Question 1

Solve for a^3 from a reciprocal linear relation: If a + 1/a = 1 (a ≠ 0), determine the value of a^3.

Accepted Answer

Introduction / Context: The relation a + 1/a = 1 leads to a quadratic equation in a. From that, powers of a can be reduced systematically. Here we are asked for a^3, which can be found without computing any decimals by using the quadratic to eliminate higher powers. Given Data / Assumptions: a + 1/a = 1 a ≠ 0 Find a^3 Concept / Approach: Multiply the given equation by a to form a quadratic: a^2 − a + 1 = 0. Then express a^2 in terms of a and substitute into a^3 = a*a^2 to reduce a^3 to a linear expression in a, which then collapses to a constant using the same quadratic again. Step-by-Step Solution: From a + 1/a = 1 ⇒ multiply by a: a^2 + 1 = a ⇒ a^2 − a + 1 = 0.Hence a^2 = a − 1.Compute a^3 = a * a^2 = a(a − 1) = a^2 − a.Replace a^2 using a^2 − a + 1 = 0 ⇒ a^2 = a − 1.Thus a^3 = (a − 1) − a = −1. Verification / Alternative check: The roots of a^2 − a + 1 = 0 are complex conjugates with magnitude 1; their cube equals −1, consistent with the algebraic reduction. Why Other Options Are Wrong: 2, 4, −2: None satisfy the derived identity; substituting back into a + 1/a = 1 fails. Common Pitfalls: Mistaking a^2 − a + 1 = 0 for the cube roots of unity equation a^2 + a + 1 = 0. Arithmetic slips when computing a^3 = a(a − 1). Final Answer: – 1

Question 2

Using orthogonal-sum pattern: Given ax + by = 6, bx − ay = 2, and x^2 + y^2 = 4, find the value of a^2 + b^2.

Accepted Answer

Introduction / Context: This problem uses a standard identity: (ax + by)^2 + (bx − ay)^2 = (a^2 + b^2)(x^2 + y^2). Recognizing this pattern allows immediate evaluation of a^2 + b^2 without solving for x or y individually. Given Data / Assumptions: ax + by = 6 bx − ay = 2 x^2 + y^2 = 4 Find a^2 + b^2 Concept / Approach: The expression (ax + by, bx − ay) behaves like the image of (x, y) under a scaled rotation. Squaring and adding corresponds to preserving x^2 + y^2 up to the scale factor a^2 + b^2. Hence compute LHS and divide by x^2 + y^2 to get the desired sum of squares of coefficients. Step-by-Step Solution: Compute LHS: (ax + by)^2 + (bx − ay)^2 = 6^2 + 2^2 = 36 + 4 = 40.By identity, LHS = (a^2 + b^2)(x^2 + y^2) = (a^2 + b^2) * 4.Thus (a^2 + b^2) * 4 = 40 ⇒ a^2 + b^2 = 10. Verification / Alternative check: Choose any (x, y) with x^2 + y^2 = 4 and construct a, b to match the given linear forms; the identity ensures the same result. Why Other Options Are Wrong: 2, 4, 5: Do not satisfy the proportionality with the computed 40 on the left and 4 on the right. Common Pitfalls: Forgetting that cross terms cancel when adding the two squares. Arithmetic error in 6^2 + 2^2. Final Answer: 10

Question 3

Circle area from two-point radius: In the xy-plane, points P(2, 0) and Q(5, 4) are given. If a circle has radius equal to the distance PQ, find its area (use π as pi).

Accepted Answer

Introduction / Context: The question combines distance formula in the coordinate plane with the area formula for a circle. First compute the segment length PQ; that length becomes the radius r. Then evaluate area = π * r^2. Given Data / Assumptions: P = (2, 0) Q = (5, 4) Area of circle = π * r^2 with r = PQ Concept / Approach: Use r = sqrt[(x2 − x1)^2 + (y2 − y1)^2]. Then square r to compute area. This avoids working with square roots at the end since r^2 is needed directly in the area formula. Step-by-Step Solution: Compute differences: Δx = 5 − 2 = 3; Δy = 4 − 0 = 4.Distance squared: r^2 = 3^2 + 4^2 = 9 + 16 = 25.Thus r = 5.Area = π * r^2 = π * 25 = 25 π. Verification / Alternative check: The 3-4-5 right triangle is a standard Pythagorean triple; distance 5 is expected, making area 25 π immediate. Why Other Options Are Wrong: 16 π, 14 π, 32 π: These correspond to incorrect computations of r^2; only 25 π matches Δx^2 + Δy^2 = 25. Common Pitfalls: Using r instead of r^2 in the area formula or mixing up Δx and Δy. Final Answer: 25 π

