Answer:
Price of each senior citizen ticket is
and one child ticket is
![\$3](https://img.qammunity.org/2020/formulas/mathematics/college/jgvuf8exmqu4hezjlbvz479mv1dy3zd2is.png)
Explanation:
Let the price of each senior citizen ticket be x.
Let the price of each Child ticket be y.
Tickets sold on first day which are given as,
5 senior citizen tickets and 14 child tickets for a total of $102.
![5x+14y = \$ 102 \ \ equation \ 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hgnmu9ao1yupq1xsjcy2o1ihtp1p04ypn4.png)
Tickets sold on first day which are given as,
$153 on the second day by selling 11 senior citizen tickets and 7 child tickets.
![11x+7y= \$153 \ \ \ equation\ 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8ne5y1r0c5lygrw8m4assa46vyr9tek182.png)
Now multiplying by 2 equation 2 we get
![2* (11x+7y)= 2 *\$153\\22x+14y = \$ 306 \ \ \ equation\ 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8tspkjeea043t2ddkzp74idnxkh40ivwpg.png)
Now Subtracting equation 1 from equation 3 we get
![17x= 204\\x=(204)/(17) = \$12](https://img.qammunity.org/2020/formulas/mathematics/middle-school/byhjewed7ntt3esapn5x602xvqat3jgtc3.png)
Substituting value of x in equation 1 we get
![5 *12+14y=\$102\\14y=\$102-42\\14y= \$42\\y=\$(42)/(14)=\$4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/napqfzpz2ezvi4e8oyq370lrsx2njbt0cc.png)
Hence Senior Citizen ticket price is
and Child ticket price is
![\$3](https://img.qammunity.org/2020/formulas/mathematics/college/jgvuf8exmqu4hezjlbvz479mv1dy3zd2is.png)