Solve this sum the lesson is simultaneous linear equation

Solve this sum the lesson is simultaneous linear equation 4. 3a + 8b 7.5 711 - 10b 8. 11x-8y=-48 15x-7y=-63S 14 11 39

Dear Student,

Please find below the solution to the asked query:

We have : $\frac{14}{x} + \frac{11}{y} = \frac{39}{10} \mathrm{and} \frac{4}{x} + \frac{3}{y} = \frac{11}{10}$ , So

We assume : $\frac{1}{x} = m \mathrm{and} \frac{1}{y} = n$ 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 ,

 $$\begin{gathered} \Rightarrow m = \frac{4}{20} \\ \mathbf{\Rightarrow }\mathbf{ }\mathit{m}\mathbf{ }\mathbf{=}\mathbf{ }\frac{{\displaystyle \mathbf{1}}}{{\displaystyle \mathbf{5}}} \end{gathered}$$
Substitute that value in equation 2 and get :

40 $\times \frac{1}{5}$ + 30 n  = 11  ,

8 + 30 n  = 11 ,

30 n  = 3 ,

$$\begin{gathered} n = \frac{3}{30} , \\ \mathit{n}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{1}}{\mathbf{10}} \end{gathered}$$

But we assume : $\frac{1}{x} = m \mathrm{and} \frac{1}{y} = n$ , So we get 

x  =  5 and y  =  10                                                                ( Ans )


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