Graphical solution of linear inequalities in two variables

From (x-y\le 2), (-y\le 2-x), and reversing the sign gives (y\ge x-2). Do not forget to reverse the sign when dividing by a negative.

Adding \(x\) to both sides of \(2y-x\le 6\) gives \(2y\le x+6\). Dividing both sides by the positive number \(2\) does not reverse the inequality sign, so \(y\le \frac{x+6}{2}\). Therefore, option A is correct. Option C incorrectly keeps \(x\) with a negative sign. Exam tip: reverse an inequality only when multiplying or dividing by a negative number.

The first inequality gives the region above the line and the second gives the region below (y=4). The common region is their overlap.

The governing concept is the graphical solution of simultaneous linear inequalities. The boundary equations are x+y=5 and x+y=2. Since both have the same coefficients of x and y but different constants, they are distinct parallel lines. The inequality x+y≤5 represents the half-plane on or below one line, while x+y≥2 represents the half-plane on or above the other. Their intersection consists of all points satisfying 2≤x+y≤5, which is the closed strip between the two parallel lines. The boundaries are included because both signs are non-strict. Therefore option A is correct. Option B gives only a boundary line, option C is false because many points satisfy both inequalities, and option D has no connection with these linear boundaries.

No value can be both less than or equal to (1) and greater than or equal to (4) at the same time. Hence there is no common solution.

The direct answer is option B: the point is on the boundary but is not included in the solution. Substitute (3,0) into the left side: 3x + y = 3(3) + 0 = 9. Thus the point satisfies the equality 3x + y = 9. However, the inequality is 3x + y < 9, and 9 < 9 is false. Therefore the point is on the boundary line, but the strict inequality excludes the boundary. Option A is wrong because the point does not satisfy the inequality. Option B is correct. Option C is wrong because an inside point would give a value strictly less than 9. Option D is wrong because no point is automatically included; it must be tested. Exam cue: a < or > sign gives a dotted boundary and excludes the line; ≤ or ≥ includes it.

From (x-3y>0), (-3y>-x), and reversing the sign gives (y<\frac{x}{3}). So the region is below the line.

A point belongs to the solution region of an inequality when its coordinates make the inequality true after substitution. For \\(2x+5y\\ge10\\), the boundary is the line \\(2x+5y=10\\), and points on the accepted side give a value at least 10. Testing the listed points is the simplest reliable method.

For \\((1,1)\\), the left side is \\(2+5=7\\), which is less than 10. For \\((0,1)\\), it is 5, also less than 10. For \\((2,0)\\), it is 4, again less than 10. For \\((3,1)\\), it is \\(2(3)+5(1)=6+5=11\\), and \\(11\\ge10\\) is true. Hence only choice D lies in the solution region.

The boundary line of an inequality is found by replacing the inequality sign with an equality. Thus the boundary of \(y\leq-2x+7\) is \(y=-2x+7\). A line written in slope-intercept form is \(y=mx+c\), where m is the slope and c is the y-intercept. Comparing the two equations gives \(m=-2\), so option C is correct.

The negative slope means that the line falls as x increases: when x moves one unit to the right, y decreases by 2 units. The number 7 is the y-intercept, not the slope. Likewise, -7 is not relevant, and 2 misses the negative sign. The inequality symbol determines which side is shaded, but it does not change the slope of the boundary line.

To obtain the boundary line, replace \(\ge\) with equality: \(y=\frac{1}{2}x-3\). For the \(y\)-intercept, put \(x=0\), giving \(y=-3\). Therefore, the correct answer is \(-3\). Here, \(\frac{1}{2}\) is the slope, not the \(y\)-intercept. Exam tip: in \(y=mx+c\), \(c\) is the \(y\)-intercept.

Solving the two oblique lines gives \(x=y=\frac{8}{3}\). This is an inner vertex of the common region.

A line segment has infinitely many points but not a two-dimensional area. It is considered a bounded linear solution.

The direct answer is option A: the line \(x=2\). Think of the two conditions separately. The statement \(x\ge 2\) allows numbers equal to 2 or greater than 2, while \(x\le 2\) allows numbers equal to 2 or less than 2. A number satisfying both must therefore be neither greater nor less than 2; it must be exactly 2. In the coordinate plane, fixing the x-coordinate at 2 gives a vertical line containing every point \((2,y)\), for all real values of \(y\). Option A is correct because it describes this complete common set. Option B, \(y=2\), is a horizontal line and does not require \(x=2\). Option C, the empty set, is wrong because points such as \((2,0)\) do satisfy both inequalities. Option D, the whole plane, is wrong because a point such as \((3,0)\) fails \(x\le2\). Memory cue: \(\ge\) and \(\le\) at the same number force equality.

