Any equation in one variable can be pictured as two graphs meeting. To solve $f(x) = g(x)$, graph $y = f(x)$ and $y = g(x)$ and look for where they cross. This generalizes line and parabola systems to any functions, including exponential ones that you cannot yet solve with algebra.
Why it works
A point $(a, b)$ where the graphs cross is on both of them, so $f(a) = b$ and $g(a) = b$. That means $f(a) = g(a)$: the input $a$ solves the equation. The shared height $b$ is just the common output. So the solutions are the $x$-coordinates of the crossing points, and a table of values can find them too: look for inputs where the two outputs agree.
Key idea
The solutions of $f(x) = g(x)$ are the $x$-coordinates of the points where the graphs of $f$ and $g$ meet. Read them from a graph or a table, then check by substituting.
Worked examples
Worked example: Read from a graph
Use the graph above to solve $2^x = x + 1$.
- The curves cross at $(0, 1)$ and $(1, 2)$.
- Check: $2^0 = 1 = 0 + 1$ and $2^1 = 2 = 1 + 1$.
Answer: $x = 0$ or $x = 1$
Worked example: Use a table
$f(x) = x^2$ and $g(x) = 2x + 3$. Make a table for $x = -2, \ldots, 4$ and solve $f(x) = g(x)$.
- $f$: $4, 1, 0, 1, 4, 9, 16$. $g$: $-1, 1, 3, 5, 7, 9, 11$.
- They agree at $x = -1$ (both $1$) and $x = 3$ (both $9$).
Answer: $x = -1$ or $x = 3$
Common mistake
Giving the $y$-coordinate of the crossing as the solution. In the first example the crossing points are $(0, 1)$ and $(1, 2)$, but the solutions are $x = 0$ and $x = 1$. The equation asks for inputs; the shared $y$ is only the value both sides equal there.