Substituting $$y = 2x - 3$$ into the second equation: $$4x - 2(2x - 3) = 6$$, which simplifies to $$4x - 4x + 6 = 6$$, giving $$6 = 6$$. This is always true, meaning the equations represent the same line and have infinitely many solutions. Choice A assumes the system has a unique solution without checking. Choice B would occur if we got a contradiction like $$6 = 5$$. Choice D is impossible for a system of two linear equations.

When you see a question about transforming systems of equations, think carefully about what operations actually change the solution versus what operations preserve it. The key insight is that multiplying every term in an equation by the same constant doesn't change which $$(x, y)$$ values satisfy the equation.

Let's say your original system was something like $$2x + y = 9$$ and $$x - y = -3$$, which has solution $$(2, 5)$$. When you multiply both equations by 3, you get $$6x + 3y = 27$$ and $$3x - 3y = -9$$. Notice that if you substitute $$x = 2$$ and $$y = 5$$ into these new equations, they still work: $$6(2) + 3(5) = 12 + 15 = 27$$ ✓ and $$3(2) - 3(5) = 6 - 15 = -9$$ ✓. The solution remains $$(2, 5)$$, so C is correct.

A) $$(6, 5)$$ represents the misconception that multiplying equations by 3 means you multiply the x-coordinate by 3. This confuses equation transformations with coordinate transformations.

B) $$(6, 15)$$ suggests multiplying both coordinates by 3, as if this were a geometric transformation like dilation rather than an algebraic operation on equations.

D) $$(2, 15)$$ implies only the y-coordinate gets multiplied by 3, which has no logical basis in how equation operations work.

Remember: multiplying an entire equation by a constant is like saying the same thing in different words—it doesn't change the underlying relationship between variables. Only operations that fundamentally alter the relationships (like changing coefficients unevenly) affect solutions.

When checking solutions to systems of equations, you substitute your found value back into one of the original equations to verify it's correct and to find the corresponding value of the other variable.

Since Tom found $$x = 4$$ and wants to check using the second equation $$3x + y = 11$$, you substitute $$x = 4$$:

$$3(4) + y = 11$$

$$12 + y = 11$$

$$y = 11 - 12$$

$$y = -1$$

So if Tom's $$x$$-value is correct, he should calculate $$y = -1$$.

Let's examine why the other options are incorrect. Choice A gives $$y = 1$$, which would result from the calculation error $$y = 12 - 11$$ instead of $$y = 11 - 12$$. Choice C, $$y = 7$$, might come from incorrectly adding: $$y = 11 - 12 = 7$$, which shows a sign error. Choice D, $$y = -7$$, could result from the mistake $$y = 12 - 11 = -7$$, combining both an order error and an incorrect final calculation.

You can verify this is correct by checking that $$(4, -1)$$ satisfies both original equations: $$5(4) - 2(-1) = 20 + 2 = 22 \neq 16$$. Wait—this suggests Tom's $$x = 4$$ might actually be wrong! But the question asks what $$y$$ he should calculate if his $$x$$ is correct.

Study tip: When checking solutions to systems, always substitute carefully and double-check your arithmetic. Sign errors are extremely common in these problems.

Substituting $$x = 3$$ into $$2x + 3y = 12$$: $$2(3) + 3y = 12$$, so $$6 + 3y = 12$$, which gives $$3y = 6$$, so $$y = 2$$. Choice B results from solving $$3y = 6$$ incorrectly as $$y = 6$$. Choice C comes from making a sign error when solving $$6 + 3y = 12$$. Choice D results from incorrectly calculating $$12 - 6 = 4$$ and forgetting to divide by 3.

When you encounter a system of linear equations, you need to determine whether the lines intersect, are parallel, or are the same line. The key is comparing their slopes and y-intercepts after converting both equations to slope-intercept form.

The first equation is already in slope-intercept form: $$y = 3x - 1$$, so the slope is 3 and the y-intercept is -1.

For the second equation $$6x - 2y = 5$$, solve for y:

$$-2y = -6x + 5$$

$$y = 3x - \frac{5}{2}$$

Now you can see that both lines have the same slope (3) but different y-intercepts: -1 and $$-\frac{5}{2}$$. When two lines have identical slopes but different y-intercepts, they are parallel lines that never intersect.

Choice A is incorrect because the lines aren't the same—they have different y-intercepts, so infinitely many solutions is impossible. Choice C is wrong because parallel lines never intersect, so there's no single solution point. Choice D identifies the correct relationship (same slope) but gives an incomplete explanation—having the same slope alone doesn't guarantee no solution; you also need different y-intercepts to confirm the lines are parallel rather than identical.

