I have two equations;

(-1/3 x + 14/3) = (2x + 2)

I require to find the value of "x" to show where two lines cross simultaneously on a graph which I have plotted.

On the graph the two lines intersect at approximately x- = 1.1.

I have tried to transpose the above equations to solve for "x", this is what I have concluded;

( - 1/3x + 14/3) = (2x + 2)

( - 1/3x) = (2x + 2 - 14/3)

Fine to here, then errors consistently creep in.
( - 1/3x

- 2x) = (2 - 14/3)

( - 1/3

- 2x^2) = (2 - 14/3)

?? If you multiply the left side by

, then you have to remember to multiply the right side by

too!

(2x^2) = (2 - 14/3 + 1/3)

(x^2) = (4/1 - 14/3 + 1/3)

(x^2) = (4/9 - 14/9 + 1/9)

(x^2) = -1

x = 1 Please review this. 1^2\neq{-1}.

has

**no real solutions** for

.

Now I don't profess to say that the above transposition is correct, but the answer appears to be very close to the intersection of the graph, where the graph is visually about x- = 1.1

Please advise if there is a more precise way to solve for "x"

Thanks

David