I keep getting the following equation wrong:
$xy=4,\quad$ and $\quad 2x-y - 7 = 0$
Firstly, I solve for y = x - 4, and substitute it in the second equation. Then once I get x from the second equation, I substite it back into the first equation, and get y.
However, my solution set shows that their are two possible answers for each x and y in this simultaneous equation. How is that possible?
Another question - can all equations that can be solved by elimination also be solved by substitution?
Will this be hard?
First, note that $$xy = 4 \implies \frac 1x\cdot xy = \frac 1x\cdot 4 \iff y = \frac 4x$$
But we can make this a bit easier:
Why not use $\;2x - y - 7 = 0\;$ and solve for $y$:
$$y = 2x - 7$$
Now, substitute $y = 2x - 7$ into $xy = 4$ to get $$x(2x - 7) = 4 \iff 2x^2 - 7x - 4 = 0$$
Now you have a quadratic in $x$ to solve. It can be factored: $$\begin{align} 2x^2 - 7x - 4 = 0 &\iff (2x +1)(x - 4) = 0 \\ \\ & \iff (2x+1) = 0,\; \text{or} \;(x -4)= 0\\ \\ &\iff x = -\frac 12 \;\text{ or } x = 4\end{align}$$
(Or, it's perfectly legitimate to use the quadratic formula to solve for $x$.)
Finally, for each solution $x$, substitute into one of the original equations to solve for $y$. And yes, you will find two ordered pairs that solve both equations.
Two solutions?
That's because $xy = 4 \iff y = \frac 4x$ is a hyperbola. If you graph both equations, you'll see that the line $y = 7x-4$ intersects the hyperbola $y = \frac 4x$ in two points, your solutions. If both equations were of lines (non-parallel, non-coincident), then you'd be certain to have exactly one point of intersection.