The line is a lens: its x-crossing solves the equation, above or below solves the inequality, two crossings solve a system.
Solving and graphing are two views of one thing. The graph of a linear function y = kx + b turns every algebra question into a place on the picture. "Solve kx + b = 0" becomes "find where the line crosses the x-axis." "Solve kx + b > 0" becomes "find where the line sits above the axis." A system of two equations becomes "find where two lines cross," and "which expression is bigger" becomes "which line is higher." This lesson welds equations and inequalities to the straight line — the heart of translating between numbers and shapes.
Pick any linear function y = kx + b and ask the plainest algebra question about it: for which x is y equal to 0? On the graph, "y = 0" means a point that sits on the x-axis — and the line meets the x-axis at exactly one place, the x-intercept. So the x there is the value that makes kx + b = 0. Reading the crossing is solving the equation.
We already know where that crossing is: set y = 0, so 0 = kx + b, which gives x = −b∕k — the very x-intercept formula we derived back in 21.2. Take y = 2x − 4. Its line meets the x-axis at (2, 0), and indeed 2·2 − 4 = 0 — so the solution of 2x − 4 = 0 is x = 2. No rearranging on paper required: the picture hands you the answer.
Solving kx + b = 0 is the same as finding where the graph of y = kx + b cuts the x-axis. That x-value is x = −b∕k.
Once you can read the equation off the line, the inequality comes free. A point on the line has height y = kx + b. Where the line rides above the x-axis its height is positive, so kx + b > 0; where it dips below, the height is negative, so kx + b < 0. The x-intercept is the boundary between the two — the one spot where the height is exactly 0.
Which side is which depends on the tilt. For k > 0 the line climbs left-to-right, so it is above the axis to the right of the crossing and below to the left: 2x − 4 > 0 ⇔ x > 2 and 2x − 4 < 0 ⇔ x < 2. For k < 0 the line falls, so the "above" side is to the left instead — the direction flips. Read it off the picture and you will never get the direction wrong.
The inequality direction flips when k < 0. Solving −2x + 4 > 0 gives x < 2, not x > 2 — because a falling line is above the axis on its left. When in doubt, look at where the drawn line sits above the axis.
Now put two lines on one plane. A point that lies on line A satisfies A's equation; a point on line B satisfies B's. A point on both satisfies both at once — and that is exactly what it means to solve a system of two equations. So the intersection of y = k₁x + b₁ and y = k₂x + b₂ is the solution of the system.
Take y = 2x − 1 and y = −x + 5. They cross at (2, 3) — and (2, 3) checks out in both: 2·2 − 1 = 3 and −2 + 5 = 3. If instead the two lines have the same slope (k₁ = k₂) they are parallel and never cross — exactly the parallel-versus-crossing test we read off the slopes in 21.3 — so the system has no solution.
One shared point on two lines is one shared solution of their two equations. Parallel lines (equal slope) share no point, so the system has no solution.
To solve "one expression is bigger than another," draw both as lines and ask which is higher. At a given x the taller point has the larger value, so where line A sits above line B is exactly where k₁x + b₁ > k₂x + b₂. The two lines swap who is on top exactly at their crossing, so the crossing splits the x-axis into an "A higher" stretch and a "B higher" stretch.
Solve 2x − 1 > −x + 5 this way. The lines cross at x = 2; to the right of it the steeper line y = 2x − 1 is on top, so the answer is x > 2. The algebra agrees: 2x − 1 > −x + 5 → 3x > 6 → x > 2. Picture and algebra tell one story.
The crossing of two lines splits the x-axis into "A is higher" and "B is higher." To solve A > B, take the stretch where line A is on top.
Every question in this lesson became a place on a graph. Here is the whole translation in one table — the unifying picture the entire stage has been building toward. Solving is finding a place on the graph.
| Algebra | On the graph of the line(s) |
|---|---|
| kx + b = 0 | where the line meets the x-axis — the x-intercept |
| kx + b > 0 | where the line is above the x-axis |
| kx + b < 0 | where the line is below the x-axis |
| a system of two equations | where the two lines cross (parallel → no solution) |
| A > B | where line A is higher than line B |
Whenever an equation or inequality is built from y = kx + b, ask "what does this say about the picture?" A solution is never just a number — it is a crossing, a side of the axis, or a stretch where one line beats another.
The graph of y = kx + b is a lens that turns algebra into geometry:
Solve 3x − 6 = 0 by finding the x-intercept of y = 3x − 6.
Set y = 0: 0 = 3x − 6 → x = 2. The line crosses the x-axis at (2, 0), so the solution is x = 2.
Use the same graph to solve 3x − 6 > 0.
Here k = 3 > 0, so the line rises and sits above the axis to the right of the crossing at x = 2. So x > 2.
Solve −2x + 4 ≥ 0 from the graph of y = −2x + 4.
The x-intercept is at −2x + 4 = 0 → x = 2. But k = −2 < 0, so the line falls and is above the axis on its left — the direction flips. So x ≤ 2 (≤ because the inequality is "≥ 0," which includes the boundary point).
The lines y = x + 1 and y = −x + 5 cross where, and what system does that crossing solve?
Set equal: x + 1 = −x + 5 → 2x = 4 → x = 2, then y = 2 + 1 = 3. They cross at (2, 3), which is the solution of the system y = x + 1, y = −x + 5.
For which x is x + 1 > −x + 5?
The lines cross at x = 2. To the right, the rising line y = x + 1 is higher, so the answer is x > 2. (Check by algebra: x + 1 > −x + 5 → 2x > 4 → x > 2.)
Does the system y = 2x + 3, y = 2x − 1 have a solution? Explain.
No. Both lines have slope 2, so they are parallel and never cross — there is no point on both lines, hence no solution. (Algebra: 2x + 3 = 2x − 1 → 3 = −1, false.)
Six questions to lock it in. Tap the answer you think is right.
The big idea. This lesson is the keystone of the stage: it unifies equations, inequalities, and systems under one picture — the graph of a line. Every algebraic question becomes a geometric reading. "Solve kx + b = 0" is "find the x-intercept"; "solve kx + b > 0" is "find where the line is above the axis"; "solve the system" is "find where two lines cross"; "which is bigger" is "which line is higher." Encourage learners to name the picture for each algebra task rather than memorizing rules. From here the stage moves on to piecewise functions (21.6), where a single graph is stitched from several lines, each ruling its own stretch of x.
The misconception to watch. The number-one slip is not flipping the inequality direction when k < 0: students reflexively answer x > 2 for both 2x − 4 > 0 and −2x + 4 > 0. The cure is to read the picture — a falling line is above the axis on its left. Two more traps: assuming every system has a solution (parallel lines don't), and reading the y-intercept when the question asks for the x-intercept (the equation kx + b = 0 lives at y = 0, on the x-axis).
Common Core alignment. This lesson targets 8.EE.C.8 (solve systems by graphing; the intersection is the solution; parallel ⇒ no solution), A-REI.D.10–12 (the graph of an equation is its solution set; solve inequalities and systems graphically), and 6.EE.B.5 / F-IF.C.7a (find and interpret solutions from a graph).