We can solve this by setting up a couple of equations.
Let's allow x to represent the number of paperbacks Tim owns, and allow y to represent the number of hardcover books he owns.
Using the information in the question, we can write the equations:
1)
![x = 4y-3](https://img.qammunity.org/2019/formulas/mathematics/middle-school/eyv2sny2lvot9cpgnje9h8etqhji34awc8.png)
2)
![x+y=447](https://img.qammunity.org/2019/formulas/mathematics/middle-school/vnbmh3hd06mzixj8odtlw4pbaxmewnwsus.png)
Let's rearrange equation 1 so that it is in standard form:
![x-4y=-3](https://img.qammunity.org/2019/formulas/mathematics/middle-school/6cz0bim3h4uekwxw8b249zbu9v1p9q5cdt.png)
And then let's multiply equation 2 by 4 so that we can cancel out y when we solve the system of equations:
![4(x+y=447)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ugdmuud9onyntrls9vglm817ey6m1q0d7m.png)
![4x+4y=1,788](https://img.qammunity.org/2019/formulas/mathematics/middle-school/z6dolq67g4sa5qkuccmxyt1yulqscojxhl.png)
Then we can add the two equations and solve for x:
1)
![x-4y=-3](https://img.qammunity.org/2019/formulas/mathematics/middle-school/6cz0bim3h4uekwxw8b249zbu9v1p9q5cdt.png)
+ 2)
![4x+4y=1,788](https://img.qammunity.org/2019/formulas/mathematics/middle-school/z6dolq67g4sa5qkuccmxyt1yulqscojxhl.png)
------------------------------------
![5x=1,785](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ikvjqddy349osm0crt4f11fh9ppfnbb0gx.png)
![x=357](https://img.qammunity.org/2019/formulas/mathematics/middle-school/nkxakwr5ymhxa2iox1p50wbl9zeecrqh14.png)
So now we now the number of paperback books Tim has is 357. Let's plug this into one of the original equations to solve for the number of hardcover books (y):
![357+y=447](https://img.qammunity.org/2019/formulas/mathematics/middle-school/2irzp3yxrgj3bpk7ju39dmi4fnj4avt8z2.png)
![y=90](https://img.qammunity.org/2019/formulas/mathematics/middle-school/qwsgd9dtlxyrek2rz6x8xqyxjf0kfkghip.png)
And now we know that
Tim owns 90 hardcover books.