Write a System of Inequalities for Each Graph: Lines, Dashes, and Shading

Here is the graph, exactly as the quiz hands it over to anyone who must write a system of inequalities for each graph.

Problem. Write a system of inequalities for each graph — the instruction, verbatim. Start with this one: a solid horizontal line, a dashed diagonal, and one shared shading where their half-planes overlap.

Two boundary lines, two line styles, one overlap region. The question — write a system of inequalities for each graph — asks you to translate that picture into two symbol sentences, and nothing on it is decoration. Every mark here is legible at a glance, which is what makes it fair to write a system of inequalities for each graph from this sketch alone.

Answer. The graph is the system $y \ge 1$ and $y < x + 1$.

Read the solid line $y = 1$ first. Solid carries the "or equal to" bar, and the shading sits above, so the first half is $y \ge 1$.

The dashed line $y = x + 1$ is strict — dashes mean the line itself is out — and the shading sits below: $y < x + 1$.

Now let test points vote. Inside the overlap, $(2,2)$: $2 \ge 1$ true, $2 < 3$ true — both halves hold. Outside at $(0,0)$: $0 \ge 1$ fails, and the point does sit off the shading. That two-way vote is the working heart of write a system of inequalities for each graph.

To write a system of inequalities for each graph, let every boundary testify and keep both verdicts at once — one verdict alone describes only a half-plane. When you can write a system of inequalities for each graph this cleanly, the symbols and the picture say the same thing twice.

Any request to write a system of inequalities for each graph surrenders to four reads:

1. Read each boundary line. Grab two clean points on the line, run the slope, and write its equation $y = mx + b$. Horizontal lines skip the algebra: $y = 1$ is just $y = 1$.
2. Read the line style. Solid means the boundary joins the solution, so its sign carries the bar ($\le$ or $\ge$); dashed means strict ($<$ or $>$). Rule of thumb: strict inequality, dashed line.
3. Read the shading side. Substitute one shaded point into the boundary equation and keep the sign that comes out true.
4. Stack and recheck. Write the halves together as a system, then confirm one point of the overlap satisfies every part.

Step 3 is older than it looks. Fine's 1904 College Algebra states it outright: for all pairs of values of x, y whose graphs lie on one side of this line, we shall have $ax + by + c > 0$. For all pairs on the other side of the line, $ax + by + c < 0$. One substitution settles which side you are on.

Run each row against its clues and write a system of inequalities for each graph becomes look-up, not guessing. Vertical boundaries fit the same table with $x$ in place of $y$. Four reads, every time: that is the whole routine behind write a system of inequalities for each graph, start to finish.

Problem. The shading forms a wedge opening to the right, fenced by the dashed lines $y = x$ and $y = 2x$. Here the four reads that write a system of inequalities for each graph run in order.

Read 1, boundaries. The lower fence passes through $(0,0)$ and $(1,1)$: $y = x$. The upper fence passes through $(0,0)$ and $(1,2)$: $y = 2x$.

Read 2, styles. Both fences are dashed, so both halves are strict.

Read 3, sides. Test $(2,3)$: $3 > 2$ true and $3 < 4$ true — the wedge sits above $y = x$ and below $y = 2x$.

Answer. $y > x$ and $y < 2x$.

Where it came from. Fine's 1904 College Algebra drills this exact pair in Exercise LIII: $y - x > 0$, $y - 2x < 0$ — the same two lines, written without solving for $y$. The origin satisfies neither half, which is why the wedge, not the axes corner, carries the shading.

One wedge, two dashes — with that, the routine to write a system of inequalities for each graph has fully played out. Read a graph like this one twice and you can write a system of inequalities for each graph on sight.

Problem. To write a system of inequalities for each graph, the quiz sometimes starts from the symbols instead. Now it runs the question backward: which graph represents the following system of inequalities — $y \le -x + 4$ and $y \ge 2$?

Draw, don't hunt. Graph $y = -x + 4$ through $(0,4)$ and $(4,0)$; the bar under $\le$ keeps the line solid, shading below. Graph $y = 2$ solid as well, shading above.

Answer. The matching graph is the one whose overlap strip sits above $y = 2$ and below $y = -x + 4$, with both boundary lines solid.

Check. $(0,3)$: $3 \le 4$ true, $3 \ge 2$ true — it lives on the strip. $(5,3)$: $3 \le -1$ false, and it sits off it. Forward or reverse, the reads that write a system of inequalities for each graph do not change.

So which graph represents the following system of inequalities is the same four reads in reverse: build each line, honor its style, and intersect the shadings.

Each of these mistakes breaks one of the four reads at a different step.

1. Swapping dashed and solid. Give a strict inequality a solid line and you hand the boundary to the solution for free. Test a point on the line itself — for a dashed line the substitution must fail.

2. Reading the side before solving for $y$. If a negative coefficient still rides with $y$, "above" and "below" lie to you. Solve for $y$ first, keep the flip honest, then read the side — that order is the system in write a system of inequalities for each graph.

3. Writing one inequality and stopping. One half-plane is not a system. The graph's overlap region is the giveaway that more than one boundary is at work — every visible line deserves a sentence.

4. Testing a boundary point. A point on a dashed line proves nothing about the shading. Pick a spot strictly inside the shaded strip, for each half and for the overlap.

Fix these four and you can write a system of inequalities for each graph without losing points — the graph itself grades your answer.

Boundary, style, side, stack — four reads, one system: that is the whole routine. Read the lines into equations, let dashes and solid strokes pick the signs, and let a shaded point vote on every half. That is all it takes, forward or in reverse. The graph already contains the answer; write a system of inequalities for each graph, and its words simply settle into place.