When you encounter an equation like this, you're dealing with a two-step process: first solve for the variable, then use that value to answer what's actually being asked.

To solve $$3x - 7 = 2x + 5$$, you need to isolate $$x$$ by getting all terms with $$x$$ on one side and all constants on the other. Subtract $$2x$$ from both sides: $$3x - 2x - 7 = 2x - 2x + 5$$, which gives you $$x - 7 = 5$$. Then add 7 to both sides: $$x = 12$$.

Now here's the crucial part—the question asks for $$x - 4$$, not just $$x$$. Since $$x = 12$$, then $$x - 4 = 12 - 4 = 8$$. The correct answer is A.

Looking at the wrong choices: B) 12 is the value of $$x$$ itself, which is a classic trap for students who solve correctly but forget to complete the final step. C) 16 would result from adding instead of subtracting: $$x + 4 = 12 + 4 = 16$$. D) -8 might come from sign errors during the solving process, such as incorrectly handling the subtraction when isolating $$x$$.

The key strategy here is to always read the question twice—once before solving and once after. Many ISEE questions will ask for an expression involving your variable rather than the variable itself. After finding $$x$$, always double-check what the question is actually asking for before selecting your answer.

When checking algebraic solutions, you substitute your answer back into the original equation to verify it's correct. This process helps catch calculation errors and confirms your solution is valid.

To find what Maria gets for the left side, substitute $$y = -9$$ into the left side of the original equation $$4(y + 3) = 2y - 6$$. The left side is $$4(y + 3)$$, so:

$$4(y + 3) = 4(-9 + 3) = 4(-6) = -24$$

This confirms that when Maria substitutes $$y = -9$$ into the left side, she gets $$-24$$.

Let's examine why the other answers are incorrect. Choice B (-6) represents the value inside the parentheses after substitution ($$-9 + 3 = -6$$), but fails to multiply by 4. Choice C (0) might come from incorrectly thinking the left and right sides should equal zero, or from calculation errors. Choice D (6) could result from a sign error, perhaps calculating $$4(-9 + 3)$$ as $$4(6)$$ instead of $$4(-6)$$.

Note that if you also substitute $$y = -9$$ into the right side $$2y - 6$$, you get $$2(-9) - 6 = -18 - 6 = -24$$. Since both sides equal $$-24$$, Maria's solution $$y = -9$$ is indeed correct.

Remember: when checking algebraic solutions, substitute carefully and follow order of operations. Work inside parentheses first, then multiply. This systematic approach prevents the sign errors and calculation mistakes that create wrong answer choices.

This equation involves fractions with variables, so you'll need to isolate $$x$$ by eliminating the fractions and combining like terms.

Start by moving all terms with $$x$$ to one side and constants to the other. Subtract $$\frac{x}{6}$$ from both sides and subtract 3 from both sides:

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

To subtract the fractions, find a common denominator. The LCD of 4 and 6 is 12:

$$\frac{3x}{12} - \frac{2x}{12} = 2$$

$$\frac{x}{12} = 2$$

Multiply both sides by 12: $$x = 24$$

You can verify: $$\frac{24}{4} + 3 = 6 + 3 = 9$$ and $$\frac{24}{6} + 5 = 4 + 5 = 9$$ ✓

Choice A ($$x = 12$$) comes from incorrectly using 12 as the final answer instead of recognizing that $$\frac{x}{12} = 2$$ means $$x = 24$$. Choice B ($$x = 18$$) results from calculation errors when finding the common denominator or combining fractions. Choice D ($$x = 30$$) might come from using an incorrect LCD or making arithmetic mistakes during the solving process.

Strategy tip: When solving equations with fractions, always find a common denominator before combining terms, and remember to multiply by the denominator in your final step to isolate the variable. Double-check your answer by substituting back into the original equation.

When you encounter word problems involving relationships between numbers, your goal is to translate the English into a mathematical equation. Here, you need to identify what "a number decreased by 8" and "3 times the number increased by 4" look like algebraically.

Let's call the unknown number $$x$$. "A number decreased by 8" becomes $$x - 8$$, while "3 times the number increased by 4" becomes $$3(x + 4)$$. Since these expressions are equal, you can write: $$x - 8 = 3(x + 4)$$.

