The system from three equations with three unknown can not have decisions, despite enough the equations. It is possible to try to solve it by means of a method of substitution or by means of Kramer's method. Kramer's method besides the solution of a system allows to estimate whether the system is solvable before finding values of unknown.

## Instruction

1. The method of substitution consists in consecutive expression of one unknown through two others and substitution of the received result in the system equations. Let the system from three equations in a general view be given: a1x + b1y + c1z = d1a2x + b2y + c2z = d2a3x + b3y + c3z = d3vyrazite from the first equation x: x = (d1 - b1y - c1z) / a1 - also substitute in the second and third equations, then from the second equation express y and substitute in the third. You receive linear expression for z through coefficients of the equations of a system. Now go ""back"": substitute z in the second equation and find y, and then z and y substitute in the first and find x. Process in a general view is displayed in the drawing before finding of z. Record in a general view too bulky, in practice will be farther, having substituted numbers, you will quite easily find all three unknown.

2. Kramer's method consists in drawing up a matrix of a system and calculation of determinant of this matrix and also three more auxiliary matrixes. The matrix of a system is formed from coefficients at unknown members of the equations. The column containing the numbers standing in the right parts of the equations is called a column of the right parts. In a system matrix it is not used, but used at the solution of a system.

3. Let, as before, the system from three equations in a general view be given: a1x + b1y + c1z = d1a2x + b2y + c2z = d2a3x + b3y + c3z = d3togda a matrix of this system of the equations will be the following matrix: | | | a2 b2 c2 | | a3 b3 c3 | First of all find a1 b1 c1 system matrix determinant. Formula of finding of determinant: |A| = a1b2c3 + a3b1c2 + a2b3c1 - a3b2c1 - a2b1c3 - a1b3c2. If it is not equal to zero, then the system is solvable and has the only decision. Now it is necessary to find determinants of three more matrixes which turn out from a system matrix by a podstavleniye of a column of the right parts instead of the first column (this matrix we will designate Ax), instead of second (Ay) and third (Az). Calculate their determinants. Then x = |Ax| / |A|, y = |Ay| / |A|, z = |Az| / |A|.