Checking answers and numerical precision

Recognize. Substitution checks the original claim, not the steps that produced a candidate.

Enter: (2*4-1)/3

Enter: 4-2

Work. At x = 4, the sides are 7/3 and 2, so the candidate fails. Multiplying the original equation by 3 gives 2x − 1 = 3x − 6, hence x = 5.

Answer: x = 5; the candidate 4 is incorrect.

Check. At x = 5, (2(5) − 1)/3 = 9/3 = 3 and 5 − 2 = 3.

Avoid the trap. Checking a candidate in a line that already contains an algebra error only repeats the error. Return to the original equation and inspect the denominator grouping.

Recognize. The square root is nonnegative, so a negative right side cannot match it.

Enter: sqrt(3+6)

Enter: sqrt(-2+6)

Work. At x = 3, the left side is 3 and the right side is 3. At x = −2, the left side is 2 while the right side is −2. Only 3 survives. The squared equation x² − x − 6 = 0 permits both signs because squaring removes the sign distinction.

Answer: 3 only.

Check. The original domain requires x ≥ −6, but equality further requires x ≥ 0. The candidate −2 passes the radicand condition yet fails the equation.

Avoid the trap. Domain checking alone is not enough here. A radicand can be valid while the two original sides still have opposite signs.

Recognize. The original expression excludes x = 3, even after the factor x − 3 cancels.

Work. For x ≠ 3, the left side simplifies to x + 3. The equation x + 3 = 6 gives the only candidate, x = 3, but that value makes the original denominator zero. Therefore no permitted input remains.

Answer: 0 real solutions.

Check. At every allowed input, x + 3 = 6 would force x = 3. Since that input is excluded, no other candidate can work.

Avoid the trap. A graph of the simplified line can appear to intersect y = 6 at (3, 6). The original function has a hole there; a plot's appearance does not restore the missing input.

Recognize. The exact answer is a short fraction and also a terminating decimal.

Enter: 3/8

Work. x = 3/8 = 0.375. Both 3/8 and .375 are suitable entries. There is no need to approximate this exact terminating value or convert it to a percentage.

Answer: Enter 3/8 or .375.

Check. 8(3/8) = 3 exactly. The answer is positive and less than 1.

Avoid the trap. An entry of 37.5 reports the percent number corresponding to the fraction, not the value of x. A calculator decimal and its interpretation are separate steps.

Recognize. The exact value is 7/12 = 0.583333..., so a two-place decimal discards available precision.

Enter: 7/12

Work. The short exact fraction 7/12 fits the positive-response field. The entry .5833 retains four decimal digits. The entry .58 would receive no credit. For a repeating decimal, fill the available answer space by truncating or rounding, for example .5833. Prefer the exact fraction when it fits.

Answer: Use 7/12 or .5833, not .58.

Check. 12(7/12) = 7. The truncated decimal gives 12(0.5833) = 6.9996, whereas 12(0.58) = 6.96.

Avoid the trap. No instruction in the problem says to round to the nearest hundredth. Do not apply a habitual two-decimal format to every SAT numeric answer.

Recognize. A negative response includes the minus sign in its character count.

Enter: -11/6

Work. The exact value is −11/6 = −1.833333.... The fraction entry -11/6 has five characters. The decimal -1.833 has six characters and uses the negative-response space appropriately.

Answer: -11/6 or -1.833.

Check. 6(−11/6) = −11. A positive response would contradict the negative right side of the equation.

Avoid the trap. The calculator can display more digits than the answer field can accept. Do not copy a long display blindly or omit the negative sign to make it fit.

Recognize. The exact value is an irrational root; the response field is not a symbolic expression editor.

Enter: sqrt(7)

Work. $\sqrt{\text{7}}$ ≈ 2.64575131. Rounding to the displayed-entry precision gives 2.646. Keep the exact radical during any intermediate work, and convert only for the final numeric response.

Answer: 2.646.

Check. 2.6² = 6.76 and 2.7² = 7.29, so the root lies between 2.6 and 2.7. The rounded entry is consistent with that interval.

Avoid the trap. Do not enter sqrt(7) in the numeric-response field or round the value to 3 merely because the prompt asks for one number. A number need not be an integer.

Recognize. The square in the area formula cancels the square root exactly.

Enter: 50*pi

Work. A = π($\sqrt{\text{50}}$)² = 50π ≈ 157.0796327, which rounds to 157.1. Rounding the radius first to 7.1 would instead give π(7.1)² ≈ 158.4, a noticeably different area.

Answer: 157.1 cm².

Check. The ratio A/π must equal r² = 50 exactly. This catches the error introduced by a rounded radius.

