Graphical solution of linear inequalities in two variables

The feasible region is a bounded triangle in the first quadrant, so A is correct. The inequalities x≥0 and y≥0 restrict points to the first quadrant, including the axes. The inequality x+y≤8 means points on or below the line x+y=8 are allowed. This line meets the x-axis at (8,0) and the y-axis at (0,8), while the axes meet at (0,0). Therefore the region has vertices (0,0), (8,0), and (0,8), a closed bounded triangle. B is wrong because the second quadrant has x<0. C is wrong because x+y≤8 cuts off part of the first quadrant. D is wrong because (0,0), for example, satisfies all inequalities. Memory cue: nonnegative x and y mean first quadrant; a ≤ line often closes a bounded region.

Below the line gives (y<), and exclusion of the line makes it strict. Check solid or dashed boundary in exams.

Putting (y=0) gives (x=3), and putting (x=0) gives (y=4). The intercept method is quick for graphing.

Both (x) and (y) are non-negative, so the region is the first quadrant. Do not miss non-negative restrictions in exams.

It is impossible for (x+y) to be at most (2) and at least (5) at the same time. Identify contradictory conditions in exams.

In the first quadrant, the region above (x+y=7) extends infinitely. Distinguish bounded and unbounded regions in exams.

The boundary is a solid line, so A is correct. To draw the boundary of 5x−2y≤10, first replace the inequality sign by equality: 5x−2y=10. Because the symbol is ≤, points on the boundary are included in the solution, so the line must be drawn solid. For comparison, a strict sign < or > excludes the boundary and uses a dashed line. B is therefore wrong because a dashed line represents exclusion. C is wrong because the boundary of a two-variable linear inequality is a complete straight line, not merely separate points. D is wrong because the equation clearly supplies a boundary line. A test point such as (0,0) can then identify the shaded side, although it does not change the solid nature of the line. Memory cue: inclusive signs ≤ and ≥ mean solid; strict signs < and > mean dashed.

Both (x) and (y) lie in bounded intervals, so a rectangle is formed. Use vertical and horizontal lines for such inequalities.

The direct answer is option A: \((2,2)\). A vertex is a corner where boundary lines meet. The two relevant boundaries are \(2x+y=6\) and \(x+2y=6\). Subtracting the second equation from the first gives \(x-y=0\), so \(x=y\). Substituting into \(2x+y=6\) gives \(3x=6\), hence \(x=2\) and \(y=2\). The point satisfies \(x\ge0\), \(y\ge0\), and both upper-bound inequalities with equality, so it is feasible and is a vertex. Option B, \((4,1)\), gives \(2x+y=9>6\), so it is outside. Option C, \((1,4)\), gives \(x+2y=9>6\), so it is outside. Option D, \((3,3)\), violates both upper bounds. The axes also form other vertices, but among the choices only A is a vertex of the feasible region.

In (x<4), (x) is less than (4), so the left side is chosen. For vertical lines, think left-right.

For (y\ge -2), the boundary is included and the region above it is taken. For horizontal lines, check above-below.

The direct answer is option A: the side containing the origin is correct. The boundary of \(-x+y\le1\) is the line \(-x+y=1\). To decide which side to shade, test the origin \((0,0)\). Substitution gives \(-(0)+0=0\), and \(0\le1\) is true. Therefore the origin satisfies the inequality, so the half-plane containing the origin is the solution side. Option A is correct. Option B is wrong because it says the origin side is false, whereas the test calculation proves it true. Option C is wrong because the inequality includes an entire half-plane, not only its boundary; the boundary alone would represent an equality. Option D is wrong because many points satisfy the inequality, for example the origin itself. When the test point lies on the boundary, choose another point, but here the origin gives a clear result. Memory cue: substitute the test point and shade the side that makes the inequality true.

The direct answer is option A: choose a point that does not lie on the boundary line. To decide shading, substitute the coordinates of a test point into the inequality. A point on the boundary makes the two sides equal, so it only tells us that the point is on the line; it does not tell us which side satisfies the inequality. Therefore choose a convenient point away from the line, such as an intercept-free point that is easy to calculate with, and test it. Option A is correct because such a point lies clearly on one side and identifies the required region. Option B is wrong because the origin is not always suitable; here it lies on the boundary. Option C is wrong because
\((1,1)\) may also lie on the boundary and cannot decide the side. Option D is wrong because a test point is needed when the side is not already obvious. Memory cue: equality means boundary; use a point off the line for shading.

For any (x), (y) cannot be both below (x+1) and above (x+4). Check parallel constraints in exams.

Both lines have the same slope and the region lies between them. Use equal slopes to identify parallel lines.

(2(5)+5(0)=10), so the point is on the boundary and included for (\ge). Check equality carefully.

Because of (y\ge), the region above the line is taken and the line is included. The direction is clear in (y)-form.

The line (x+y=6) cuts the axes at ((6,0)) and ((0,6)). Write triangle vertices with axes in exams.

The direct answer is option A,
\((2,4)\). A feasible region contains only points that satisfy every inequality at the same time. Test option A step by step:
\(x=2\ge1\) is true,
\(y=4\ge2\) is true, and
\(x+y=2+4=6\le7\) is true. Thus it belongs to the feasible region. Option B,
\((0,4)\), satisfies the y-condition and the sum condition, but fails
\(x\ge1\) because
\(0<1\). Option C,
\((3,5)\), satisfies the first two conditions, but
\(3+5=8>7\), so it fails the third. Option D,
\((4,1)\), satisfies
\(x\ge1\), but fails
\(y\ge2\) because
\(1<2\). Therefore only A works. Exam cue: for a feasible point, check every inequality, not just most of them.

At origin, (0\le 10) is true, and the solid line indicates (\le). Match both shading and line type.

From \(2x-y<0\), we get \(-y<-2x\). Multiplying both sides by \(-1\) reverses the inequality sign, giving \(y>2x\). The form \(y<2x\) reverses the sign incorrectly, while \(y=2x\) represents only the boundary line and is not included in a strict inequality. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

Since the sign is \(<\), the boundary \(2x+y=4\) is not included, so it must be dashed. Testing the origin gives \(2(0)+0=0<4\); therefore, shade the side containing the origin. Exam tip: decide dashed versus solid before choosing the region.

The boundary-line rule depends on whether equality is included. For x≤5, the boundary is x=5, and the symbol ≤ includes equality, so x=5 must be drawn as a solid line. For y>1, the boundary is y=1, but the strict symbol > excludes equality, so y=1 must be drawn as a dashed line. Hence option A is correct. The shading for x≤5 is on the left side of the vertical line x=5, while the shading for y>1 is above the horizontal line y=1; their intersection is the solution region. Option B incorrectly treats the strict inequality as inclusive. Option C incorrectly excludes the boundary for x≤5. Option D reverses the two line types. Remember that the line style is determined by the inequality symbol, not by whether the boundary is vertical or horizontal.

A solid boundary means points satisfying \(ax+by=c\) are included. The shaded side has smaller values, so the inequality is \(ax+by\le c\). With \(<\), the boundary would be dashed. Exam tip: a solid line indicates equality.

The small bounded region with axes is in the first quadrant and below the line. Use the origin test in exams.