Let the cost function be C(x)=c1x+c2 and the revenue function be R(x)=r1x+r2, where the subscripted letters are constants and x is the number of items.
Profit P(x)=r1x+r2-(c1x+c2)=(r1-c1)x+r2-c2=0 at break-even.
Solving for x we get: x=(c2-r2)/(r1-c1), the quantity of items at break-even.
C(x)=R(x)=c1(c2-r2)/(r1-c1)+c2,
C(x)=R(x)=(c1c2-c1r2+c2r1-c1c2)/(r1-c1)=(c2r1-c1r2)/(r1-c1), the cost and revenue at break-even.
So the ordered pair is ((c2-r2)/(r1-c1),(c2r1-c1r2)/(r1-c1)).
That's the algebraic solution which should be the same as the graphical solution.
To draw the graphs, use the intercepts to provide two points for C(x) and two points for R(x).
For C(x) the two intercepts are (-c2/c1,0) and (0,c2) and for R(x) they are (-r2/r1,0) and (0,r2). Draw the two lines through their respective intercepts (don't just join the intercepts). Make sure that the two lines are long enough to intercept. They should intercept at ((c2-r2)/(r1-c1),(c2r1-c1r2)/(r1-c1)). When the relevant constants (numbers) are plugged into these the ordered pair will look a lot simpler.
EXAMPLE
If the overhead cost (machinery, labour, etc.) for producing an item is $100 and the cost of making each individual item is $2 then C(x)=2x+100.
If the revenue has an overhead of $10 and each item sells for $3 then R(x)=3x+10, profit=x-90, so break-even occurs when x=90. C(x)=R(x)=$280.
c1=2, c2=100, r1=3, r2=10; ((c2-r2)/(r1-c1),(c2r1-c1r2)/(r1-c1))=((100-10)/(3-2),(300-20)/(3-2))=(90,280).
The scales of the axes need to accommodate the intercept and because x is a positive quantity and only the positive parts of the lines are relevant, the origin of the graph is best placed in the bottom left-hand corner of the page.
In real life you may be dealing with large numbers for overheads, fixed costs and charges, etc., and quantities so the intercept may not be as accurate because of the resolution of the graph. Algebra and arithmetic will always yield accurate answers.