Avoid the trap. A direction to round the answer does not authorize rounding every intermediate measurement. Preserve the exact relationship until the final output.

Recognize. The time is three times as long, so the distance must also triple.

Enter: (14/9)*27

Work. The exact rate calculation is (14/9)(27) = 42. Equivalently, 27/9 = 3, giving 14(3) = 42. Retyping the rate as 1.56 kilometers per minute would produce 42.12, which is not the exact answer implied by the given values.

Answer: 42 kilometers.

Check. The units cancel: (km/min)(min) = km. The direct tripling argument independently confirms the result.

Avoid the trap. More digits in a final display do not make an answer more accurate if the input rate was already rounded incorrectly.

Recognize. The number of vans is an integer, and total capacity must be at least 450.

Enter: 450/16

Work. The quotient is 28.125. Twenty-eight vans are insufficient, so choose the next integer, 29. This is a minimum-capacity problem, not ordinary rounding to the nearest integer.

Answer: 29.

Check. 28(16) = 448 < 450, while 29(16) = 464 ≥ 450. The neighboring integer check proves minimality.

Avoid the trap. Rounding 28.125 to 28 leaves two people without seats. The inequality determines the direction of adjustment.

Recognize. The fraction is 18/48, but the variable p names the percent number.

Enter: 100*(18/48)

Work. 18/48 = 3/8 = 0.375. Multiplying by 100 gives p = 37.5. The fraction of tiles and the percent number are different numerical representations of the same comparison.

Answer: Enter 37.5.

Check. 37.5% of 48 is 0.375(48) = 18.

Avoid the trap. An entry of .375 would be correct if the question asked for the probability of selecting a blue tile, but not when it asks for p in p%. Do not include the percent sign in the field.

Recognize. The requested result is an expression of the solution, not either coordinate alone.

Enter: y=2*x+1

Enter: x+y=19

Enter: 2*6+13

Work. The intersection is (6, 13). Substitute these coordinates into the requested expression: 2x + y = 2(6) + 13 = 25. Record the point before closing the graph so its coordinates are not confused.

Answer: 25.

Check. 13 = 2(6) + 1 and 6 + 13 = 19. Then 2x + y = (x + y) + x = 19 + 6 = 25.

Avoid the trap. Both 6 and 13 are useful intermediate results, but neither answers the question. Read the final target again after the graph has done its work.

Recognize. The constant is close to 1 but is not 1. Rounding it changes the solution count.

Enter: y=(x-1)^2-0.000001

Work. Completing the square gives (x − 1)² = 0.000001, so x = 1 ± 0.001. The two distinct roots are 0.999 and 1.001. A narrow window around x = 1 can show their separation, but the exact equation establishes it.

Answer: 2 distinct real solutions.

Check. For either root, (x − 1)² = 0.001² = 0.000001. In contrast, (x − 1)² = 0 has only the one distinct root 1.

Avoid the trap. An apparently touching graph in a broad window is not proof of a repeated root. Preserve all given digits before making an exact count.

Recognize. Two numerical matches can be intersections of different functions rather than evidence of an identity.

Enter: 0^2+1

Enter: 3*0-1

Work. At x = 0, the values are 1 and −1, so the claim is false. Their difference is x² − 3x + 2 = (x − 1)(x − 2), which is zero exactly at the two tested inputs, not everywhere.

Answer: No; x = 0 is a counterexample.

Check. The algebraic difference explains both the agreements and the failure. It is not the zero polynomial.

Avoid the trap. Testing answers is useful for eliminating a false identity. Proving an identity requires a valid general argument, not a handful of successful substitutions.

Recognize. Closeness must be judged relative to the target and the mathematical task. Exact equality is not the same as a small residual.

Enter: sqrt(0.0001)

Work. The positive root is 0.01 because (0.01)² = 0.0001. At x = 0, the output is 0, missing the entire target output. The residual −0.0001 is small in absolute size only because the problem itself uses a small scale.

Answer: The claim is false; the positive solution is 0.01.

Check. 0.01 × 0.01 = 0.0001. Both 0.01 and −0.01 solve the equation, but only the first is positive.

Avoid the trap. Do not apply an arbitrary residual threshold such as “less than 0.001 means correct.” A fixed absolute tolerance is not a proof of exact equality and can erase all the meaningful detail in a small-scale problem.

Check the scale, not just the last few digits

Transfer habit

The residual test identifies both signed roots, but the word positive selects only one. A check must retain the problem’s qualifier as well as the equation. Numerical closeness, exact equality, and satisfaction of every condition are different claims.