Algebraic Solution of Linear Inequalities in One Variable

The governing concept is interpreting inequality notation from a number-line graph. An open point at 3 means that 3 itself is not included in the solution set; this corresponds to a strict inequality, either < or >. Shading to the left represents numbers smaller than 3, so the correct relation is x < 3. Therefore option A is correct. Option B has the correct leftward direction but incorrectly includes 3, which would require a closed point. Options C and D represent values to the right of 3, so they contradict the shading direction; D would also include the endpoint. In interval notation, the graph represents (−∞, 3). A test value such as x = 2 lies in the shaded region and satisfies 2 < 3, while x = 3 is excluded exactly as the open circle indicates.

The governing language rule is that “at least” includes the stated minimum value. Since the passing requirement is 33 marks, Ravi passes when his marks are exactly 33 or any number greater than 33. Mathematically, this is written as m ≥ 33. The equality sign is essential because m = 33 must be accepted. Option B, m > 33, would incorrectly reject a student scoring exactly 33. Options C and D describe marks no greater than or below 33, which represent the opposite condition and would include failing scores. Thus option A correctly translates the verbal condition into an inequality. This is a one-variable linear inequality because the variable m appears only to the first power.

This is solved using the distributive property followed by the rules for a linear inequality. Expand the left side: 3(2x-1)-2(x+5)=6x-3-2x-10=4x-13. The inequality therefore becomes 4x-13\le x+4. Subtracting x gives 3x-13\le4, and adding 13 gives 3x\le17. Since 3 is positive, division preserves the sign and yields x\le17/3. Thus option C is correct. The boundary value 17/3 is included because the original sign is ‘less than or equal to’. Option A stops before dividing by the coefficient 3, option B reverses the direction and uses the boundary as a lower bound, and option D has both an incorrect direction and an incorrect scale.

The union \((-\infty,1)\cup(1,\infty)\) includes all real numbers less than 1 or greater than 1. Thus, the only excluded value is 1, so it represents \(x\ne 1\). Options C and D describe only one side of 1, whereas this interval includes both sides. Exam tip: Open endpoints at 1 indicate that 1 is not included.

The boundary equation for this inequality is \(2x-y=4\). Since the symbol \(<\) represents a strict inequality, points on the boundary are not included in the solution set, so the boundary is drawn as a dashed line. A solid line is used for inequalities such as \(\leq\) or \(\geq\). Exam tip: use a dashed line for < or > and a solid line for ≤ or ≥.

In \(y\geq 3\), the value of \(y\) must be 3 or greater. The boundary is the line \(y=3\), and because the inequality is \(\geq\), the boundary line is included in the solution. Hence, the region is above and including the line. Exam tip: use a solid boundary line for \(\geq\) or \(\leq\), and a dashed line for strict inequalities.

In \(y<-2\), the value of \(y\) must be less than \(-2\). The boundary \(y=-2\) is a horizontal line, so all points below it satisfy the inequality. Because the inequality is strict, the boundary line itself is not included and is usually drawn dashed. Exam tip: when \(y\) is compared with a constant, draw a horizontal boundary and choose the region above or below it.

In \(x>-1\), the value of \(x\) must be greater than \(-1\). Therefore, all points to the right of the vertical boundary line \(x=-1\) satisfy the inequality. Since the inequality is strict, the boundary line itself is not included. Exam tip: For \(x>a\), select the region to the right of \(a\) on the number line.

Substituting the coordinates of \((0,0)\) gives \(0+0\leq 4\), or \(0\leq 4\), which is true. Therefore, the point satisfies the inequality. It is not on the boundary \(x+y=4\); it lies inside the solution region. In an exam, test a point by substituting its coordinates directly into the inequality.

An open circle means that the boundary value is excluded, as in \(x>a\) or \(x<a\). A filled circle is used for \(\geq\) or \(\leq\), where the endpoint is included. Exam tip: check the inequality sign first.

Every point on the x-axis has \(y=0\). Substituting \(y=0\) in \(x+y=5\) gives \(x=5\). Therefore, the x-coordinate of the intercept is 5, and the intercept point is \((5,0)\). For exams, remember to set \(y=0\) when finding the x-intercept.

