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Posted on Jan 02, 2017

There is nothing to solve here.You just have an expression with two letters; you have no equation to solve because you need an = sign.

Sep 07, 2013 | Office Equipment & Supplies

Your scientific calculator is unable to solve complex equation with complex coefficients. You should try to solve by hand directly using the quadratic formula or by factoring the polynomial in z

failing that, another way would be to set z=x+iy, substitute this for z, carry out the algebra and try to separate real and imaginary parts. But your two equations will constitute a system of two quadratic equations. I am not aware of any general method to solve coupled nonlinear equations.

Good luck.

failing that, another way would be to set z=x+iy, substitute this for z, carry out the algebra and try to separate real and imaginary parts. But your two equations will constitute a system of two quadratic equations. I am not aware of any general method to solve coupled nonlinear equations.

Good luck.

Dec 20, 2012 | Casio FX-115ES Scientific Calculator

Thhe Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Mar 17, 2012 | Casio FX-9860G Graphic Calculator

The Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Mar 17, 2012 | Casio 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...

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

This expression 5x^2+15x+3 is not an equation, therefore it can have any value depending on the value of x. You have to make it an equation before looking for the particular value of x that satisfy your equation **5x^2+15x+3 =0**

This calculator cannot solve equations because it does not have an equation solver.

This calculator cannot solve equations because it does not have an equation solver.

Dec 15, 2010 | Sharp ELW535 Calculator

And your question is? If you just want to know how to solve this system, you only need to substitute one equation into the other and solve for the remaining unknown.

y=3x (equation 2), so plug it into equation 1

7x-2y=1

7x-2(3x)=1

7x-6x=1

x=1

from equation 2 then, you know y=3x=3

y=3x (equation 2), so plug it into equation 1

7x-2y=1

7x-2(3x)=1

7x-6x=1

x=1

from equation 2 then, you know y=3x=3

Oct 05, 2010 | Linear (HDMICAT5EXT) Repeater

Hi joanmae jmeh;

You can add these equations just like you would numbers

3x + 2y = 7

5x - 2y = 1

-----------------------

8x =8

x=1

Now plug x = 1 back into either equation and you can solve for y

Y = 2

When you add the 2 equations together the +2y and the -2y cancel

out

Hope this helps Loringh Please leave a rating for me Thks

You can add these equations just like you would numbers

3x + 2y = 7

5x - 2y = 1

-----------------------

8x =8

x=1

Now plug x = 1 back into either equation and you can solve for y

Y = 2

When you add the 2 equations together the +2y and the -2y cancel

out

Hope this helps Loringh Please leave a rating for me Thks

Dec 01, 2008 | Bagatrix Algebra Solved! 2005 (105101) for...

Just plug in one side of the equation and find the answer. Then plug in the other side and compare answers.

Feb 19, 2008 | Texas Instruments TI-89 Calculator

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