Now solve: $$x - 8 = 3x + 12$$. Subtract $$x$$ from both sides: $$-8 = 2x + 12$$. Subtract 12 from both sides: $$-20 = 2x$$. Divide by 2: $$x = -10$$. You can verify this by substituting back: $$-10 - 8 = -18$$ and $$3(-10 + 4) = 3(-6) = -18$$. Both sides equal $$-18$$, confirming our answer.

Choice (A) -2 results from incorrectly setting up the equation as $$x - 8 = 3x - 4$$, missing the parentheses around $$(x + 4)$$. Choice (B) -6 comes from algebraic errors, likely sign mistakes when moving terms. Choice (C) -8 might result from confusing the "decreased by 8" portion as the answer itself, rather than solving the complete equation.

Remember to carefully translate each phrase into mathematical expressions, paying special attention to parentheses when dealing with "increased by" or "decreased by" phrases that follow multiplication or division operations.

This question tests your ability to solve linear equations and then evaluate expressions using the solution. When you see an equation with variables on both sides, your goal is to isolate the variable by collecting like terms.

Start by expanding both sides: $$5(x - 2) = 3(x + 4)$$ becomes $$5x - 10 = 3x + 12$$. Now collect all $$x$$ terms on one side and constants on the other. Subtract $$3x$$ from both sides: $$2x - 10 = 12$$. Add $$10$$ to both sides: $$2x = 22$$, so $$x = 11$$.

The question asks for $$2x + 1$$, not just $$x$$. Substitute $$x = 11$$: $$2(11) + 1 = 22 + 1 = 23$$.

Looking at the wrong answers: Choice (A) gives 11, which is the value of $$x$$ itself—this catches students who stop solving too early and don't evaluate the requested expression. Choice (B) gives 21, which equals $$2x - 1$$ when $$x = 11$$; this traps students who make a sign error in the final calculation. Choice (D) gives 45, which doesn't correspond to any reasonable mistake in this problem.

The key strategy here is to always read carefully what the question is asking for. Many algebra problems on standardized tests will ask you to find an expression involving the variable rather than the variable itself. After solving for the variable, take that extra step to substitute into the requested expression. This two-step process prevents careless errors and ensures you're answering the actual question.

When you encounter a linear equation with a parameter like this, you need to determine when the equation behaves normally (one solution) versus when it becomes degenerate (no solutions or infinitely many solutions).

Let's solve $$3x + c = cx + 9$$ by collecting like terms. Subtract $$cx$$ from both sides: $$3x - cx + c = 9$$. Factor out $$x$$ on the left: $$(3-c)x + c = 9$$. Now subtract $$c$$ from both sides: $$(3-c)x = 9-c$$.

For a linear equation in the form $$ax = b$$, we get exactly one solution when $$a \neq 0$$ (specifically $$x = \frac{b}{a}$$). Here, that means we need $$3-c \neq 0$$, which gives us $$c \neq 3$$. When $$c \neq 3$$, we get the unique solution $$x = \frac{9-c}{3-c}$$.

But what happens when $$c = 3$$? Substituting back: $$(3-3)x = 9-3$$, which becomes $$0x = 6$$. This is impossible—no value of $$x$$ makes this true, so there are no solutions.

Choice A is wrong because $$c = 0$$ works fine—it gives us $$3x = 9$$, so $$x = 3$$. Choice C is backwards—when $$c = 3$$, we get no solutions, not one solution. Choice D is incorrect because $$c = 9$$ gives us $$-6x = 0$$, so $$x = 0$$, which is exactly one solution.

Remember: for parametric linear equations, the coefficient of the variable cannot be zero if you want exactly one solution. Always check what happens when that coefficient equals zero.

This is a classic linear equation problem where you have a fixed cost plus a variable cost. When you see taxi fares, phone bills, or rental fees, you're usually dealing with the format: Total Cost = Fixed Fee + (Rate × Quantity).

Let's set up the equation. The taxi charges $3 flat fee plus $0.50 per mile, and the total trip costs $8.50. If we call the number of miles $$m$$, then:

$$8.50 = 3.00 + 0.50m$$

To solve, subtract the flat fee from both sides:

$$8.50 - 3.00 = 0.50m$$

$$5.50 = 0.50m$$

Divide both sides by 0.50:

$$m = \frac{5.50}{0.50} = 11$$

So the trip was 11 miles, making A correct.

