This problem tests your ability to solve linear equations involving fractions, a fundamental algebra skill that appears regularly on quantitative reasoning exams.

To solve $$\frac{3}{4}(x-2)=\frac{5}{6}$$, you need to isolate $$x$$ by undoing the operations in reverse order. First, eliminate the fraction coefficient by multiplying both sides by $$\frac{4}{3}$$ (the reciprocal of $$\frac{3}{4}$$):

$$\frac{4}{3} \cdot \frac{3}{4}(x-2) = \frac{4}{3} \cdot \frac{5}{6}$$

The left side simplifies to $$(x-2)$$, and the right side becomes $$\frac{4 \cdot 5}{3 \cdot 6} = \frac{20}{18} = \frac{10}{9}$$:

$$x-2 = \frac{10}{9}$$

Now add 2 to both sides: $$x = \frac{10}{9} + 2 = \frac{10}{9} + \frac{18}{9} = \frac{28}{9}$$

Answer A ($$\frac{28}{9}$$) is correct. Answer B ($$\frac{26}{9}$$) results from incorrectly subtracting 2 instead of adding it, giving $$\frac{10}{9} - 2 = -\frac{8}{9}$$ then making sign errors. Answer C ($$\frac{14}{9}$$) comes from incorrectly multiplying by $$\frac{3}{4}$$ instead of its reciprocal, then adding 2. Answer D ($$\frac{7}{3}$$) represents $$\frac{21}{9}$$, which results from adding 1 instead of 2 to $$\frac{10}{9}$$.

When solving equations with fractions, always multiply by the reciprocal to eliminate fraction coefficients, and double-check your fraction arithmetic by finding common denominators carefully.

When you encounter a quadratic equation like $$2x^2-7x-4=0$$, you need to find the values of $$x$$ that make the equation true. The most reliable approach is using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$.

For this equation, $$a=2$$, $$b=-7$$, and $$c=-4$$. Substituting these values:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-4)}}{2(2)} = \frac{7 \pm \sqrt{49+32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4}$$

This gives us two solutions: $$x = \frac{7+9}{4} = \frac{16}{4} = 4$$ and $$x = \frac{7-9}{4} = \frac{-2}{4} = -\frac{1}{2}$$

Therefore, the solution set is $$\left\{-\frac{1}{2}, 4\right\}$$, which matches choice A.

Choice B gives $${-2, 4}$$ - this keeps the correct positive solution but incorrectly gives $$-2$$ instead of $$-\frac{1}{2}$$. Choice C gives $$\left\{\frac{1}{2}, -4\right\}$$ - this has sign errors on both solutions. Choice D gives $${2, -4}$$ - this also has incorrect values for both solutions, possibly from calculation errors in applying the quadratic formula.

Always double-check your solutions by substituting them back into the original equation. For quadratic equations, the quadratic formula is foolproof when factoring seems difficult, so master this formula and be careful with your arithmetic.

When you encounter an inequality with parentheses, you need to distribute first, then isolate the variable while carefully handling the inequality direction.

Starting with $$-3(2-x)>9$$, distribute the $$-3$$ to both terms inside the parentheses: $$-3(2) + (-3)(-x) = -6 + 3x$$. So the inequality becomes $$-6 + 3x > 9$$.

Next, add 6 to both sides: $$3x > 15$$. Finally, divide both sides by 3: $$x > 5$$. Since we divided by a positive number, the inequality direction stays the same.

Looking at the wrong answers: Answer B ($$x < 5$$) represents the most common error - flipping the inequality sign when dividing by a positive number. Remember, you only flip the inequality when multiplying or dividing by a negative number. Answer C ($$x > -5$$) likely comes from incorrectly distributing the negative sign, perhaps getting $$-6 - 3x > 9$$, which would lead to $$x < -5$$, then incorrectly flipping it. Answer D ($$x < -5$$) results from the same distribution error but without the extra incorrect flip.

You can verify the answer by testing a value: if $$x = 6$$, then $$-3(2-6) = -3(-4) = 12$$, and indeed $$12 > 9$$.

The key strategy here is to work systematically through algebraic steps and remember that inequality direction only changes when multiplying or dividing by negative numbers - not positive ones.

When you encounter absolute value equations, remember that the absolute value represents distance from zero on the number line. The equation $$|2x-7|=3$$ means the expression $$2x-7$$ is exactly 3 units away from zero, which gives us two possibilities.

