Review question

# When do these simultaneous equations have a positive solution? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8158

## Solution

There are positive real numbers $x$ and $y$ which solve the equations $2x + ky = 4, \qquad x+y = k$ for

1. all values of $k$;

2. no values of $k$;

3. $k=2$ only;

4. only $k>-2$.

The second equation tells us that $x=k-y$.

We can substitute this into the first equation to give $2(k-y)+ky=4$, or $y(k-2) +2k-4=0$.

Note that $2k-4=2(k-2)$, so the equation factorises to $(k-2)(y+2)=0$.

So either $k = 2$, or $y = -2$, or both.

Clearly if $y = -2$, we will never have a solution to the pair of equations that gives positive values for both $x$ and $y$.

Our only hope for such a solution is therefore $k = 2$.

This gives us $2x+2y = 4$ and $x+y=2$. In this case, the lines representing each equation coincide, and we have infinitely many solutions for the pair of equations.

In particular we have the solution $x = 1, y = 1$, and so a solution where both $x$ and $y$ are positive is possible.