Respuesta :
Given the equations:
[tex]\begin{gathered} 20x+12y=140\text{ \lparen1\rparen} \\ 45x+20y=280\text{ \lparen2\rparen} \end{gathered}[/tex]Where,
x= Number of adults.
y= Number of children.
Isolating y in the equation (1):
[tex]\begin{gathered} 20x+12y=140 \\ 12y=140-20x \\ y=\frac{140-20x}{12}\text{ \lparen3\rparen} \end{gathered}[/tex]Replacing the equation (3) in equation (2).
[tex]\begin{gathered} 45x+20y=280 \\ 45x+20*(\frac{140-20x}{12})=280 \end{gathered}[/tex]Simplifying:
[tex]\begin{gathered} 45x+\frac{20*140-20*20x}{12}=280 \\ \\ 45x+\frac{2800-400x}{12}=280 \\ \\ \frac{45*12x+2800-400x}{12}=280 \end{gathered}[/tex][tex]540x+2800-400x=12*280[/tex]Finally, solving for x:
[tex]\begin{gathered} 140x=3360-2800 \\ 140x=560 \\ x=\frac{560}{140}=\frac{56}{14}=4 \end{gathered}[/tex]The number the adults is x=4.
And the number of children is:
[tex]\begin{gathered} y=\frac{140-20(4)}{12}=5 \\ \end{gathered}[/tex]
