Instruction

1

Method of addition.

You need to write two equations strictly under each other:

2 –5Y=61

-9x+5Y=-40.

Next, fold each term of the equations, respectively, considering their signs:

2+(-9x)=-7x, -5U+5U=0, 61+(-40)=21. As a rule, one of the sums containing the unknown amount, will be zero.

Write the equation of the resulting members:

-7x+0=21.

The unknown: -7x=21, h=21:(-7)=-3.

Already substitute the value found in any of the original equations and obtain a second unknown, solving a linear equation:

2x–5Y=61, 2(-3)–5Y=61, -6-5Y=61, -5Y=61+6, -5Y=67, y=-13,4.

The response of the system of equations: x=-3, y=-13,4.

You need to write two equations strictly under each other:

2 –5Y=61

-9x+5Y=-40.

Next, fold each term of the equations, respectively, considering their signs:

2+(-9x)=-7x, -5U+5U=0, 61+(-40)=21. As a rule, one of the sums containing the unknown amount, will be zero.

Write the equation of the resulting members:

-7x+0=21.

The unknown: -7x=21, h=21:(-7)=-3.

Already substitute the value found in any of the original equations and obtain a second unknown, solving a linear equation:

2x–5Y=61, 2(-3)–5Y=61, -6-5Y=61, -5Y=61+6, -5Y=67, y=-13,4.

The response of the system of equations: x=-3, y=-13,4.

2

The method of substitution.

One equation should Express any of the required members:

x–5Y=61

-9x+4U=-7.

x=61+5Y, x=61+5U.

Substitute the resulting equation into the second instead of the number "x" (in this case):

-9(61+5Y)+4U=-7.

Then deciding

the linear equation to find the number of "y":

-549+45у+4U=-7, 45у+4U=549-7, 49у=542, y=542:49,≈11.

In randomly selected (from the system) to insert the equation is already found "y" the number 11 and to compute the second unknown:

X=61+5*11 x=61+55, x=116.

Answer the given system of equations: x=116, y=11.

One equation should Express any of the required members:

x–5Y=61

-9x+4U=-7.

x=61+5Y, x=61+5U.

Substitute the resulting equation into the second instead of the number "x" (in this case):

-9(61+5Y)+4U=-7.

Then deciding

the linear equation to find the number of "y":

-549+45у+4U=-7, 45у+4U=549-7, 49у=542, y=542:49,≈11.

In randomly selected (from the system) to insert the equation is already found "y" the number 11 and to compute the second unknown:

X=61+5*11 x=61+55, x=116.

Answer the given system of equations: x=116, y=11.

3

A graphical way.

Is to practice finding the coordinates of a point which intersect the lines mathematically written in the system of equations. You should draw graphs of both direct separately in the same coordinate system. The General form of straight line equation: – y=KX+b. To build direct, it is sufficient to find the coordinates of the two points, and x is chosen arbitrarily.

Given the system: 2x – y=4

y=-3x+1.

Video is based on the first equation, for convenience, it should be written: y=2x-4. To invent (easier) the values for x, substituting into equation, solving it, find y. You get two points, which is based on video. (see pic.)

x 0 1

at -4 -2

Video is based on the second equation: y=-3x+1.

To build a direct. (see pic.)

x 0 2

1 to 5

Find the coordinates of the intersection point of two constructed lines on the chart (if the lines do not intersect, then the system of equations has no solution – it happens).

Is to practice finding the coordinates of a point which intersect the lines mathematically written in the system of equations. You should draw graphs of both direct separately in the same coordinate system. The General form of straight line equation: – y=KX+b. To build direct, it is sufficient to find the coordinates of the two points, and x is chosen arbitrarily.

Given the system: 2x – y=4

y=-3x+1.

Video is based on the first equation, for convenience, it should be written: y=2x-4. To invent (easier) the values for x, substituting into equation, solving it, find y. You get two points, which is based on video. (see pic.)

x 0 1

at -4 -2

Video is based on the second equation: y=-3x+1.

To build a direct. (see pic.)

x 0 2

1 to 5

Find the coordinates of the intersection point of two constructed lines on the chart (if the lines do not intersect, then the system of equations has no solution – it happens).

Useful advice

If the same system of equations solved in three different ways, the answer would be the same (if the solution is correct).

# Advice 2: How to solve a system of three equations with three unknowns

A system of three equations with three unknowns may not have solutions, in spite of a sufficient number of equations. You can try to solve it using the method of substitution or by the method of Kramer. The method of Kramer in addition to solving the system allows to assess whether the system is solvable, to how to find unknown values.