A common solution must satisfy both inequalities at the same time. The statement \(y>1\) describes all points whose vertical coordinate is above 1, while \(y<1\) describes all points whose vertical coordinate is below 1. A number cannot be strictly greater than 1 and strictly less than 1 simultaneously. The boundary value \(y=1\) is excluded from both inequalities because both signs are strict.

To see this algebraically, a common value would have to obey \(1<y<1\). No real number can lie strictly between the same number and itself. Therefore the intersection of the two solution regions is empty, represented by \(\varnothing\). The line \(y=1\) is not a solution, and neither a whole strip nor the first quadrant describes the intersection. Hence option C is correct.

To find the \(y\)-intercept, put \(x=0\). Then \(3y=12\), so \(y=4\). Therefore, the line meets the \(y\)-axis at \((0,4)\). The point \((4,0)\) lies on the \(x\)-axis, so it is not the \(y\)-intercept. Exam tip: always set \(x=0\) to find a \(y\)-intercept.

To find the \(x\)-intercept, use the boundary line \(4x+y=20\) and put \(y=0\). This gives \(4x=20\), so \(x=5\). Hence, the \(x\)-intercept is \((5,0)\). The point \((0,5)\) lies on the \(y\)-axis, so it is not an \(x\)-intercept. Exam tip: for an \(x\)-intercept, always set \(y=0\).

Dividing the whole inequality by (2) gives (x+y<5). Dividing by a positive number does not change the sign.

To obtain the boundary, write \(2x+y=4\). Since \(>\) is a strict inequality, points on the line are not solutions, so the boundary is dashed. Exam tip: use a solid line for \(\ge\) or \(\le\).

For the point \((5,0)\), \(x=5\), which does not satisfy \(x\le 4\). Hence, this point is not in the solution region. For \((4,1)\), \(x=4\) and \(x+y=5\), so it satisfies all the inequalities. Exam tip: Substitute the coordinates of a point into every inequality separately.

The line is (x+y=5), and the origin does not satisfy (0\ge 5). Since the line is solid, equality is included.

In the first quadrant, the region above both lines extends infinitely. Therefore the solution region is unbounded.

Vertices occur at intersections of boundary lines. Solving \(y=0\) and \(y=x-2\) gives \(0=x-2\), so \(x=2\). Therefore, \((2,0)\) is a vertex of the region. For \((6,6)\), \(y\le x-2\) is false because \(6\le4\) is not true. Exam tip: Substitute each option into all the inequalities to check it quickly.

The boundary line of the inequality is \(3x+y=15\). For the \(x\)-intercept, put \(y=0\): \(3x=15\), so \(x=5\). For the \(y\)-intercept, put \(x=0\): \(y=15\). Hence, the intercepts are \((5,0)\) and \((0,15)\). In option B, the two values are interchanged. Exam tip: To find intercepts of a line, set one variable to zero at a time.

For a linear inequality in two variables, the equation 2x-3y=12 first gives the boundary line. To identify the required half-plane, test a convenient point not on that line, usually the origin. Substitution gives 2(0)-3(0)=0, and 0<12 is true. Therefore the origin satisfies the inequality, so the half-plane containing the origin is the solution region. Because the sign is strict (<), the boundary line itself is not included and should normally be drawn dashed. Option B is correct. Option A reverses the test result, option C incorrectly includes only the line, and option D is false because many points, including (0,0), satisfy the inequality.

Two graphical conventions govern this question. First, a strict inequality such as > does not include its boundary, so the equation x+4y=8 must be drawn as a dashed line rather than a solid line. Second, use a test point to choose the half-plane. The origin gives 0+4(0)=0, and 0>8 is false. Thus the origin side is not the solution side; the region opposite the origin must be shaded. Equivalently, points with x+4y greater than 8 are selected. Therefore option C is correct. Options A and B incorrectly use a solid boundary, which would represent ≥ or ≤. Option D uses the dashed line correctly but shades the wrong side because the origin fails the inequality.