Question 4

In the xy-coordinate system, the points (a, b) and (a + 3, b + k) both lie on the line defined by x = 3y − 7. Determine the value of k that must hold for the second point to remain on the same line.

Accepted Answer

Introduction / Context: In coordinate geometry, if two points lie on the same straight line, each point must satisfy the line’s equation. This question tests your ability to substitute coordinates into a linear equation written in the form x = 3y − 7 and solve for an unknown increment k that keeps a translated point on the same line. Given Data / Assumptions: Line: x = 3y − 7 Point 1: (a, b) lies on the line Point 2: (a + 3, b + k) also lies on the line a, b, k are real numbers Concept / Approach: If (x, y) lies on x = 3y − 7, then x equals 3y − 7. Apply this for each point. Use the fact that the first point already satisfies a = 3b − 7, then impose the second point to find k by simple algebraic elimination. Step-by-Step Solution: From (a, b) on the line: a = 3b − 7.For (a + 3, b + k) on the same line: a + 3 = 3(b + k) − 7.Expand: a + 3 = 3b + 3k − 7.Substitute a = 3b − 7 into the left side: (3b − 7) + 3 = 3b + 3k − 7.Simplify left side: 3b − 4 = 3b + 3k − 7.Subtract 3b both sides: −4 = 3k − 7.Add 7 both sides: 3 = 3k.Therefore, k = 1. Verification / Alternative check: Pick any b (say b = 3), then a = 3*3 − 7 = 2. The second point should be (5, 4) when k = 1. Check: x = 5 and 3y − 7 = 3*4 − 7 = 5, which matches. So k = 1 is consistent. Why Other Options Are Wrong: k = 3 leads to a + 3 = 3b + 9 − 7, which contradicts a = 3b − 7. k = 9 and k = 7/3 similarly fail the equality. k = 0 would give a + 3 = 3b − 7, which is false since a = 3b − 7 implies a + 3 = 3b − 4, not 3b − 7. Common Pitfalls: Mixing the roles of x and y in the equation x = 3y − 7; forgetting to use the first point’s relation to substitute for a; or incorrectly distributing 3 over (b + k). Final Answer: 1

Question 5

Area of the trapezium formed in the first quadrant by the x-axis, the y-axis, and the two lines 3x + 4y = 12 and 6x + 8y = 60.

Accepted Answer

Introduction / Context: This geometry problem asks for the area enclosed by the coordinate axes and two parallel lines in the first quadrant. Recognizing intercepts and using triangle area helps compute the area of the resulting trapezium efficiently. Given Data / Assumptions: Lines: 3x + 4y = 12 and 6x + 8y = 60 (they are parallel). Region: bounded by x-axis (y = 0), y-axis (x = 0), and the two given lines in the first quadrant. Area unit: square units. Concept / Approach: Find intercepts with axes to determine vertices. Each line with the axes forms a right triangle with the origin. The trapezium in question is the difference between the larger triangle (from the line farther from the origin) and the smaller triangle (from the line closer to the origin). Step-by-Step Solution: For 3x + 4y = 12: intercepts are x = 4 (set y = 0) and y = 3 (set x = 0). Area of small triangle = 1/2 * 4 * 3 = 6.For 6x + 8y = 60, divide by 2 → 3x + 4y = 30: intercepts are x = 10 and y = 7.5. Area of large triangle = 1/2 * 10 * 7.5 = 37.5.The trapezium area = large triangle area − small triangle area = 37.5 − 6 = 31.5. Verification / Alternative check: Vertices on axes are (0,3), (4,0) for the first line and (0,7.5), (10,0) for the second line. The trapezium is bounded by these four axis points, reinforcing the area difference approach. Why Other Options Are Wrong: 48 sq. unit and 36.5 sq. unit ignore the correct intercepts or the subtraction step. 37.5 sq. unit is the large triangle alone, not the difference. 24 sq. unit is an arbitrary miscalculation. Common Pitfalls: Not noticing that the lines are parallel; forgetting to reduce 6x + 8y = 60; mixing up intercepts; or adding instead of subtracting the triangle areas. Final Answer: 31.5 sq. unit

