# Apply matrix to solve simultaneous equations x + y - 5 = 0 2x - y -1 = 0

*print*Print*list*Cite

### 3 Answers

x + y - 5 = 0

==> x + y = 5...........(1)

2x - y - 1 = 0

==> 2x - y = 1.........(2)

We will use the elimination method to solve:

Let us add (1) t0 (2):

==> 2x + x + y - y = 5 + 1

Reduce similar:

==> 3x = 6

Divide by 3:

==> x = 6/3 = 2

**==> x = 2**

Now to find y , let us substitute in (1):

x + y = 5

2 + y = 5

==> Subtract 2 from both sides:

==> y= 5 - 2 = 3

**==> y= 3**

**Then the solution for the system is the point ( 2,3)**

We'll re-write the equations:

x + y = 5

2x - y = 1

We'll use the matrix to solve the system. We'll form the matrix of the system, using the coefficients of x and y:

[1 1]

A =

[2 -1]

We'll calculate the determinant of the system:

detA = -1 - 2 = -3

Since det A is not zero, the system is determinated and it will have only one solution.

x = det X/detA

|5 1|

det X =

|1 -1|

detX = -5 - 1 = -6

x = det X/detA

x = -6/-3

**x = 2**

**We'll calculate y:**

|1 5|

det Y =

|2 1|

det Y = 1 - 10

det Y = -9

y = detY/detA

y = -9/-3

**y = 3**

**The solution of the system is: (2 , 3). **

To solve the equaations;

x+y-5 = 0...(1)

2x-y-1 = 0....(2).

Adding the equations, we get:

3x-6 = 0

3x= 6.

x = 6/3 = 2.

We substitute x= 2 in eq (2): 2x-y-1 = 0:

2*2 -y -1 = 0

4-y -1 = 0.

Add y to both sides:

4-1 = y.

So 3 = y.

Therefore x = 2 and y = 3 are the solutions:

Tally :

Put x= 2 and y = 3 in x+y - 5 = 0 ... (1)and 2x-y-1 = 0...(2)

First equation: LHS 3+2-5 = 0= RHS

Second equation: LHS : 2*2-3-1 = 0 = RHS.