80 adult tickets were sold that day.
This is a two equation problem.
First, let:
a = adult tickets
c = child tickets
![6c + 9.6a = 1350](https://img.qammunity.org/2021/formulas/mathematics/college/v3zsim8wxk49b0thlcic4a63wf1dsu9qzf.png)
![c + a = 177](https://img.qammunity.org/2021/formulas/mathematics/college/62go1voi01cr8208usbogd8voytsf8y5mo.png)
Now, solve for either variable (c or a). We're going to isolate "c" in the second equation and plug it into the first.
![c + a = 177](https://img.qammunity.org/2021/formulas/mathematics/college/62go1voi01cr8208usbogd8voytsf8y5mo.png)
![c = 177 - a](https://img.qammunity.org/2021/formulas/mathematics/college/f9rnneia321qhnbq4e0mmlb4anf9vgpibs.png)
Plug it in:
![6(177 - a) + 9.6a = 1350](https://img.qammunity.org/2021/formulas/mathematics/college/jq3qsxkefzmw58yk515gb4qcgz12f5fdn9.png)
![1062 - 6a + 9.6a = 1350](https://img.qammunity.org/2021/formulas/mathematics/college/a8u2yu9z83scivfyardhu5dp4u7tcmou75.png)
![1062 + 3.6a = 1350](https://img.qammunity.org/2021/formulas/mathematics/college/fkc5o28rshem5s2wh7o5l5saekd2b15cbx.png)
![3.6a = 288](https://img.qammunity.org/2021/formulas/mathematics/college/8eonkpzs5w8hqrkklx33ttxdlvfu3c41c3.png)
![a = 80](https://img.qammunity.org/2021/formulas/mathematics/college/ul0nmfvqurbtej5bhmy2w3v0y5f584lxdq.png)
Now, plug in 80 for "a" in either of the equations to find "c."
![c + (80) = 177](https://img.qammunity.org/2021/formulas/mathematics/college/agekluc396od8j42xkzjt73c7a3z0249p3.png)
![c = 97](https://img.qammunity.org/2021/formulas/mathematics/college/a94w6gfczsw3uuj65ksd2sq2xwvtifhhsg.png)
Rather than solving algebraically, as shown above, you can also graph the two equations, and wherever they intersect will be the solution.