Answer:
a.
![\left\{\begin{array}{l}x+y=40\\ \\7x+4y=193\end{array}\right.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kyy4olo1rd2a3m5lrjr51mrq4dbzb1js7m.png)
b. 11 red bricks
Explanation:
Let x be the number of red bricks and y be the number of grey bricks.
1. Monica purchased 40 bricks, so
![x+y=40](https://img.qammunity.org/2020/formulas/mathematics/college/z905281vka3lzcbxvpwkl0z1d3h4ehoaiu.png)
2. If she purchased the red bricks for $7 each, then x red bricks cost $7x. If the gray bricks are for $4 each, then y bricks cost $4y. In total, all bricks cost $(7x + 4y) that is $193. Hence,
![7x+4y=193](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wjbw3kuzj3yp14luhydekklgfszzehqvn7.png)
a. The system of two equations is
![\left\{\begin{array}{l}x+y=40\\ \\7x+4y=193\end{array}\right.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kyy4olo1rd2a3m5lrjr51mrq4dbzb1js7m.png)
b. From the first equation:
![y=40-x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/msbm59pzuh9m1zb0jx778hhqqt63zs88wa.png)
Substitute it into the second equation:
![7(40-y)+4y=193\\ \\280-7y+4y=193\\ \\-7y+4y=193-280\\ \\-3y=-87\\ \\3y=87\\ \\y=29\\ \\x=40-29=11](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dgp8uqnvlu8s5s7nd1t242aymp4qbtkmmk.png)
Monica purshased 11 red bricks and 29 grey bricks.