Question 6

Let x = 1 + √2 + √3. Compute the exact value of the expression 2x^4 − 8x^3 − 5x^2 + 26x − 28.

Accepted Answer

Introduction / Context: This problem evaluates a higher-degree polynomial at x = 1 + √2 + √3. Such expressions often simplify to a neat surd multiple (like √6), provided the coefficients are crafted to cancel many terms. The goal is to compute the exact value without resorting only to decimal approximations. Given Data / Assumptions: x = 1 + √2 + √3 Target expression: 2x^4 − 8x^3 − 5x^2 + 26x − 28 We use algebraic expansion and rational-surd simplifications. Concept / Approach: Although direct expansion is possible, a structured approach helps: set t = √2 + √3 so x = 1 + t. Compute needed powers of t using (√2 + √3)^2 = 5 + 2√6 and recognize patterns where conjugate pairs or symmetric terms cancel. After obtaining x^2, x^3, and x^4, substitute into the polynomial and simplify surd parts. Step-by-Step Solution: Let t = √2 + √3 ⇒ t^2 = 5 + 2√6.Then x = 1 + t, so compute x^2, x^3, x^4 via binomial expansions.Substitute x^2, x^3, x^4 into 2x^4 − 8x^3 − 5x^2 + 26x − 28.Collect rational terms and surd terms (multiples of √6). Many terms cancel by design.The final simplified form evaluates exactly to 6√6. Verification / Alternative check: Numerically, √2 ≈ 1.4142, √3 ≈ 1.7321, so x ≈ 4.1463. Evaluating the polynomial gives approximately 14.6969. Since √6 ≈ 2.4495, 6√6 ≈ 14.6969, confirming the exact value matches. Why Other Options Are Wrong: 0, 3√6, 2√6, and √6 are either too small or do not match the computed numerical value. Only 6√6 aligns with the precise simplification and the numerical check. Common Pitfalls: Mistakes occur when expanding powers of (√2 + √3) or forgetting that √2 * √3 = √6. Sign errors in the polynomial substitution also commonly lead to incorrect constants or surd coefficients. Final Answer: 6√6

Question 7

Given 4x = 18y, determine the exact value of (x / y) − 1.

Accepted Answer

Introduction / Context: This quick algebra question checks proportional reasoning and manipulation of ratios. From a simple linear relation between x and y, you are asked to find (x / y) − 1 exactly as a rational number. Given Data / Assumptions: Equation: 4x = 18y. Both x and y are real numbers (y ≠ 0 for x/y to be defined). Concept / Approach: Transform the linear relation into a ratio for x/y. Divide both sides by 4y to isolate x/y directly, then subtract 1. Keep fractions in simplest terms to avoid arithmetic errors. Step-by-Step Solution: Start with 4x = 18y.Divide both sides by 4y: (4x)/(4y) = (18y)/(4y) ⇒ x/y = 18/4.Reduce the fraction: x/y = 9/2.Compute (x/y) − 1 = (9/2) − 1 = (9/2) − (2/2) = 7/2. Verification / Alternative check: Pick numbers satisfying 4x = 18y. For example, let y = 2 ⇒ 18y = 36 ⇒ 4x = 36 ⇒ x = 9. Then x/y = 9/2 and (x/y) − 1 = 7/2, confirming the result. Why Other Options Are Wrong: 5/2, 3/2, 2, and 9/2 do not represent (x/y) − 1. Note 9/2 is x/y itself, not the requested expression. The others are arbitrary mis-subtractions or reductions. Common Pitfalls: Forgetting to reduce 18/4 to 9/2; subtracting 1 incorrectly; or computing y/x by mistake. Always check which ratio is asked for and simplify carefully. Final Answer: 7/2

Question 8

Let x = 2 + √3 and y = 2 − √3. Find the exact value of (x^2 + y^2) / (x^3 + y^3).

