Graphical solution of linear inequalities in two variables

Given \(x-y=4\), moving \(x\) to the other side gives \(-y=4-x\). Multiplying both sides by \(-1\) gives \(y=x-4\). Hence, option D is the boundary line. Option A has the wrong sign for the constant term. Exam tip: When isolating \(y\) with a negative coefficient, change the signs carefully.

To obtain the boundary line of \(y\ge -x+6\), replace \(\ge\) with equality: \(y=-x+6\). Substituting \(x=2\) gives \(y=-2+6=4\). Therefore, the correct answer is \(4\). Note that \(6\) is the value of \(y\) only when \(x=0\). Exam tip: For a boundary line, first replace the inequality sign with \(=\).

Substituting \((4,0)\) into \(3x+y\le 9\) gives \(3(4)+0=12\). Since \(12\le 9\) is false, the point lies outside the solution region. A point would lie on the boundary only if \(3x+y=9\), which is not true here. Exam tip: test a point by substituting its coordinates into the inequality.

The governing concept is intersecting two half-planes represented by linear inequalities. The boundary equations are x+2y=10 and x+2y=4. Their coefficients are identical, so the lines have the same slope and are distinct parallel lines. Combining the inequalities gives 4≤x+2y≤10. Thus a point is accepted when it lies on or between these two boundaries, producing the closed strip between them. The equality signs show that both parallel boundary lines are included. Option D is therefore correct. The common region is not empty; for example, (0,5) gives x+2y=10 and satisfies both conditions. It is not one line, because values such as x+2y=6 also work, and it is not the whole plane because points with x+2y<4 or x+2y>10 fail at least one inequality.

The direct answer is option B: only \((0,0)\). Since \(x\ge0\) and \(y\ge0\), both variables are non-negative. Their sum is therefore also non-negative. The third condition says \(x+y\le0\). A non-negative sum can be at most zero only when it is exactly zero. For two non-negative numbers, \(x+y=0\) is possible only if \(x=0\) and \(y=0\). Thus the solution region contains just the origin. Option A is wrong because the whole first quadrant contains points such as \((1,1)\), whose sum is 2 and does not satisfy the third condition. Option B is correct. Option C is wrong because the origin satisfies every inequality. Option D is wrong because the entire line \(x+y=0\) includes points with negative coordinates, which violate \(x\ge0\) or \(y\ge0\). Exam cue: non-negative quantities adding to a non-positive number must all be zero.

Substituting ((2,3)) gives (3=2\times 2-1). Equality places the point on the boundary and it is included because of (\le).

A point belongs to the required region only when it satisfies every inequality at the same time. The first two conditions place the point on or to the right of the y-axis and on or above the x-axis. The third condition keeps it on or below the line x+y=8, while the fourth keeps its y-coordinate at most 2.

For (6,2), both coordinates are nonnegative, 6+2=8, and 2≤2, so it is included. For (7,1), 7+1=8 and its y-coordinate is allowed. The origin also satisfies all conditions. For (3,3), although the coordinates are nonnegative and 3+3≤8, its y-coordinate is 3, which is greater than 2. Therefore, (3,3) is outside the common region, so choice C follows.

Dividing the whole equation by (3) gives (2x+y=6). Simplify the equation before drawing the graph.

For ((3,-3)), (y=-3), and (-3>-4) is true. In a strict inequality, boundary points are not included.

In \(x<-2\), every solution point has an \(x\)-coordinate less than \(-2\). Hence, the required half-plane lies to the left of the vertical boundary line \(x=-2\). Since the inequality is strict, the boundary line is not included and is drawn as a dashed line. Option B includes the right side and the boundary, which would represent \(x\ge -2\). Exam tip: for inequalities involving \(x\), check left or right of the vertical line; use a dashed boundary for \(<\) or \(>\).

The inequality x≥1 places the region on or to the right of the vertical line x=1, while x≤5 places it on or to the left of x=5. Together these give the vertical range from x=1 to x=5. Similarly, y≥0 gives the region on or above the horizontal line y=0, and y≤4 gives the region on or below y=4. Their combination has both a fixed horizontal width and a fixed vertical height.

The boundary lines meet at the four points (1,0), (5,0), (1,4), and (5,4). Since all four boundary inequalities are inclusive, the boundary is included, producing a closed rectangle. It is not a triangle because there are four sides, not an unbounded strip because both coordinates are bounded, and not empty because points such as (3,2) satisfy every inequality. Therefore option C is correct.

Substituting the origin \((0,0)\) gives \(2(0)-0=0\). Thus, \(0\le 5\) is true, so the origin lies in the solution region and the half-plane containing it is shaded. Option C is incorrect because a point lies on the boundary only if \(2x-y=5\); at the origin, the left-hand side is \(0\), not \(5\). Exam tip: If the origin is not on the boundary line, substitute it to identify the correct shaded side quickly.

From \(x-y=2\), we get \(y=x-2\). Substituting this in \(2x+y=7\) gives \(2x+x-2=7\), so \(3x=9\) and \(x=3\). Then \(y=3-2=1\). Therefore, the intersection point is \((3,1)\). Option \((4,1)\) does not satisfy the first equation because \(2(4)+1=9\), not 7. Exam tip: Verify an intersection point by substituting it into both equations.

For the same point, (x+y) cannot be both less than (4) and greater than (4). Hence the common solution is empty.

Dividing the whole inequality by (3) gives (x+y\ge 7). Dividing by a positive number does not change the sign.

Since the sign is \(>\), the boundary \(x+2y=4\) is excluded, so it is dashed. Testing the origin gives \(0+2(0)>4\), or \(0>4\), which is false; hence its side is not shaded. Exam tip: use a solid line only when equality is included.

Since \(\ge\) includes equality, the boundary \(2x-y=3\) must be solid. Substituting \((0,0)\) gives \(0\ge3\), which is false, so shade the opposite side. Exam tip: decide solid or dashed before testing a point.

To find the \(x\)-intercept, put \(y=0\). The boundary line \(x+3y=9\) then gives \(x=9\). Therefore, the point on the \(x\)-axis is \((9,0)\). \((0,3)\) is the \(y\)-intercept, not the \(x\)-intercept. Exam tip: for an \(x\)-intercept, always set \(y=0\).

The direct answer is option A: \((5,2)\). To test a point in \(x+5y\ge15\), substitute its coordinates and compare the result with 15. For A, \(5+5(2)=5+10=15\), so the inequality is true because equality is allowed. Thus A lies on the boundary and is included in the solution region. For B, \((4,1)\), the expression is \(4+5=9\), and \(9\ge15\) is false. For C, \((0,2)\), it is \(0+10=10\), also less than 15, so it is outside. For D, \((10,0)\), it is \(10+0=10\), again outside. Therefore only A works. The boundary is \(x+5y=15\); the solution side contains values giving at least 15. Exam cue: substitute carefully and remember equality is included for \(\ge\).

Where (y) is greater than or equal to (x), the inequality is (y\ge x). If the boundary (y=x) is included, equality is also used.

Solving (x+2y=6) and (2x+y=6) gives (x=2) and (y=2). This is an important vertex of the common region.

The vertices are ((1,0)), ((4,0)), ((4,1)), and ((1,4)). With parallel sides (4) and (1) and distance (3), the area is (\frac{1}{2}(4+1)3=\frac{15}{2}).

(0+0\le 6) is true, so the side containing origin is shaded. In exams, always use a test point.

Since the inequality has (>), the boundary line is not included. Remember dashed lines for (<) and (>).

To get the boundary line, replace the inequality sign by equality. First draw the boundary line in exams.