To solve $$|2x-7|=3$$, you need to consider both cases where the expression inside equals positive and negative 3:

Case 1: $$2x-7=3$$ Adding 7 to both sides: $$2x=10$$ Dividing by 2: $$x=5$$

Case 2: $$2x-7=-3$$ Adding 7 to both sides: $$2x=4$$ Dividing by 2: $$x=2$$

You can verify: when $$x=5$$, we get $$|2(5)-7|=|3|=3$$ ✓, and when $$x=2$$, we get $$|2(2)-7|=|-3|=3$$ ✓.

Looking at the answer choices: Choice A gives $${2,5}$$, which matches our solutions perfectly. Choice B $${1,3}$$ would give us $$|2(1)-7|=5$$ and $$|2(3)-7|=1$$, neither of which equals 3. Choice C $${4,7}$$ yields $$|2(4)-7|=1$$ and $$|2(7)-7|=7$$, again missing our target of 3. Choice D $${5}$$ only includes one of the two correct solutions.

Remember that absolute value equations typically have two solutions unless the expression inside the absolute value bars equals zero at the boundary. Always set up both the positive and negative cases, and don't forget to check your work by substituting back into the original equation.

When you encounter absolute value inequalities, remember that the absolute value represents distance on a number line. The inequality $$|x+1|\le4$$ asks: "For which values of x is the distance from x to -1 equal to or less than 4?"

To solve this systematically, use the definition that $$|A| \le B$$ is equivalent to $$-B \le A \le B$$ when B is positive. Here, we have $$|x+1| \le 4$$, so this becomes:

$$-4 \le x+1 \le 4$$

Subtracting 1 from all parts: $$-4-1 \le x \le 4-1$$ $$-5 \le x \le 3$$

You can verify this by testing boundary values: when $$x = -5$$, we get $$|-5+1| = |-4| = 4 \le 4$$ ✓, and when $$x = 3$$, we get $$|3+1| = |4| = 4 \le 4$$ ✓.

Choice A gives $$-5 \le x \le 3$$, which matches our solution perfectly. Choice B represents the solution to $$|x+1| \ge 4$$, the opposite inequality. Choice C shifts the interval incorrectly—this would be the solution if the original expression were $$|x-1| \le 4$$. Choice D uses strict inequalities and represents values where $$|x+1| > 4$$.

Remember this key pattern: $$|expression| \le number$$ always gives you a single interval (AND condition), while $$|expression| \ge number$$ gives you two separate intervals (OR condition). Don't confuse the direction of the inequality when converting from absolute value form.

When you encounter a system of equations where one equation is linear and the other is quadratic, you're looking for intersection points. The solution requires setting the equations equal to each other since both expressions equal $$y$$.

Setting $$x + 1 = x^2 - 3$$, you need to rearrange this into standard quadratic form. Moving all terms to one side gives you $$x^2 - x - 4 = 0$$. Now apply the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = 1$$, $$b = -1$$, and $$c = -4$$.

Substituting these values: $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}$$

This gives you the two solutions: $$\frac{1 - \sqrt{17}}{2}$$ and $$\frac{1 + \sqrt{17}}{2}$$, which matches choice A.

Choice B uses the correct discriminant ($$\sqrt{17}$$) but incorrectly shows $$-1$$ in the numerator instead of $$+1$$, which would result from mishandling the sign when applying the quadratic formula. Choices C and D both incorrectly calculate the discriminant as $$13$$ instead of $$17$$, likely from computational errors when evaluating $$1 + 16$$. Choice D compounds this error with the same sign mistake as choice B.

Remember: when solving systems with linear and quadratic equations, always double-check your algebra when rearranging to standard form, and be extra careful with signs when applying the quadratic formula—these are the most common sources of error.

This question tests your understanding of exponential growth and requires finding when a power of 2 exceeds a given threshold. When you encounter problems asking for the "least integer n such that..." you're looking for the smallest value that satisfies the inequality.

To solve $$2^n > 1000$$, you need to systematically check powers of 2 around the target value. Since $$2^{10} = 1024$$ is a common benchmark you should memorize, start there. We have $$2^{10} = 1024 > 1000$$, so $$n = 10$$ satisfies the inequality. But is this the smallest such integer?

Check $$n = 9$$: $$2^9 = 512 < 1000$$. Since 512 is less than 1000, $$n = 9$$ doesn't work.

Therefore, $$n = 10$$ is indeed the least integer where $$2^n > 1000$$, making choice (A) correct.

Looking at the wrong answers: (B) $$n = 9$$ fails because $$2^9 = 512 < 1000$$. (C) $$n = 11$$ gives $$2^{11} = 2048 > 1000$$, which satisfies the inequality but isn't the least such integer. (D) $$n = 8$$ fails because $$2^8 = 256 < 1000$$.

