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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?

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    $\begingroup$ The first equation gives $y=\frac{4}{x}$. $\endgroup$ – Joe Johnson 126 Jun 10 '14 at 11:15
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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.

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An alternate solution would be to solve for $y$ in the second equation:

$$y=2x-7$$

Then substitute the result into the first equation:

$$x(2x-7)=4$$

$$2x^2-7x-4=0$$

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The problem in your answer is that $xy=4$ does not imply $y=x-4$. Instead, you should get, from the first equation, that

$$y=\frac{4}{x}$$

Or, if it is easier for you to solve from there, you could also use $x=\frac{4}{y}$ in the second equation. However, don't try to substitute both in.

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