# For this linear system what values of m and n is (5,-3) the solutionmx+ny=12 mx-ny=18

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Since the poinmt ( 5, -3 ) is a solution for the system, then the pint should verify both equations:

==> mx + ny = 12

==> mx - ny = 12

We will substitute in both equations:

m*5 + n*-3 = 12

==> 5m - 3n = 12............(1)

5x + 3n = 18

==> 5m + 3n = 18...........(2)

Now we will solve the system:

We will add (1) and (2):

==> 10m = 30

**==> m = 3**

Now we will subtract (1) from (2):

==> 6n = 6

**==> n = 1**

We have to find for what values of m and n is (5, -3) the solution for the two equations.

mx+ny=12...(1)

mx-ny=18...(2)

Now we substitute x with 5 and y with -3 in the two equations (1) and (2).

=> 5m - 3n = 12...(3)

and 5m + 3n = 18...(4)

We now solve (3) and (4) for m and n

(4) - (3)

=> 5m + 3n - 5m + 3n = 18 - 12

=> 6n = 6

=> n = 1

Substitute n = 1 in (3)

=> 5m - 3 = 12

=> 5m = 15

=> m = 15/ 5 = 3

**Therefore the required values of m and n are 3 and 1 respectively.**

To find the value of m and n for which the lines pass through.

mx+ny=12..(1)

mx-ny=18...(2).

So (5,-3) shoild be the point of intersection of the lines as given.

Adding the two equations, we get:

2mx = 12+18 = 30.

x = 30/2m = 15/m.

Eq(1)-eq(2):

2ny = 12-18= -6.

y = -6/2n = -3/n.

Therefore (x,y) = (15/m , (-3/n) is the point of intersection which should be (5, -3).

Therefore (15/m ,-3/n) = (5,-3)....(3)

Therefore equating the x cordinates on both sides, we get:

15/m = 5, or 5m = 15, m = 15/5 = 3.

Similarly, equatiing the y cordintes in (3), we get:

-3/n = -3. So -3n = -3, or n = 1.

Therefore, m= 3 and n=1 for which the given lines intersect at (5,-3).