Plotting $y=f(x)$ and $y=g(x)$ on the same axes, any point where the two graphs cross satisfies both equations at once — so its coordinates are the solution to the simultaneous system. Solving graphically is really just finding where $f(x)=g(x)$, but visually.

• Linear–linear: exactly one intersection point (unless the lines are parallel).
• Linear–quadratic: zero, one (if the line is tangent to the curve), or two intersection points — because a straight line can cross a parabola at most twice.

Rearrange every equation into the form $y=f(x)$ first. A sign slip here (e.g. turning $2x-y=3$ into $y=2x-3$ incorrectly) invalidates the entire plot even if your plotting technique is flawless — so check this step carefully before drawing anything.

For straight lines, 2–3 points (ideally spread across the grid, including $x=0$) plus a ruler are enough. For curves, you need a denser table of values (five or more points), joined with a smooth freehand curve — never straight segments.

1. Mark every point where the graphs cross.
2. Read the $x$- and $y$-coordinates of each intersection, to the precision implied by the grid (often 1 d.p.).
3. State solutions as clearly paired $(x,y)$ values — if there are two intersections, don't mix up which $x$ goes with which $y$.

If you're given the graph of $y=x^2+3x+1$ but asked to solve $x^2+3x-4=0$, manipulate the target equation until the plotted expression appears on one side:

Since $y=x^2+3x+1$ is already drawn, this becomes "solve $y=5$": draw the horizontal line $y=5$ and read off where it meets the curve.

Worked Example

Solve the system $y=x^2+3x+1$ and $y=2x+1$ graphically.

1. Rearrange (already in $y=f(x)$ form for both).
2. Plot the parabola using a table of values, and the line using two points.
3. The line crosses the curve at $(-1,-1)$ and $(0,1)$.

Solutions: $x=-1, y=-1$ and $x=0, y=1$.

Algebraically, you'd substitute one equation into the other and solve the resulting quadratic exactly. Graphically, you're reading the same intersection points off a drawing, which means your answer is only as precise as your plotting and reading allow — hence the tolerance built into the mark scheme. The two methods aren't really separate skills: a graphical question is testing whether you can translate the same underlying algebra into an accurate physical picture, and then read it back out again correctly. If a question doesn't specify a method, solving algebraically first and using the graph only to double-check is often the safest combination for securing full marks.