The equation in itself with three unknown has a set of decisions therefore most often it is complemented with two more equations or conditions. Depending on what basic data, the decision course will depend in many respects.

## It is required to you

- - system from three equations with three unknown.

## Instruction

1. If two of three equations of a system have only two unknown from three, try to express some variables through others and to set up them in the equation with three unknown. Your purpose at the same time is to turn it into the usual equation from one unknown. If it worked well, the further decision is quite simple – substitute the found value in other equations and find everything other unknown.

2. Some systems of the equations can be solved subtraction from one equation of another. Look whether there is no opportunity to increase one of expressions by number or a variable so that at subtraction two unknown were reduced at once. If such opportunity is, use it, most likely, the subsequent decision is possible. Do not forget that at multiplication by number it is necessary to multiply both the left part, and right. In the same way, at subtraction of the equations it is necessary to remember that the right part has to be subtracted also.

3. If the last ways did not help, use the general way of solutions of any equations with three unknown. For this purpose rewrite the equations in a look a11kh1+a12kh2+ a13x3=b1, a21kh1+ a22kh2+ a23x3=b2, a31kh1+ a32kh2+ a33x3=b3. Now make a matrix of coefficients at x (A), a matrix of unknown (X) and a matrix of free members (C). Pay attention, multiplying a matrix of coefficients by a matrix of unknown, you receive the matrix equal to a matrix of free members, that is A*X=B.

4. Find a matrix And in degree (-1) previously having found matrix determinant, pay attention, it should not be equal to zero. After that increase the received matrix by a matrix In, as a result you receive a required matrix of X, with the indication of all values.

5. It is possible to find a solution of a system from three equations also by means of Kramer's method. For this purpose find determinant of the third order ∆, corresponding to a system matrix. Then consistently find three more determinants ∆1, ∆2 and ∆3, substituting instead of values of the corresponding columns of value of free members. Now find x: h1= ∆ 1 / ∆, h2= ∆ 2 / ∆, h3= ∆ 3 / ∆.