In our previous post about expanding Step-by-step solutions, we introduced a revamped equation solver. I’m proud to say that it has now been extended to solve systems of linear equations. In addition, you have four different methods to choose from when looking for a solution! These methods are elimination, substitution, Gaussian elimination, and Cramer’s rule. Let’s look at x + y = 5, xy = 1 to see all four methods in action.

First we look at elimination:

Wolfram|Alpha Step-by-step Solution

Now we solve using substitution:

Wolfram|Alpha Step-by-step Solution

Expressing our system in matrix form, we can perform Gaussian elimination:

Wolfram|Alpha Step-by-step Solution

Lastly, the determinant can play a fundamental role in solving linear systems. This is called Cramer’s rule:

Wolfram|Alpha Step-by-step Solution

The number of unknowns is not a problem for Wolfram|Alpha; here w + x + y + z = 1, 2wx + yz = 0, w + 3x + y + z = 5, w + xy + 2z = 2 is an example of a 4×4 linear system. We have even added this new functionality when solving word problems.

Wolfram|Alpha Step-by-step Solution

Having various methods to choose from is a great way to look at a problem from all angles. Stay tuned for more and more step-by-step math in the future!

5 Comments

Can this be used to solve systems of DE’s?

Posted by shane toreki March 20, 2013 at 11:12 pm Reply

How to solve this system?
3x-2y+z=1;
x+3y-z=2;
x-8y+3z=-3;

Or this one?
3x-2y+z-2w-u=3,
-9x+6y-3z+6w+10u=-2,
3x-y+3z-w-u=4,

Above systems have parametric solutions. But wolfram don’t say which variables are parametric and which are static.
This systems you can solve with matrix, so step by step solution by Gauss elimination would be nice.

Posted by Joe March 19, 2014 at 6:15 pm Reply

did not help at all maybe adding a calculator to solve will help1

Posted by billy joe September 15, 2014 at 3:30 pm Reply

For the initial value problem

y’ = xy^3 / ?1+x^2 y (0) = 1

(a) Find the solution in the explicit form
(b) Plot the graph of the solution
(c) Determine (at least approximately) the interval in which the solution is defined.

Posted by enes kalay November 3, 2014 at 7:44 am Reply
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