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2x-y=5, (1)
4x^2-y^2=15. (2)
Express y = 2x-5 from (1) and substitute it into (2), replacing y. You will get
4x^2 - (2x-5)^2 = 15.
Simplify and solve this equation.
4x^2 - 4x^2 + 20x - 25 = 15,
20x = 15 + 25,
20x = 40,
x = 2.
Then from (1) y = 2x-5 = 2*2-5 = -1.
There is only one solution (x,y) = (2,-1) which means that the straight line (1) touches the hyperbola (2).
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".
