIS does not have the capability of solving equations. It can, however, calculate expressions using logarithms. Use the LOG key for common logarithms and the LN key for natural logarithms. In both cases, press the function key, enter the value or expression, and then ) to match the ( the function automatically enters.

The LOG and LN keys are the second and third keys from the top in the leftmost column of the keyboard.

The LOG and LN keys are the second and third keys from the top in the leftmost column of the keyboard.

Nov 13, 2010 | Office Equipment & Supplies

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 assume you are speaking of solving a system of equations with a number of unknowns. If not, please correct me. Here's an example in practice:

If you have a system of 3 equations with 3 unknowns, you would set up your matrix so that the coefficients of each variable for a particular equation are on one row. So, given equations x + y + z = 0, 2x + 3y - 4z = 1, x + -z = -1 you would type the following into your calculator: [[1,1,1,0][2,3,-4,1][1,0,-1,-1]] and press enter to make sure you typed it correctly. notice that in the third row there is a zero, since we have zero time y for the third equation. Then row-reduce the matrix (2nd > 5 > 4 > 4 or in the CATALOG as rref). You should get out the matrix [[1,0,0,-1][0,1,0,1][0,0,1,0]]. This says that x=-1 y = 1 z=0 since my first column contained the coefficients for the x variable, the second for the y variable, and the third for the z variable. The last column contains the solution, the part on the other side of the equals sign.

Hope this helps! For more reading (from someone else; I just made this one up), check out the Wikipedia articles on Gaussian elimination and Systems of linear equations

If you have a system of 3 equations with 3 unknowns, you would set up your matrix so that the coefficients of each variable for a particular equation are on one row. So, given equations x + y + z = 0, 2x + 3y - 4z = 1, x + -z = -1 you would type the following into your calculator: [[1,1,1,0][2,3,-4,1][1,0,-1,-1]] and press enter to make sure you typed it correctly. notice that in the third row there is a zero, since we have zero time y for the third equation. Then row-reduce the matrix (2nd > 5 > 4 > 4 or in the CATALOG as rref). You should get out the matrix [[1,0,0,-1][0,1,0,1][0,0,1,0]]. This says that x=-1 y = 1 z=0 since my first column contained the coefficients for the x variable, the second for the y variable, and the third for the z variable. The last column contains the solution, the part on the other side of the equals sign.

Hope this helps! For more reading (from someone else; I just made this one up), check out the Wikipedia articles on Gaussian elimination and Systems of linear equations

May 03, 2009 | Texas Instruments TI-84 Plus Calculator

To get an exponent press 2nd then LN or LOG

LN is natural logs, so entering 1 then 2nd LN gives you the natural exponent of 1 which is.2.71828....pressing LN gives you the natural log of 2.71828.... which is 1

LOG is logs to base 10, so entering 1 then 2nd LN gives you the 10 to the power 1 which is.10, pressing LOG gives you the log of 10 which is 1 again

LN is natural logs, so entering 1 then 2nd LN gives you the natural exponent of 1 which is.2.71828....pressing LN gives you the natural log of 2.71828.... which is 1

LOG is logs to base 10, so entering 1 then 2nd LN gives you the 10 to the power 1 which is.10, pressing LOG gives you the log of 10 which is 1 again

Dec 04, 2008 | Texas Instruments TI-30XA Calculator

Jun 10, 2013 | Casio FX-7400G Plus Calculator

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