Unraveling the Mystery Behind Quadratic and Linear Simultaneous Equations

[david18]

I'm currently going through some questions and came across quadratic and linear simultaneous equations.

Solve the equation: x + y = 1
x^2 + y^2 = 16

I am not interested in the question itself but rather the explanation the book gives me which says after a couple of steps shows the equation:

2x^2 - 2x -15 = 0

Then it tells me to divide the equation by 2 and says when you divide the equation by 2 you will get:

x^2 - x - 15 = 0

I'm confused because i thought the -15 would also have to divide by two...
is it the book or am i missing something serious here?

 

[neutrino]

If the equation you have provided is right, then it is an error on the part of the author/editor/printer.

 

[Architeuthis Dux]

I would like to clarify that the book's explanation is correct. When solving simultaneous equations, it is important to manipulate the equations in a way that makes it easier to find the solution. In this case, dividing the equation by 2 allows us to simplify the equation and make it easier to solve for x and y. The constant term, -15, does not need to be divided by 2 because it does not have a coefficient attached to it. Only the terms with coefficients need to be divided by 2. Therefore, the equation x^2 - x - 15 = 0 is equivalent to 2x^2 - 2x -15 = 0. This manipulation does not change the solution to the equation, it simply makes it easier to find. I hope this helps clarify any confusion you may have had about the solution process.