Study tip: Memorize key powers of 2 up to $$2^{10} = 1024$$. These appear frequently on quantitative reasoning exams. When finding "least" or "greatest" values satisfying inequalities, always check the boundary cases to ensure you haven't found a value that works but isn't optimal.

When you encounter linear inequalities involving fractions, your goal is to isolate the variable while preserving the inequality's direction. The key is treating inequalities like equations, except when multiplying or dividing by negative numbers.

Starting with $$\frac{5-2x}{3} \le x+1$$, first eliminate the fraction by multiplying both sides by 3: $$5-2x \le 3(x+1)$$. This gives you $$5-2x \le 3x+3$$.

Next, collect all terms with $$x$$ on one side and constants on the other. Subtract $$3x$$ from both sides: $$5-2x-3x \le 3$$, which simplifies to $$5-5x \le 3$$. Then subtract 5 from both sides: $$-5x \le -2$$.

Here's the crucial step: when dividing by $$-5$$ (a negative number), you must flip the inequality sign. Dividing both sides by $$-5$$ gives $$x \ge \frac{2}{5}$$.

Looking at the wrong answers: Choice B ($$x \le \frac{2}{5}$$) results from forgetting to flip the inequality sign when dividing by a negative. Choice C ($$x \ge -\frac{2}{5}$$) comes from an arithmetic error, likely incorrectly handling the signs during algebraic manipulation. Choice D ($$x \le -\frac{2}{5}$$) combines both errors—wrong arithmetic and failing to flip the inequality.

The correct answer is A: $$x \ge \frac{2}{5}$$.

Strategy tip: Always remember that dividing or multiplying an inequality by a negative number flips the inequality sign. This is the most common mistake students make with linear inequalities, so double-check this step every time.

When you encounter exponential equations like this one, the key strategy is to express both sides using the same base, which allows you to equate the exponents directly.

Notice that both 9 and 27 can be written as powers of 3: $$9 = 3^2$$ and $$27 = 3^3$$. Substituting these into the original equation gives us $$(3^2)^x = 3^3$$. Using the power rule for exponents, $$(3^2)^x = 3^{2x}$$, so our equation becomes $$3^{2x} = 3^3$$.

Since the bases are now equal, the exponents must be equal: $$2x = 3$$. Solving for $$x$$ gives us $$x = \frac{3}{2}$$, which is answer choice A.

Let's examine why the other options don't work. Choice B ($$x = 2$$) would give us $$9^2 = 81$$, which is far too large. Choice C ($$x = \frac{4}{3}$$) yields $$9^{4/3} = (3^2)^{4/3} = 3^{8/3} = 3^{2.67...}$$, which equals approximately 15.6, still too small. Choice D ($$x = 1$$) simply gives us $$9^1 = 9$$, which is much smaller than our target of 27.

You can verify: $$9^{3/2} = (3^2)^{3/2} = 3^3 = 27$$ ✓

Study tip: When solving exponential equations, always look for a common base first. Most DAT problems will use bases like 2, 3, 5, or 10, so practice recognizing powers of these numbers quickly.

When you encounter a quadratic equation, you need to find the values of x that make the equation true. For $$\frac{1}{2}x^2 - 3x + 4 = 0$$, you can solve this using the quadratic formula or by testing the given options through substitution.

Let's verify by substituting the values from option A into the original equation. For x = 2: $$\frac{1}{2}(2)^2 - 3(2) + 4 = \frac{1}{2}(4) - 6 + 4 = 2 - 6 + 4 = 0$$ ✓

For x = 4: $$\frac{1}{2}(4)^2 - 3(4) + 4 = \frac{1}{2}(16) - 12 + 4 = 8 - 12 + 4 = 0$$ ✓

Both values satisfy the equation, confirming that A is correct.

Option B gives negative values that, when substituted, don't satisfy the equation. For x = -2: $$\frac{1}{2}(-2)^2 - 3(-2) + 4 = 2 + 6 + 4 = 12 \neq 0$$. Option C fails because x = 1 gives $$\frac{1}{2}(1) - 3(1) + 4 = 2.5 \neq 0$$. Option D is incorrect since x = -1 yields $$\frac{1}{2}(1) + 3 + 4 = 7.5 \neq 0$$.

For quadratic equations on standardized tests, substitution is often faster than using the quadratic formula, especially when answer choices are provided. Always check both solutions when given a pair, as partial credit traps are common where only one value from a pair might work.