Recognizing useful algebraic structure

Recognize. The requested quantity is already grouped as one factor on the left.

Work. Divide both sides by 7: x − 3 = $\frac{\text{42}}{\text{7}}$ = 6.

There is no need to find x and then subtract 3.

Answer: 6.

Check. If desired, x = 9 makes the original left side 7(9 − 3) = 42. The target is 6, not 9.

Avoid the trap. Many routine equations train you to stop at x = something. Here that habit produces a correct intermediate value but the wrong response.

Recognize. Both variable coefficients in the target are three times those in the known expression.

Work. Rewrite the target: 9a + 6b − 5 = 3(3a + 2b) − 5 = 3(17) − 5 = 46.

Answer: 46.

Check. Choose one valid pair, such as a = 3, b = 4. It satisfies 3a + 2b = 17 and gives 27 + 24 − 5 = 46.

Avoid the trap. There are infinitely many possible pairs (a, b), but that does not make the target undetermined. The entire target depends only on their fixed combination.

Recognize. Adding the equations gives equal coefficients of x and y, creating a multiple of the target.

Work. Add left sides and right sides:

7x + 7y = 56 x + y = 8.

Answer: 8.

Check. Subtracting the equations gives 3x − 3y = 6, so x − y = 2. Together with x + y = 8, this gives x = 5, y = 3, which satisfy both originals.

Avoid the trap. Solving for both variables is valid, but unnecessary. Once the target is established, additional work is a check, not part of the required solution.

Recognize. The numerator is a difference of squares. Its factors are easy even though the squares themselves are large.

Work. Factor before calculating:

$\frac{\text{(501 − 499)(501 + 499)}}{\text{1000}}$ = $\frac{\text{2(1000)}}{\text{1000}}$ = 2.

Answer: 2.

Check. The numerator must be 2,000. Direct calculation gives 251,001 − 249,001 = 2,000.

Avoid the trap. The difference of squares is not the square of the difference. (501 − 499)² = 4 is a different expression.

Recognize. The repeated group is x − 4. Let u = x − 4 to expose a simple quadratic.

Work. The equation becomes u² − 5u − 14 = 0, or (u − 7)(u + 2) = 0. Thus u = 7 or u = −2. Returning to x = u + 4 gives x = 11 or x = 2.

Answer: 11.

Check. At x = 11, the repeated group is 7, so 7² − 5(7) = 14.

Avoid the trap. The greater value of u is 7, but the question asks for x. A temporary substitution must be translated back unless the target is the substituted group itself.

Recognize. Rearranging gives x² + 11 = 8x. Dividing by x creates exactly the requested expression.

Work. First, x = 0 cannot satisfy the original equation because it would give 11 = 0. Therefore division by x is valid: $\frac{{\text{x}}^{\text{2}}\text{+ 11}}{\text{x}}$ = 8 x + $\frac{\text{11}}{\text{x}}$ = 8.

Answer: 8.

Check. The quadratic has discriminant 64 − 44 = 20 > 0, so real solutions do exist. Both give the same target value.

Avoid the trap. Finding a decimal root first adds work and can obscure that the target is identical for both roots.

Recognize. A reciprocal sum equals the ordinary sum divided by the product, provided neither root is zero.

Work. Comparing coefficients gives r + s = 12/3 = 4 and rs = 5/3. Hence

$\frac{\text{1}}{\text{r}}$ + $\frac{\text{1}}{\text{s}}$ = $\frac{\text{r + s}}{\text{rs}}$ = $\frac{\text{4}}{\text{5/3}}$ = $\frac{\text{12}}{\text{5}}$.

Answer: 12/5.

Check. The product is nonzero, so both reciprocals are defined. The discriminant is 144 − 60 = 84 > 0, confirming two real roots.

Avoid the trap. The reciprocal of a sum is not the sum of reciprocals. Here 1/(r + s) = 1/4, which answers a different question.

Recognize. The zeros are 2 and 8, so the axis of symmetry is their midpoint, x = 5. The positive leading coefficient makes the vertex a minimum.

Work. Evaluate at the midpoint: f(5) = 3(3)(−3) = −27.

Answer: −27.

Check. Rewriting gives f(x) = 3[(x − 5)² − 9] = 3(x − 5)² − 27. The square term is nonnegative, so no output is below −27.

Avoid the trap. The vertex input is 5, but the minimum value is its output, −27. The two coordinates serve different roles.

Recognize. Squaring the known sum produces the target plus 2ab.