Accepted Answer

Introduction / Context: This problem involves conjugate surds x = 2 + √3 and y = 2 − √3. Sums and products of such pairs often simplify nicely because cross terms cancel. The task is to evaluate the ratio (x^2 + y^2) / (x^3 + y^3) exactly. Given Data / Assumptions: x = 2 + √3 and y = 2 − √3 We need x^2 + y^2 and x^3 + y^3 Use identities with conjugates to simplify. Concept / Approach: Use symmetric identities: x + y and xy are rational for conjugates. Specifically, (x + y) = 4 and xy = (2 + √3)(2 − √3) = 4 − 3 = 1. Then apply formulas: x^2 + y^2 = (x + y)^2 − 2xy and x^3 + y^3 = (x + y)^3 − 3xy(x + y). Step-by-Step Solution: Compute x + y = (2 + √3) + (2 − √3) = 4.Compute xy = (2 + √3)(2 − √3) = 4 − 3 = 1.x^2 + y^2 = (x + y)^2 − 2xy = 4^2 − 2*1 = 16 − 2 = 14.x^3 + y^3 = (x + y)^3 − 3xy(x + y) = 4^3 − 3*1*4 = 64 − 12 = 52.Therefore, (x^2 + y^2)/(x^3 + y^3) = 14/52 = 7/26. Verification / Alternative check: Numerically, x ≈ 3.732 and y ≈ 0.268. Compute x^2 + y^2 ≈ 13.93 + 0.072 ≈ 14.002 and x^3 + y^3 ≈ 51.96 + 0.036 ≈ 51.996 → ratio ≈ 0.2692 ≈ 7/26 (≈ 0.26923). Why Other Options Are Wrong: 7/38, 7/40, and 7/19 come from incorrect substitution or algebra. 1/2 is a common guess but does not match the exact identity-based result. Common Pitfalls: Forgetting identities for sums of squares and cubes; mixing up xy and x + y; or approximating too early and losing exactness. Work symbolically first, then check numerically. Final Answer: 7/26

Question 9

Find the area of the triangle formed by the lines 5x + 7y = 35 and 4x + 3y = 12 together with the x-axis.

Accepted Answer

Introduction / Context: Two nonparallel lines intersect above the x-axis, and each meets the x-axis at a positive x-intercept. Along with the x-axis, these form a triangle. We need the triangle’s area using intercepts and the intersection point. Given Data / Assumptions: Lines: 5x + 7y = 35 and 4x + 3y = 12. x-axis: y = 0. Area unit: square units. Concept / Approach: Find the x-intercepts by setting y = 0. Then find the intersection of the two lines. With a base on the x-axis, the triangle’s area equals (1/2) * base * height, where the base is the distance between the two x-intercepts and the height is the y-coordinate of the intersection point (vertical distance to the x-axis). Step-by-Step Solution: For 5x + 7y = 35 with y = 0 ⇒ x = 7. First intercept: (7, 0).For 4x + 3y = 12 with y = 0 ⇒ x = 3. Second intercept: (3, 0).Base length on x-axis = 7 − 3 = 4.Solve the system: 5x + 7y = 35 and 4x + 3y = 12 ⇒ intersection P(−21/13, 80/13).Height (vertical) = y of P = 80/13.Area = (1/2) * base * height = (1/2) * 4 * (80/13) = 160/13. Verification / Alternative check: Plotting sketches reveal a base segment from x = 3 to x = 7 on the x-axis, with the third vertex above the axis, matching the computed height and area formula. Why Other Options Are Wrong: 150/13 and 140/13 arise from arithmetic slips. 10 and 12 are rough guesses that ignore the exact intersection height 80/13. Common Pitfalls: Using y-intercepts instead of x-intercepts for the base; sign mistakes when solving the system; or forgetting that area must be positive and uses the vertical height to the base line. Final Answer: 160/13 sq. unit

Question 10

Coordinate geometry application: Find the area (in square units) of the triangle formed by the three lines x = 4, y = 3, and 3x + 4y = 12. Clearly identify each pairwise intersection point and then compute the triangle's area.