To obtain the boundary line of an inequality, replace the inequality sign with an equality sign. Thus, the boundary line of \(3x-y>6\) is \(3x-y=6\). Options A and C describe regions, not the boundary line itself. Exam tip: for strict inequalities such as \(>\) or \(<\), the boundary is generally drawn as a dashed line.

Substituting the coordinates of \\(1,2\\) into \\(x+y\\leq 3\\) gives \\(1+2\\leq 3\\), or \\(3\\leq 3\\). This is true because equality is included in the symbol \\(\\leq\\). Hence, the point lies on the boundary line \\(x+y=3\\). Exam tip: To test a point in an inequality, substitute its coordinates and check whether the resulting statement is true.

For the point \((2,3)\), \(x+y=2+3=5\). Thus, the inequality becomes \(5<5\), which is false. Since this is a strict inequality, equality is not included, so the point does not satisfy it. In exams, carefully distinguish between \(<\) and \(\leq\).

Substituting the origin \,\((0,0)\) into the inequality gives \(0+0\leq 7\), or \(0\leq 7\), which is true. Therefore, the solution region lies on the side containing the origin. Option B incorrectly reverses the inequality sign. Exam tip: In graphing a linear inequality, choose a test point that does not lie on the boundary line.

The inequality \(x \ge 0\) means that the \(x\)-coordinate of every point is zero or positive. Hence, it represents the entire region to the right of the \(y\)-axis and includes the \(y\)-axis because \(x=0\) is allowed. In contrast, \(x \le 0\) represents the region to the left of the \(y\)-axis. Exam tip: an inequality involving \(x\) determines the horizontal side, while one involving \(y\) determines the vertical side.

The inequality \(y\geq 0\) means that the y-coordinate of every point is zero or positive. Therefore, it includes the points on the x-axis as well as all points above it. Option C represents the region where \(y<0\), so it is incorrect. Exam tip: The sign of y indicates vertical position—\(y\geq 0\) lies on or above the x-axis.

The inequality \(x\geq 0\) places a point on or to the right of the y-axis, while \(y\geq 0\) places it on or above the x-axis. Hence, the common region is the first quadrant, including both axes and the origin. Exam tip: the symbol \(\geq\) means that the boundary lines are included.

In the equation (x=4), the x-coordinate is fixed at 4, while y can take any value. Therefore, its graph is a vertical line parallel to the y-axis. A horizontal line generally has the form (y=c). Exam tip: remember that (x=c) represents a vertical line, whereas (y=c) represents a horizontal line.

In the equation \(y=-3\), the \(y\)-coordinate remains fixed at \(-3\), while \(x\) can take any real value. Therefore, its graph is a horizontal line parallel to the \(x\)-axis. A vertical line has a constant \(x\)-coordinate, so option A is incorrect. Exam tip: \(y=c\) represents a horizontal line, whereas \(x=c\) represents a vertical line.

The inequality \(y\leq x+1\) means that the value of \(y\) is less than or equal to the corresponding value on the line \(y=x+1\). Therefore, the solution region lies below the line, and the boundary line is included because of \(\leq\); it is drawn as a solid line. In an exam, remember that \(\leq\) and \(\geq\) include the boundary.

The boundary line is \(y=2x-1\). In \(y>2x-1\), the value of \(y\) must be greater than the corresponding value on the boundary line, so the solution region lies above the line. Since the inequality is strict (\(>\)), the boundary line itself is not included. As an exam tip, you can verify the shaded side by testing a point such as \((0,0)\) in the inequality.

For (4, 0), \(4+0=4\), and for (0, 4), \(0+4=4\); hence both points lie on the line. In option B, the point (1, 1) gives a sum of 2, so it does not lie on the line. Exam tip: Put \(x=0\) and \(y=0\) in the equation to obtain the two intercepts quickly.

At the x-intercept, the point lies on the x-axis, so \(y=0\). Substituting gives \(2x+0=6\), hence \(x=3\). Therefore, the x-coordinate is 3, and the complete intercept is \((3,0)\). Exam tip: To find an x-intercept, set \(y=0\).

To find the y-intercept, set x=0 because every point on the y-axis has an x-coordinate of 0. Then 0+3y=12, giving y=4. Therefore, the y-intercept is 4. Option 3 is only the coefficient of y, not the intercept. Exam tip: set y=0 for the x-intercept and x=0 for the y-intercept.