Finding solutions of a system of equations

A solution to a system is a point that makes all of the equations true simultaneously. On a graph, that is exactly where the curves cross, since a crossing point lies on both graphs at once.

The figure shows a line and a parabola. They may cross at two points (two solutions), touch at exactly one point where the line is tangent (one solution), or never meet (no solution).

To solve algebraically when both equations give $y$, set them equal. For $y = x^2$ and $y = x + 2$: $x^2 = x + 2$, so $x^2 - x - 2 = 0$, which factors as $(x - 2)(x + 1) = 0$, giving $x = 2$ or $x = -1$.

Finding $x$ is only half the answer — put each $x$ back into either equation to get $y$. At $x = 2$, $y = 2 + 2 = 4$; at $x = -1$, $y = -1 + 2 = 1$. So the solutions are the points $(2, 4)$ and $(-1, 1)$.

Solve the system $y = x^2 - 6x + 8$ and $y = x + 2$.
Set the right sides equal: $x^2 - 6x + 8 = x + 2$, so $x^2 - 7x + 6 = 0$, which factors as $(x - 1)(x - 6) = 0$.
Then $y = x + 2$: the solutions are $(1, 3)$ and $(6, 8)$.

A line and a parabola form a system with exactly one solution. What does that mean graphically?
Exactly one solution means the graphs meet at exactly one point — the line is tangent to the parabola, just touching it.

The system $5y - 4 = 4x$ and $y = -3x^2 + 16x - 8$ has solutions at $x = \frac{2}{3}$ and $x = 4.4$. Which one satisfies $x > y$?
Substitute into $5y - 4 = 4x$: at $x = \frac{2}{3}$, $y = \frac{4}{3}$, so $x < y$. At $x = 4.4$, $y = 4.32$, so $x > y$.
The answer is $x = 4.4$.