Accepted Answer

Introduction / Context: This problem uses straight-line graphs to form a triangle. The goal is to find the three vertices by intersecting the given lines and then compute the triangle's area. It tests core coordinate geometry skills: finding intersections and applying a basic area formula for a right triangle. Given Data / Assumptions: Lines: x = 4 (vertical), y = 3 (horizontal), and 3x + 4y = 12 (a slanted line). All intersections are within the real plane. Area is requested in square units. Concept / Approach: The intersections of x = 4 and y = 3 with 3x + 4y = 12 give two vertices on the axes. The third vertex is simply the intersection of x = 4 and y = 3. The area of a right triangle is (1/2) * base * height; here the legs will be horizontal and vertical, making the calculation straightforward. Step-by-Step Solution: Intersect x = 4 with 3x + 4y = 12 ⇒ 3(4) + 4y = 12 ⇒ 12 + 4y = 12 ⇒ y = 0, so A(4, 0).Intersect y = 3 with 3x + 4y = 12 ⇒ 3x + 12 = 12 ⇒ x = 0, so B(0, 3).Intersect x = 4 with y = 3 ⇒ C(4, 3).Segment BC is horizontal (length 4), segment CA is vertical (length 3), so triangle ABC is right-angled at C.Area = (1/2) * 4 * 3 = 6 square units. Verification / Alternative check: Plotting quickly shows the right angle at (4, 3) with legs along x and y directions. Shoelace formula also yields 6 if applied to (4,0), (0,3), (4,3). Why Other Options Are Wrong: 12 and 8 correspond to using full rectangle areas (4*3) or mis-halving; 10 is an arbitrary miscalculation. The only correct area is 6. Common Pitfalls: Confusing which intersections lie on axes, or forgetting to take half the product of the perpendicular legs when computing the area. Final Answer: 6

Question 11

Algebraic identity with radicals (minimal repair for clarity): Let x = √a + 1 and y = √a − 1 with a > 0. Evaluate the expression (x^4 + y^4 − 2x^2 y^2) / [ (√a)(√a) ].

Accepted Answer

Introduction / Context: Expressions like x^4 + y^4 − 2x^2 y^2 simplify dramatically using the identity x^4 + y^4 − 2x^2 y^2 = (x^2 − y^2)^2. With x and y written in terms of √a, the task reduces to computing a simple difference and then dividing by (√a)(√a) = a. Given Data / Assumptions: x = √a + 1, y = √a − 1, with a > 0. Compute (x^4 + y^4 − 2x^2 y^2) / a. Concept / Approach: First compute x^2 − y^2, because the target expression equals (x^2 − y^2)^2 / a. Squaring binomials with radicals requires care but is routine. Finally, divide by a to obtain a constant. Step-by-Step Solution: x^2 = (√a + 1)^2 = a + 2√a + 1y^2 = (√a − 1)^2 = a − 2√a + 1x^2 − y^2 = (a + 2√a + 1) − (a − 2√a + 1) = 4√aTherefore, x^4 + y^4 − 2x^2 y^2 = (x^2 − y^2)^2 = (4√a)^2 = 16aDivide by a: (16a)/a = 16 Verification / Alternative check: Recognize symmetry: replacing 1 by −1 in y changes only the middle term of (√a ± 1)^2, making the difference double that middle term; squaring then removes the radical. Why Other Options Are Wrong: 5, 10, and 20 stem from arithmetic slips like squaring 4√a incorrectly (e.g., writing 4a or 8a) or forgetting to divide by a. Common Pitfalls: Miscomputing (√a ± 1)^2 or neglecting that (√a)(√a) = a leads to wrong constants. Final Answer: 16

Question 12

Use the “sum to zero” cubes identity: If a + b + c = 8, evaluate (a − 4)^3 + (b − 3)^3 + (c − 1)^3 − 3(a − 4)(b − 3)(c − 1).

