Algebra 1 Linear Inequalities in Two Variables and Systems Quiz: 10 Questions with Explanations

A \(y > x – 2\)B \(y \ge x – 2\)C \(y < x - 2\)D \(y > x + 2\)

The line is dashed, so the symbol is strict, and the shading is above, so y is greater.

A \((0, 2)\)B \((2, 3)\)C \((1, 3)\)D \((-1, 0)\)

For (2, 3): 3 < 5 is true. For (1, 3): 3 < 3 is false because the inequality is strict.

A \(y \ge 3x – 2\)B \(y > 3x – 2\)C \(y < 3x - 2\)D \(y > 3\)

Only inequalities with ≤ or ≥ have solid boundaries.

A Yes, it is in the overlapB No, it is on the dashed lineC No, it is below both linesD No, it is above the solid line

Check y ≤ -x + 4: 3 ≤ 1 is false, so (3, 3) is not a solution even though it satisfies y > x – 1.

A \(y < 2x - 3\)B \(y > 2x – 3\)C \(y < -2x + 3\)D \(y > -2x – 3\)

Subtract 4x: -2y > -4x + 6. Dividing by -2 reverses the symbol: y < 2x - 3.

A \((0, 1)\)B \((1, 2)\)C \((0, 0)\)D \((2, 3)\)

The points (0, 1), (1, 2), and (2, 3) all lie on the line y = x + 1, so they cannot decide the side. (0, 0) does.

A \(x \ge 0, y \ge 0, x + y \ge 6\)B \(x \ge 0, y \ge 0, x + y \le 6\)C \(x \le 0, y \le 0, x + y \le 6\)D \(x \ge 0, y \ge 0, x – y \le 6\)

The triangle lies in the first quadrant and below the line x + y = 6.

A \(6x + 3y \ge 30\)B \(3x + 6y \le 30\)C \(6x + 3y = 30\)D \(6x + 3y \le 30\)

At most means less than or equal to, and each ticket price multiplies its own count.

A \(y < 2\)B \(y \le 2\)C \(y \ge 2\)D \(x \le 2\)

The line is solid, so the symbol includes equality, and the shading is below, so y is less than or equal to 2.

A No, because 8 > 8B No, because 6 > 8C Yes, because 8 ≤ 8D Yes, because 8 < 8

Substitute: 4 + 2(2) = 8, and 8 ≤ 8 is true. Points on a solid boundary are solutions.