# Solve this sum the lesson is simultaneous linear equation

Dear Student,

Please find below the solution to the asked query:

We have : , So

We assume : and get our equations , As:

14 m  +  11 n  = $\frac{39}{10}$ ,

140 m  + 110 n  = 39                                                  --- ( 1 )

And

4 m  + 3 n  = $\frac{11}{10}$ ,

40 m  + 30 n  = 11                                                      --- ( 2 )

Now we multiply by 3 in equation 1 and by 11 in equation 2 and get :

420 m  + 330 n  = 117                                                  --- ( 3 )

And

440 m  + 330 n  = 121                                                   --- ( 4 )

Now we subtract equation 3 from equation 4 and get :

20 m  = 4 ,

Substitute that value in equation 2 and get :

40 $×\frac{1}{5}$ + 30 n  = 11  ,

8 + 30 n  = 11 ,

30 n  = 3 ,

But we assume : , So we get

x  =  5 and =  10                                                                ( Ans )

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• 0
X=5,y=10
• -3
x=5
• -2
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