.
4x - y = 7, (1)
xy = 15. (2)
From (1), express y = 4x-7 and substitute it into equation (2), replacing y. You will get a single quadratic equation
x*(4x-7) = 15.
Simplify:
4x^2 - 7x - 15 = 0,
= = = .
= = 3 ---> y = = = 5.
= = = -1.25. ---> y = = = -12.
Answer. There are two solutions: (x,y) = (3,5) and (x,y) = (-1.25,-12).
For many other similar solved problems see the lesson
- Solving systems of algebraic equations of degree 2 and degree 1
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of equations that are not linear".