Accepted Answer

Introduction / Context: This is a direct application of the identity p^3 + q^3 + r^3 − 3pqr = (p + q + r)(p^2 + q^2 + r^2 − pq − qr − rp). Whenever p + q + r = 0, the entire expression simplifies to 0. The shifts by 4, 3, and 1 are designed precisely to make the sum vanish. Given Data / Assumptions: a + b + c = 8. Let p = a − 4, q = b − 3, r = c − 1. Concept / Approach: Compute p + q + r and use the cubes identity. If p + q + r = 0, then p^3 + q^3 + r^3 − 3pqr must be 0, regardless of the actual individual values of a, b, and c. Step-by-Step Solution: p + q + r = (a − 4) + (b − 3) + (c − 1) = (a + b + c) − (4 + 3 + 1) = 8 − 8 = 0Therefore, (a − 4)^3 + (b − 3)^3 + (c − 1)^3 − 3(a − 4)(b − 3)(c − 1) = 0 Verification / Alternative check: Plug in any specific numbers satisfying a + b + c = 8, e.g., a = 4, b = 3, c = 1. Then each bracket becomes 0, making the expression 0 immediately. Why Other Options Are Wrong: Any nonzero value would contradict the identity when p + q + r = 0. Common Pitfalls: Forgetting the shift and computing cubes of a, b, c directly; or using incorrect variations of the identity. Final Answer: 0

Question 13

Minimal repair (clarity): Find the minimum value of the expression x^2 + 1 − 3 for real x. State the least possible value.

Accepted Answer

Introduction / Context: The original stem is garbled; applying the Recovery-First Policy, we interpret it minimally as finding the minimum value of x^2 + 1 − 3. This preserves the intended skill: identifying minima of a simple quadratic expression. Given Data / Assumptions: Expression: x^2 + 1 − 3 = x^2 − 2. x is any real number. Concept / Approach: For any real x, x^2 ≥ 0. A quadratic of the form x^2 + k reaches its minimum when x = 0. Therefore, the least value occurs at the vertex of the parabola, here at x = 0. Step-by-Step Solution: Rewrite the expression: x^2 + 1 − 3 = x^2 − 2Since x^2 ≥ 0 for all real x, the minimum of x^2 is 0Thus, minimum of x^2 − 2 is 0 − 2 = −2 Verification / Alternative check: Plug x = 0 into the expression: 0^2 − 2 = −2. For any nonzero x, x^2 > 0, so the value becomes greater than −2, confirming it is the minimum. Why Other Options Are Wrong: −3 and −1 are values that cannot be achieved by x^2 − 2 since x^2 cannot be negative; 0 is achievable only if x^2 = 2, which is larger than the minimum. Common Pitfalls: Missing the simplification to x^2 − 2 or thinking x^2 can be negative. Remember x^2 ≥ 0 for all real x. Final Answer: − 2

Question 14

Standard cube-sum identity with cyclic differences (minimal repair): Let x = a − b, y = b − c, z = c − a. Evaluate x^3 + y^3 + z^3 − 3xyz.

Accepted Answer

Introduction / Context: This classic identity says x^3 + y^3 + z^3 − 3xyz = (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx). When x + y + z = 0, the entire expression equals 0. With cyclic differences x = a − b, y = b − c, z = c − a, the sum vanishes automatically. Given Data / Assumptions: x = a − b, y = b − c, z = c − a. a, b, c are real numbers. Concept / Approach: Compute x + y + z: it telescopes to 0. Then apply the identity to conclude the target expression is 0, without needing the individual values of a, b, c. Step-by-Step Solution: x + y + z = (a − b) + (b − c) + (c − a) = 0Thus, x^3 + y^3 + z^3 − 3xyz = (x + y + z)(...) = 0 * (...) = 0 Verification / Alternative check: Choose specific values, e.g., a = 3, b = 2, c = 1 ⇒ x = 1, y = 1, z = −2. Compute 1 + 1 − 8 − 3*(1*1*−2) = −6 + 6 = 0. Why Other Options Are Wrong: a + b + c, 3abc, or scaled sums require nonzero (x + y + z); here it is 0, forcing the whole expression to 0. Common Pitfalls: Forgetting to test x + y + z first; the identity simplifies dramatically once you notice the telescoping sum. Final Answer: 0

Elementary Algebra Questions