Looking at the wrong answers: B (11.5 miles) would cost $3.00 + (11.5 × $0.50) = $8.75, which is too much. This might trick you if you made a small arithmetic error. C (17 miles) would cost $3.00 + (17 × $0.50) = $11.50 — this could result from forgetting to subtract the flat fee before dividing. D (23 miles) would cost $3.00 + (23 × $0.50) = $14.50, which might come from dividing the total cost by the rate without accounting for the flat fee at all.

Remember: always isolate the variable portion first by subtracting any fixed costs, then solve for the variable. This two-step approach prevents the most common errors on these linear cost problems.

When you encounter problems about changing dimensions and comparing areas, focus on setting up equations that accurately represent the relationship between original and new measurements.

The original rectangle has area $$lw$$. After increasing both length and width by 3 units, the new dimensions become $$(l + 3)$$ and $$(w + 3)$$, giving a new area of $$(l + 3)(w + 3)$$. Since this new area is 39 square units more than the original area, you can write: new area = original area + 39, which translates to $$(l + 3)(w + 3) = lw + 39$$.

Looking at the wrong answers: Choice B incorrectly adds the new dimensions $$(l + 3) + (w + 3)$$ instead of multiplying them. This gives you a perimeter-related expression, not area. Choice C simplifies the left side of choice B to $$l + w + 6$$, but this still represents a linear measurement, not area. Choice D rearranges the correct relationship incorrectly—it states $$lw + 6 = (l + 3)(w + 3) - 39$$, which would mean the original area plus 6 equals the new area minus 39. This doesn't match the problem statement.

To verify choice A works, expand $$(l + 3)(w + 3) = lw + 3l + 3w + 9$$. The increase in area is $$3l + 3w + 9 = 3(l + w + 3)$$, which equals 39 according to the problem.

Study tip: For area problems involving dimension changes, always write "new area = original area ± change" first, then substitute the appropriate formulas for each area.

When you encounter an equation with a parameter that could result in "no solution," you're dealing with a scenario where the algebraic manipulation leads to a contradiction.

Let's solve $$ax + 6 = 4x - 2$$ by collecting like terms. First, subtract $$4x$$ from both sides: $$ax - 4x + 6 = -2$$. Factor out $$x$$: $$(a - 4)x + 6 = -2$$. Subtract 6 from both sides: $$(a - 4)x = -8$$.

Now here's the key insight: if $$a - 4 = 0$$ (meaning $$a = 4$$), our equation becomes $$0 \cdot x = -8$$, or $$0 = -8$$. This is impossible—no value of $$x$$ can make zero equal negative eight. Therefore, when $$a = 4$$, the equation has no solution.

Let's check why the other options are wrong. For choice A, when $$a = -4$$: $$(-4 - 4)x = -8$$ becomes $$-8x = -8$$, so $$x = 1$$. This gives us a valid solution. For choice B, when $$a = 0$$: $$-4x = -8$$, so $$x = 2$$. Again, a valid solution exists. For choice C, when $$a = 2$$: $$-2x = -8$$, so $$x = 4$$. This also works.

Only choice D creates the impossible situation where we need $$0 = -8$$.

Strategy tip: When an equation has "no solution," look for cases where you end up with a coefficient of zero on the variable side but a non-zero constant on the other side. This creates a mathematical impossibility.

When you encounter word problems involving relationships between quantities, your first step is to translate the words into mathematical expressions. This question describes Jamie's stickers in terms of Alex's stickers, so you need to work backwards from Jamie's known quantity.

Let's call Alex's number of stickers $$x$$. The phrase "3 more than twice the number" translates to $$2x + 3$$. Since Jamie has 17 stickers, you can write the equation: $$2x + 3 = 17$$

To solve for $$x$$, subtract 3 from both sides: $$2x = 14$$. Then divide by 2: $$x = 7$$. Therefore, Alex has 7 stickers.

Let's check why the other answers don't work. Choice B (10 stickers) would give Jamie $$2(10) + 3 = 23$$ stickers, not 17. Choice C (14 stickers) would result in $$2(14) + 3 = 31$$ stickers for Jamie. Choice D (37 stickers) is what you'd get if you mistakenly calculated $$2(17) + 3$$, thinking you needed to find how many stickers Jamie would have if Alex had 17 — this reverses the relationship described in the problem.

Choice A (7 stickers) is correct because $$2(7) + 3 = 17$$, which matches Jamie's actual count.

Study tip: In "more than" problems, always identify which person or quantity is being described in terms of the other. Set up your variable for the unknown quantity, write the equation based on the given relationship, then solve systematically. Double-check by substituting your answer back into the original relationship.