Let C be the number of chidren and A be the number of adults. Then, we have
[tex]\begin{gathered} 4C+5.40A=1298\ldots(1) \\ C+A=286\ldots(2) \end{gathered}[/tex]Then, we have 2 equations in 2 unknonws.
Solving by elimination method
If we multiply equation (2) by -4, we get an equivalent system of equation:
[tex]\begin{gathered} 4C+5.40A=1298\ldots(1^{\prime}) \\ -4C-4A=-1144\ldots(2^{\prime}) \end{gathered}[/tex]By adding both equations, we have
[tex]5.40A-4A=1298-1144[/tex]because 4C - 4C =0. This last expression gives
[tex]1.40\text{ A=}154[/tex]By moving the coefficient of A to the right hand side, we get
[tex]A=\frac{154}{1.40}[/tex]and A is equal to 100, that is A=110.
Now, we can substitute this result into equation (2). It yields
[tex]C+110=286[/tex]By moving +110 to the right hand side, we have
[tex]C=286-110[/tex]then, C is equal to 176.
Therefore, there are 176 children and 110 adults.