Choice B correctly identifies that the lines are parallel with different y-intercepts, which means the system has no solution.

Study tip: Always convert both equations to slope-intercept form when analyzing systems. Same slope + same y-intercept = infinitely many solutions; same slope + different y-intercepts = no solution; different slopes = exactly one solution.

Adding the equations: $$(x + 2y) + (3x - 2y) = 8 + 4$$. This simplifies to $$x + 3x + 2y - 2y = 12$$, which gives $$4x = 12$$. Choice B incorrectly keeps the $$y$$ terms instead of recognizing they cancel. Choice C results from adding only the $$x$$ coefficients without the constants. Choice D comes from subtracting instead of adding the equations.

When you have a system of equations and know that a specific point is the solution, you can substitute those coordinates into either equation to find unknown coefficients.

Since $$(3, 1)$$ is the solution to this system, it must satisfy both equations. Let's use the first equation $$ax + 3y = 12$$ to find $$a$$. Substituting $$x = 3$$ and $$y = 1$$:

$$a(3) + 3(1) = 12$$

$$3a + 3 = 12$$

$$3a = 9$$

$$a = 3$$

You can verify this works by checking that $$(3, 1)$$ also satisfies the second equation: $$2(3) - 1 = 6 - 1 = 5$$... wait, that's not right! Let me recalculate: $$2(3) - 1 = 6 - 1 = 5 \neq 4$$. Actually, let me check the point again in the second equation: $$2(3) - (1) = 6 - 1 = 5$$, but we need 4. This suggests I should double-check by substituting back into both equations with $$a = 3$$: First equation becomes $$3x + 3y = 12$$, so $$3(3) + 3(1) = 9 + 3 = 12$$ ✓. For the second equation: $$2(3) - 1 = 5$$, not 4, which indicates there might be an error in the problem setup. However, assuming the given information is correct, $$a = 3$$.

Looking at the wrong answers: A) $$a = 1$$ would give $$1(3) + 3(1) = 6 \neq 12$$. B) $$a = 4$$ would give $$4(3) + 3(1) = 15 \neq 12$$. C) $$a = 2$$ would give $$2(3) + 3(1) = 9 \neq 12$$.

Strategy tip: When finding unknown coefficients, always substitute the given solution into the equation containing that unknown, then solve algebraically.

When solving systems of equations by substitution, you're replacing one variable with an equivalent expression from another equation. The key is understanding what "before simplifying" means - you want the equation immediately after substitution, not after combining like terms.

Sarah correctly rewrote $$x - y = 2$$ as $$x = y + 2$$. Now she substitutes this expression for $$x$$ in the first equation $$3x + 4y = 24$$. Since $$x = y + 2$$, she replaces the $$x$$ with $$(y + 2)$$, giving her $$3(y + 2) + 4y = 24$$. This is exactly what choice D shows.

Choice A, $$3(x + 2) + 4y = 24$$, incorrectly substitutes $$x + 2$$ instead of $$y + 2$$ for the variable $$x$$. This shows confusion about which variable to substitute.

Choice B, $$3y + 6 + 4y = 24$$, shows the equation after simplifying by distributing the 3. The question specifically asks for the equation before simplifying.

Choice C, $$3x + 4(y + 2) = 24$$, makes the mistake of substituting the expression for $$y$$ instead of $$x$$. Sarah solved for $$x$$ in terms of $$y$$, so she should substitute for $$x$$, not $$y$$.

Remember: in substitution, carefully track which variable you solved for and substitute that exact expression. The phrase "before simplifying" means show the equation with parentheses intact, before distributing or combining like terms.

This system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). The second equation directly gives x = 6, making this a simple substitution problem: substitute x = 6 into the first equation to get $6 + y = 10$, which gives $y = 4$. The solution is $(6, 4)$, verified by checking: $6 + 4 = 10$ ✓ and $x = 6$ ✓. Common mistakes include reversing the coordinates to get (4, 6) or misreading which variable is given. When one equation provides a variable's value directly, always use substitution for efficiency.

This system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). The second equation can be rewritten as $ y = 2x + 3 $ by dividing both sides by 2, which is identical to the first equation. Since both equations represent the same line, every point on the line $ y = 2x + 3 $ is a solution, giving infinitely many solutions. This can be verified algebraically: the second equation $ 2y = 4x + 6 $ is exactly twice the first equation $ y = 2x + 3 $. Students might confuse this with no solution (parallel lines) or think there's only one solution. When equations are multiples of each other, the system has infinitely many solutions.