91影视

We are given that the cost of adults, student and child tickets is$\$10,\$8$and$\$5$. The total number of ticket sold is$600$.

Let $x,y,z$be the number of tickets sold of adults, students and child.

Total number of ticket sold is $600$, so

$x+y+z=600-\left(1\right)$

The total amount of ticket is $\$4,900$, so

$10x+8y+5z=4900-\left(2\right)$

Also, it is given that the number of adult tickets is twice the number of child tickets, so

$x=2z-\left(3\right)$

Putting $x=2z$in first and second equations, we get

role="math" localid="1644594039768" $$\begin{gathered} 2z+y+z=600 \\ y+3z=600-\left(4\right) \\ 10\left(2z\right)+8y+5z=4900 \\ 8y+20z+5z=4900 \\ 8y+25z=4900-\left(5\right) \end{gathered}$$

Multiplying by role="math" localid="1644594051514" $8$in fourth equation, we get

role="math" localid="1644594084338" $8y+24z=4800-\left(6\right)$

Now, subtracting fifth and sixth equation, we get

role="math" localid="1644594233009" $$\begin{gathered} 8y-8y+25z-24z=4900-4800 \\ z=100 \end{gathered}$$

Putting the value of $z$in third equation, we get

$$\begin{gathered} x=2z \\ x=2脳100 \\ x=200 \end{gathered}$$

Now, putting the value of $x,z$in first equation, we get

$$\begin{gathered} x+y+z=600 \\ 200+y+100=600 \\ y=600-300 \\ y=300 \end{gathered}$$

Putting the value of $x,y,z$in the equations, we get

$$\begin{gathered} x+y+z=600 \\ 200+300+100=600 \\ 600=600 \\ 10x+8y+5z=4900 \\ 10\left(200\right)+8\left(300\right)+5\left(100\right)=4900 \\ 2000+2400+500=4900 \\ 4900=4900 \end{gathered}$$

This is true, hence the solution is correct.