Systems Of Equations - GMAT Quantitative Reasoning

1. If $\frac{x + y}{2} = 5$ and $\frac{x - y}{3} = 2$, what is the value of $x^2 - y^2$?
   Answer: 60 From the given equations: $x + y = 10$ and $x - y = 6$. Adding these equations: $2x = 16$, so $x = 8$. Subtracting the second from the first: $2y = 4$, so $y = 2$. Now, $x^2 - y^2 = (x+y)(x-y) = (10)(6) = 60$. Alternatively, we could calculate $x^2 - y^2 = 8^2 - 2^2 = 64 - 4 = 60$. Choice B might result from incorrectly calculating $(x+y)(x-y)$ as $10 \times 7 = 70$. Choice C might come from $8 \times 10 = 80$. Choice D might come from $9 \times 10 = 90$.
2. In the system $3x + 2y = k$ and $6x + 4y = 2k + 3$, for which value of $k$ does the system have no solution?
   Answer: All values of $k$ The second equation can be rewritten as $2(3x + 2y) = 2k + 3$, or $3x + 2y = \frac{2k + 3}{2}$. For the system to be consistent, we need $k = \frac{2k + 3}{2}$. Solving: $2k = 2k + 3$, which simplifies to $0 = 3$, a contradiction. Since the coefficients of the second equation are exactly twice those of the first equation, but the constant terms are never proportional (regardless of the value of $k$), the system has no solution for any value of $k$.
3. For what value of $k$ does the system $kx + 3y = 6$ and $4x + 6y = 12$ have infinitely many solutions?
   Answer: $k = 2$ For infinitely many solutions, one equation must be a scalar multiple of the other. Notice that the second equation $4x + 6y = 12$ can be simplified by dividing by 2: $2x + 3y = 6$. For the first equation $kx + 3y = 6$ to be identical to this, we need $k = 2$. Let's verify: when $k = 2$, the system becomes $2x + 3y = 6$ and $4x + 6y = 12$. The second equation is exactly twice the first equation, so they represent the same line, giving infinitely many solutions. Choice A gives different coefficients for $x$. Choice C makes the coefficient of $x$ too large. Choice D also makes the coefficient of $x$ too large.
4. In a system of equations $ax + by = 12$ and $cx + dy = 18$, if $a = 2c$ and $b = 2d$, what must be true about the relationship between the two equations?
   Answer: The system has no solution since $12 \neq 2 \times 18$ Given $a = 2c$ and $b = 2d$, the first equation becomes $2cx + 2dy = 12$, which simplifies to $2(cx + dy) = 12$, or $cx + dy = 6$. But the second equation states $cx + dy = 18$. Since $6 ≠ 18$, we have a contradiction: the same expression ($cx + dy$) cannot equal both 6 and 18. Therefore, the system has no solution. Choice A is wrong because proportional coefficients with non-proportional constants lead to no solution, not one solution. Choice C is wrong because the equations are inconsistent. Choice D has the wrong calculation ($2 \times 9 = 18$, but we need $2 \times 18 = 36$ to match 12, which it doesn't).

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