Work. Use (a + b)² = a² + 2ab + b²:

a² + b² = 9² − 2(14) = 81 − 28 = 53.

Answer: 53.

Check. The pair a = 7, b = 2 satisfies both given conditions and gives 49 + 4 = 53. Swapping the values does not change the target.

Avoid the trap. (a +b)² contains a cross term. Omitting 2ab is the same error whether the symbols represent numbers, side lengths, or function values.

Recognize. Square the known combination. The cross term simplifies because x(1/x) = 1.

Work. Squaring gives (x + $\frac{\text{1}}{\text{x}}$)² = x² + 2 + $\frac{\text{1}}{{\text{x}}^{\text{2}}}$ = 25.

Subtracting 2 yields the requested value 23.

Answer: 23.

Check. Multiplying the original equation by x gives x² − 5x + 1 = 0, which has nonzero real roots. The result does not depend on which root is used.

Avoid the trap. Squaring a sum is not the same as squaring each term and adding. The missing cross term would incorrectly give 25.

Recognize. The models have the same nonzero growth factor at every input. The target is a ratio, so that factor cancels.

Work. $\frac{\text{P(t)}}{\text{Q(t)}}$ = $\frac{\text{240(1.}{\text{07)}}^{\text{t}}}{\text{360(1.}{\text{07)}}^{\text{t}}}$ = $\frac{\text{2}}{\text{3}}$.

Answer: 2/3.

Check. At t = 0 the ratio is 240/360 = 2/3. Both populations are multiplied by the same positive number at every later time, preserving the ratio.

Avoid the trap. Equal percentage growth does not mean the difference stays constant. The difference is 120(1.07)^t, which grows even while the ratio remains fixed.

Recognize. Equal outputs at distinct inputs of a quadratic occur symmetrically about its axis. Algebra can show the required sum directly.

Work. From p² − 6p + 9 = q² − 6q + 9, obtain

(p − q)(p + q) − 6(p − q) = (p − q)(p + q − 6) = 0.

Since p ≠ q, the first factor is nonzero, so p + q = 6.

Answer: 6.

Check. The function is (x − 3)², with symmetry axis x = 3. Inputs equally spaced around 3 sum to 6.

Avoid the trap. Without the distinctness condition, p = q would satisfy the equation for any real value, and the sum would not be fixed.

Recognize. The powers are x⁴ and x². Setting u = x² creates a quadratic, with the restriction u ≥ 0.

Work. Factor u² − 13u + 36 = (u − 4)(u − 9) = 0. Thus x² = 4 or x² = 9, giving x = ±2, ±3. The positive solutions are 2 and 3.

Answer: 5.

Check. Both 2 and 3 satisfy the original equation. The values 4 and 9 are solutions for u, not for x.

Avoid the trap. A quadratic in x² can have four distinct real solutions in x. Count and filter only after translating the substituted variable back.

Recognize. The squared difference can be built directly from the sum of squares and the square of the sum.

Work. Since (x + y)² = x² + 2xy + y², we have 121 = 65 + 2xy, so 2xy = 56. Therefore

(x − y)² = x² + y² − 2xy = 65 − 56 = 9.

Answer: 9.

Alternative. Use (x − y)² = 2(x² + y²) − (x + y)² = 130 − 121 = 9 as an alternative identity.

Avoid the trap. The data allow x − y = 3 or x − y = −3, depending on the order. The squared difference is fixed even though the signed difference is not.

Recognize. Divisibility by x − r means that P(r) = 0. The input x = 1 exposes a + b directly. The second factor supplies additional information but is not needed for this target.

Work. Because x − 1 is a factor, P(1) = 1 + a + b + 6 = 0.

Thus a + b = −7.

Answer: −7.

Check. The other factor gives P(2) = 8 + 4a + 2b + 6 = 0, or 2a + b = −7. Combining this with a + b = −7 yields a = 0 and b = −7. Indeed,

P(x) = x³ − 7x + 6 = (x − 1)(x − 2)(x + 3).

Both stated factors are consistent with the result.

Avoid the trap. You do not have to use every piece of information in the main calculation. But you must not ignore a condition that could invalidate a candidate. Here the direct result is forced by one factor, and the optional check confirms the other condition is compatible.

Transfer. Compare three routes: coefficient matching after an unknown factorization, solving the two root equations, and evaluating only at 1. All are valid; the last is shortest because the target is a sum. A different target, such as a − b, would need more information.