# Solve the system using matrix x+3y=7=0 2x+y+4=0 3x+2y+11=0

william1941 | Student

This system has one unique solution if the three lines meet at a common point.

x+3y+7=0 ...(1)

2x+y+4=0 ...(2)

3x+2y+11=0 ...(3)

Now 2*(1) - (2)

=> 2x + 6y + 14 - 2x - y - 4 =0

=> 5y + 10 =0

=> y = -2

substituting in (1)

x = -3y - 7 = 6 -7 = -1

Now we have to see if (3) passes through (-1,-2)

3x+2y+11

= 3*-1 -2*2 +11 =-3 -4 +11 = 4 which is not equal to one.

Therefore the three lines do not meet at a single point.

giorgiana1976 | Student

To solve the system, we'll have to re-write the equations:

x+ 3y +7 =0

We'll subtract 7 both sides:

x + 3y = -7 (1)

3x+2y +11=0

We'll subtract 11 both sides:

3x + 2y = -11 (2)

2x+y+4=0

We'll subtract 4 both sides:

2x + y = -4 (3)

We'll determine the matrix of the system. The determinant is formed from the coefficients of x and y.

1    3

A =  3    2

2    1

We'll take the first2 lines and columns of the matrix and we'll calculate the minor of the matrix A.

1    3

d =

3     2

d = 1*2 - 3*3 = 2 - 9 = -7

Now, we'll calculate the determinant that we'll tell us if the system has solution or not.

This determinant will be formed from the minor, the last row and the column of the terms from the right side of equal:

1    3    -7

C =  3     2   -11

2    1    -4

C = -2*4 - 7*3 - 3*2*11 + 2*2*7 + 11 + 3*3*4

C = - 8 - 21 - 66 + 28 + 11 + 36

C = 20  - 1 - 30

C = -11

Since C is not zero, the system has no solutions.

The system doesn't have solutions.