Instruction

1

The substitution method is a consistent expression of one unknown in two others, and the substitution of the result into the equations of the system. Let given a system of three equations in a General form:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

Express from the first equation for x: x = (d1 - b1y - c1z)/a1 - and the substitute second and third equations, then from the second equation to Express y and substitute in the third. You will get a linear expression for z in the coefficients of the equations of the system. Now go "back": substitute z into the second equation and find y, and then substitute y and z into the first and find x. The process is generally shown in the figure to find z. On the record in General would be too cumbersome in practice, substituting numbers, you quite easily find all three unknowns.

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

Express from the first equation for x: x = (d1 - b1y - c1z)/a1 - and the substitute second and third equations, then from the second equation to Express y and substitute in the third. You will get a linear expression for z in the coefficients of the equations of the system. Now go "back": substitute z into the second equation and find y, and then substitute y and z into the first and find x. The process is generally shown in the figure to find z. On the record in General would be too cumbersome in practice, substituting numbers, you quite easily find all three unknowns.

2

The method of Kramer is the establishment of a system matrix and calculating the determinant of this matrix, as well as three auxiliary matrices. The matrix system is composed of coefficients of the unknown members of the equations. The column that contains the integers in the right parts of the equations, called the column of right-hand parts. In the matrix of the system is not in use, but is used in the solution of the system.

3

Let, as before, given a system of three equations in a General form:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

Then the matrix of this system of equations is the following matrix:

| a1 b1 c1 |

| a2 b2 c2 |

| a3 b3 c3 |

First of all, find the determinant of the system matrix. The formula for finding the determinant: |A| = a1b2c3 + a3b1c2 + a2b3c1 - a3b2c1 - a2b1c3 - a1b3с2. If it is not zero, then the system is solvable and has a unique solution. Now we need to find the determinants of three matrices, which are obtained from the matrix system by turning the column of right-hand parts instead of the first column (this matrix will be denoted by Ax), instead of the second (Ay) and the third (Az). Calculate their determinants. Then x = |Ax|/|A|, y = |Ay|/|A|, z = |Az|/|A|.

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

Then the matrix of this system of equations is the following matrix:

| a1 b1 c1 |

| a2 b2 c2 |

| a3 b3 c3 |

First of all, find the determinant of the system matrix. The formula for finding the determinant: |A| = a1b2c3 + a3b1c2 + a2b3c1 - a3b2c1 - a2b1c3 - a1b3с2. If it is not zero, then the system is solvable and has a unique solution. Now we need to find the determinants of three matrices, which are obtained from the matrix system by turning the column of right-hand parts instead of the first column (this matrix will be denoted by Ax), instead of the second (Ay) and the third (Az). Calculate their determinants. Then x = |Ax|/|A|, y = |Ay|/|A|, z = |Az|/|A|.

# Advice 3: How to solve an equation with three unknowns

In itself

**an equation**with three**unknowns**has many solutions, so often it is complemented by two more equations or conditions. Depending on what the source data will depend largely on the progress of the solution.You will need

- - a system of three equations with three unknowns.

Instruction

1

If two of the three equations of the system have only two unknown of the three, try to Express some variables through others and substitute them in

**the equation**with three**unknown**. Your goal is to turn it into a normal**equation**with one unknown. If this is successful, then the solution is quite simple – substitute the value found into the other equation and find all the other unknown.2

Some systems of equations can be solved by subtracting one equation from the other. Let's see if there are opportunities to multiply one expression by the number or variable so that when subtraction was reduced from two unknown. If so, use it, most likely, a subsequent decision is not difficult. Don't forget that when multiplying by a number, multiply both the left and right. Similarly, when you subtract the equations you need to remember that the right side must also be deducted.

3

If the previous methods did not help, use the General method of solutions of any equations with three

**unknowns**. To do this, rewrite the equation in the form а11х1+а12х2+а13х3=b1, а21х1+а22х2+а23х3=b2, а31х1+а32х2+а33х3=b3. Now make a matrix of the coefficients x (A), the unknown matrix (X) matrix and free members (In). Please note, multiplying the coefficient matrix by a matrix is unknown, you will get a matrix equal to the matrix free members, that is, A*X=B.4

Find the matrix a of degree (-1) after finding the determinant of a matrix, note that it should not be zero. Then, multiply the resulting matrix on the matrix, as a result you get the desired matrix X, showing all the values.

5

Find the solution of the system of three equations using the method of Kramer. To do this, find the determinant of the third order ∆, the corresponding matrix system. Then find three more of the determinant ∆1, ∆2 and ∆3, by substituting values of the corresponding columns of the values of the free members. Now find x: x1=∆1/∆, x2=∆2/∆, X3=∆3/∆.