krit.club logo

Algebra and Graphs - Linear inequalities

Grade 11A Level

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Linear inequalities involve comparing expressions using signs like <<, >>, ≤\leq, and ≥\geq. Solving a linear inequality is similar to solving an equation, except that when multiplying or dividing by a negative number, the direction of the inequality sign must be reversed.

Number line showing x > 0 with an open circle and arrow to the right.
•

On a coordinate plane, the line y=mx+cy = mx + c acts as a boundary. Use a solid line for ≤\leq or ≥\geq (indicating the boundary is included) and a dashed/broken line for << or >> (indicating the boundary is excluded).

A coordinate plane showing a solid boundary line for a linear inequality.
•

The solution to a system of linear inequalities is the overlapping shaded region where all individual inequalities are satisfied simultaneously.

A shaded triangular feasible region satisfying multiple inequalities.
•

To identify the correct region, pick a test point not on the line (often (0,0)(0,0)). If the point satisfies the inequality, shade the side containing that point. If not, shade the opposite side.

📐Formulae

ax+b<cax + b < c

If ax>b and a<0, then x<ba (Sign Reversal Rule)If \ ax > b \ and \ a < 0, \ then \ x < \frac{b}{a} \text{ (Sign Reversal Rule)}

y>mx+c (Region above the line)y > mx + c \text{ (Region above the line)}

y<mx+c (Region below the line)y < mx + c \text{ (Region below the line)}

x=k (Vertical boundary line)x = k \text{ (Vertical boundary line)}

y=k (Horizontal boundary line)y = k \text{ (Horizontal boundary line)}

💡Examples

Problem 1:

Solve the inequality: 5−2x≤115 - 2x \leq 11.

Solution:

−2x≤6  ⟹  x≥−3-2x \leq 6 \implies x \geq -3

Explanation:

Subtract 5 from both sides to get -2x ≤ 6. When dividing by -2, the inequality sign must be flipped from ≤ to ≥.

Problem 2:

Represent the region defined by y≥2x−4y \geq 2x - 4 on a graph.

Solution:

Draw the line y=2x−4y = 2x - 4 as a solid line. Shade the area above the line.

Explanation:

The line is solid because of the 'equal to' part of the symbol (≥). Testing (0,0): 0≥2(0)−40 \geq 2(0) - 4 results in 0≥−40 \geq -4, which is true, so the side containing the origin is shaded.

Problem 3:

Find the integer values of xx that satisfy: −3<2x+1≤7-3 < 2x + 1 \leq 7.

Solution:

−4<2x≤6  ⟹  −2<x≤3-4 < 2x \leq 6 \implies -2 < x \leq 3. Integers: {−1,0,1,2,3}\{-1, 0, 1, 2, 3\}.

Explanation:

Subtract 1 from all parts of the inequality, then divide all parts by 2. The solution includes integers greater than -2 and up to (and including) 3.

Problem 4:

Identify the region on a graph that satisfies the inequality 2x+3y<122x + 3y < 12 for the first quadrant (x≥0,y≥0x \geq 0, y \geq 0).

Graph of 2x + 3y < 12 showing the shaded region in the first quadrant.

Solution:

  1. Find intercepts for the boundary line 2x+3y=122x + 3y = 12.
  • If x=0x = 0, 3y=12⇒y=43y = 12 \Rightarrow y = 4. Point: (0,4)(0, 4).
  • If y=0y = 0, 2x=12⇒x=62x = 12 \Rightarrow x = 6. Point: (6,0)(6, 0).
  1. Draw a dashed line through (0,4)(0, 4) and (6,0)(6, 0) because the inequality is strict (<<).
  2. Test (0,0)(0,0): 2(0)+3(0)=0<122(0) + 3(0) = 0 < 12. This is true, so shade the region including the origin.
  3. Restrict to the first quadrant.

Explanation:

The boundary line intercepts the axes at (6,0)(6,0) and (0,4)(0,4). Since it is a strict inequality, we use a dashed line and shade the area below it toward the origin.

Problem 5:

Solve the system of inequalities graphically and find the vertices of the enclosed region: y≤4y \leq 4, x≤3x \leq 3, and y≥xy \geq x.

A graph showing the region bounded by y=4, x=3, and y=x.

Solution:

  1. Plot y=4y = 4 (horizontal line). Shade below.
  2. Plot x=3x = 3 (vertical line). Shade to the left.
  3. Plot y=xy = x (diagonal line through origin). Shade above (y≥xy \geq x).
  4. The intersection forms a triangle with vertices at (0,0)(0,0), (3,3)(3,3), and (0,4)(0,4) if we consider the y-axis, or (3,4)(3,4) if we use the given constraints. Vertices: (0,0)(0,0), (3,3)(3,3), (3,4)(3,4), and (0,4)(0,4).

Explanation:

The region is bounded by a horizontal line at y=4y=4, a vertical line at x=3x=3, and the identity line y=xy=x. The feasible region is the quadrilateral formed by these constraints in the positive quadrant.