This, I think is pre-algebra, not pre-calculus...
The line V passes through the points (-5,3) and (7,-3) and the line W passes through the points (2,-4) and (4,2). The lines V and W intersect the point A. Work out the coordinates of the point A.
How do I do this? Please do step by step. I have worked out the equation of both lines in the form ax + by + c = 0.
V ->
W ->
EDIT: I've been thinking it through some more. I know that both x and y have to be the same for both lines. Would I be right in thinking of using simultaneous equations to work this out?
Hmm, the simultaneous equation is confusing me a little. I can see how it works, and that it provides the correct answer. The problem is, I thought you had to subtract (2) from (1), not add them together. Is there a reason for the subtraction, and do you always subtract for simultaneous equations?
This method is called 'elimination' because you need to eliminate one variable from consideration in order to solve for the other.
If you'll notice, equation (1) is
(1) x + 2y = 1
and equation (2) is
(2) 3x - y = 10
My decision to eliminate the y variable caused me to make the y values in each equation additive inverses of each other. Then, when you ADD them, the result is 0.
So, I multipled (2) by 2 to get
(2) 6x - 2y = 20
Now, you can see that the y terms add to zero leaving just
7x = 21
and that leads to
x = 3
You could've done it a number of different ways. This just seemed like the simplest way to me. We can discuss other ways if you like.
It definitely makes sense, but it's not something we've covered yet (therefore I'd rather not go this way). What other methods are there (then I can say which I have learned and go from there)?
EDIT: I've tried substitution and got the right answer that way, thanks a lot for your help .
Hi viral,
You could use matrices to solve the set of equations, however, simultaneous equations and substitution are by far the most basic methods.
smghost
SciCalculator.com - The Scientific Calculator
Well, there's the 'substitution method'. Usually the 'elimination' and 'substitution' methods are taught at about the same time.
You could use a matrix equation or Cramer's Rule, but I'm thinking you haven't covered that either.
So let's try the 'Substitution Method'
(1) x + 2y = 1
(2) 3x - y = 10
Solve (1) for x and substitute it into (2)
(1) x = 1 - 2y
(2) 3(1 - 2y) - y = 10
Continuing with (2), we simplify to
3 - 6y - y = 10
3 - 7y = 10
-7y = 7
y = -1
Using y = -1 into (1) we get
(1) x + 2(-1) = 1
x - 2 = 1
x = 3