Pretty Basic: Determining the intersection.

I have to do the following problems, however my question is basicly: How would you go about interpreting these problems? At first I thought it might be something like this (for problem 1.A)

Determine the point of intersection of the graphs of each system

x + y = 10

x - y = 2

So I decided to solve for it. So therefore.

x *must* equal 6
y *must* equal 4


Am I close?

-
(x,y) = (6,4) is called solution to the system. It is such a point such that it satisfies both equation. If you plot this in graph, then two lines must meet in at this point. –  Santosh Linkha Jun 26 '12 at 17:36

The graphs of $x+y=10$ and $x-y=2$ are straight lines in the $x,y$ plane. By it's interesecton, we mean the point $P=(x,y)$ where the equations $x+y=10$ and $x-y=2$ are simultaneously satisfied.

What we want in this case is that $x+y-10=0$ and that $x-y-2=0$. This means, we want that

$$x+y-10=x-y-2$$

$$y-10=-y-2$$

$$2y=8$$

$$y=4$$

By substituting $y=4$ in any of our original equations we get that $x=6$. In general, the system $$S:\begin{cases} ax+by+c=0 \cr dx+ey+f=0 \end{cases}$$

has a unique solution $P=(X,Y)$ if and only if $$\frac{a}{b} - \frac{d}{e} \ne 0$$ which simply states that the straight lines aren't parallel: The slope of the first line is $-a/b$ and that if the other is $-d/e$ so we want

\eqalign{ & - \frac{a}{b} - \left( { - \frac{d}{e}} \right) \ne 0 \cr & - \frac{a}{b} + \frac{d}{e} \ne 0 \cr & \frac{a}{b} - \frac{d}{e} \ne 0 \cr}

You'll understand this from an algebraic point of view, rather than from a geometric one when you read about determinants. If $$\frac{a}{b} - \frac{d}{e}=0$$

Then $$\left(\begin{matrix} a & b \\ d & e \end{matrix}\right)$$

has no inverse, so the matrix equation

$$\left(\begin{matrix} a & b \\ d & e \end{matrix}\right)\left(\begin{matrix} x \\ y \end{matrix}\right)= \left(\begin{matrix} c \\ f \end{matrix}\right)$$

has no solution.

-
Your answer is correct; you can verify this by substituting your values for $x$ and $y$ into the two equations and seeing that they are indeed both satisfied.
I don’t know how you got it $-$ possibly by inspection, since these numbers are easy to work with $-$ but here’s perhaps the easiest systematic approach. If you add the two equations together, you get (x+y)+(x-y)=12$, or$2x=12$, and dividing both sides by$2$immediately tells you that$x=6$. Once you have that, you can substitute it into either of the original equations to find$y$:$6+y=10$, so$y=10-6=4$, or$6-y=2$, so$6=y+2$,$4=y\$.