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1. Solve the system:
- \(2x + 3y = 11\)
- \(3x - y = 11\)
Use substitution or elimination. Solve the second equation for \(y\): \(y = 3x - 11\). Substitute into the first equation: \(2x + 3(3x - 11) = 11\), which simplifies to \(2x + 9x - 33 = 11\), so \(11x = 44\), giving \(x = 4\). Then \(y = 3(4) - 11 = 1\). Solution: \((4, 1)\).

2. Solve the system:
- \(\frac{2}{x} + \frac{3}{y} = 11\)
- \(\frac{3}{x} - \frac{1}{y} = 11\)
Let \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\). The system becomes:
- \(2u + 3v = 11\)
- \(3u - v = 11\)
Solve the second equation for \(v\): \(v = 3u - 11\). Substitute into the first: \(2u + 3(3u - 11) = 11\), so \(2u + 9u - 33 = 11\), giving \(11u = 44\), so \(u = 4\). Then \(v = 3(4) - 11 = 1\). Thus, \(x = \frac{1}{4}\), \(y = 1\). Solution: \(\left(\frac{1}{4}, 1\right)\).

3. Solve the system:
- \(2x^2 + 3y^2 = 11\)
- \(3x^2 - y^2 = 11\)
Use elimination. Multiply the second equation by 3: \(9x^2 - 3y^2 = 33\). Add to the first: \(2x^2 + 3y^2 + 9x^2 - 3y^2 = 11 + 33\), so \(11x^2 = 44\), giving \(x^2 = 4\), so \(x = \pm 2\). Substitute into the second equation: \(3(4) - y^2 = 11\), so \(12 - y^2 = 11\), so \(y^2 = 1\), so \(y = \pm 1\). Check in first equation: \(2(4) + 3(1) = 8 + 3 = 11\). Valid. Solutions: \((2, 1), (2, -1), (-2, 1), (-2, -1)\).

4. For what \(k\) does \(kx + 3y = 11\) and \(3x - y = 11\) have no solution?
The system has no solution if the lines are parallel (same slope, different intercepts). Solve \(3x - y = 11\) for \(y\): \(y = 3x - 11\), slope 3. For \(kx + 3y = 11\), solve for \(y\): \(3y = -kx + 11\), so \(y = -\frac{k}{3}x + \frac{11}{3}\), slope \(-\frac{k}{3}\). Set slopes equal: \(-\frac{k}{3} = 3\), so \(k = -9\). Check if different intercepts: when \(k = -9\), the first equation is \(-9x + 3y = 11\), so \(y = 3x + \frac{11}{3}\), which has different intercept from \(y = 3x - 11\). So no solution when \(k = -9\).

5. For what \(k\) does \(2x + ky = 11\) and \(3x - y = 11\) have no solution?
Solve \(3x - y = 11\) for \(y\): \(y = 3x - 11\), slope 3. For \(2x + ky = 11\), solve for \(y\): \(ky = -2x + 11\), so \(y = -\frac{2}{k}x + \frac{11}{k}\), slope \(-\frac{2}{k}\). Set equal to 3: \(-\frac{2}{k} = 3\), so \(k = -\frac{2}{3}\). Check intercepts: when \(k = -\frac{2}{3}\), the equation is \(2x - \frac{2}{3}y = 11\), so \(y = 3x - \frac{33}{2}\), different from \(y = 3x - 11\). So no solution when \(k = -\frac{2}{3}\).

6. Given \(2a + b = 19\), \(2c + d = 37\), and \(b + d = 24\), find \(a + b + c + d\).
From \(2a + b = 19\), \(b = 19 - 2a\). From \(2c + d = 37\), \(d = 37 - 2c\). Substitute into \(b + d = 24\): \((19 - 2a) + (37 - 2c) = 24\), so \(56 - 2a - 2c = 24\), so \(2a + 2c = 32\), so \(a + c = 16\). Then \(a + b + c + d = (a + c) + (b + d) = 16 + 24 = 40\).

7. Given \(a + b = 29\) and \(ab = 204\), find \(a^2 + b^2\).
Use identity: \(a^2 + b^2 = (a + b)^2 - 2ab = 29^2 - 2(204) = 841 - 408 = 433\).

8. Solve the linear systems:
a. \(5x + 6y = 7\), \(8x + 9y = 10\)
Use elimination. Multiply first by 3: \(15x + 18y = 21\). Multiply second by 2: \(16x + 18y = 20\). Subtract: \((16x + 18y) - (15x + 18y) = 20 - 21\), so \(x = -1\). Substitute into first: \(5(-1) + 6y = 7\), so \(-5 + 6y = 7\), so \(6y = 12\), \(y = 2\). Solution: \((-1, 2)\).

b. \(x + 2y = 3\), \(4x + 5y = 6\)
Multiply first by 4: \(4x + 8y = 12\). Subtract second: \((4x + 8y) - (4x + 5y) = 12 - 6\), so \(3y = 6\), \(y = 2\). Substitute into first: \(x + 4 = 3\), so \(x = -1\). Solution: \((-1, 2)\).

9. a. For \(x^2 - y^2 = n\), find integer solutions.
Factor: \((x - y)(x + y) = n\). Let \(d_1 = x - y\), \(d_2 = x + y\), so \(d_1 d_2 = n\), and \(x = \frac{d_1 + d_2}{2}\), \(y = \frac{d_2 - d_1}{2}\). For \(x, y\) integers, \(d_1\) and \(d_2\) must have same parity (both even or both odd).

i. \(x^2 - y^2 = 48\)
Factor pairs of 48: (1,48), (2,24), (3,16), (4,12), (6,8), and negatives. Check same parity:
- (2,24): both even, \(x = 13\), \(y = 11\), \(13^2 - 11^2 = 169 - 121 = 48\). Solution: (13,11), etc.
- (4,12): both even, \(x = 8\), \(y = 4\), \(64 - 16 = 48\). Solution: (8,4), etc.
- (6,8): both even, \(x = 7\), \(y = 1\), \(49 - 1 = 48\). Solution: (7,1), etc.
Solutions exist.

ii. \(x^2 - y^2 = 23\)
23 is prime, factor pairs: (1,23), (-1,-23). Both odd, same parity. \(x = 12\), \(y = 11\), \(144 - 121 = 23\). Solution: (12,11), etc. Solutions exist.

iii. \(x^2 - y^2 = 45\)
Factor pairs: (1,45), (3,15), (5,9), and negatives. All odd, same parity.
- (1,45): \(x = 23\), \(y = 22\), \(529 - 484 = 45\).
- (3,15): \(x = 9\), \(y = 6\), \(81 - 36 = 45\).
- (5,9): \(x = 7\), \(y = 2\), \(49 - 4 = 45\).
Solutions exist.

iv. \(x^2 - y^2 = 90\)
Factor pairs: (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Check parity:
- (1,90): odd, even → different parity → no.
- (2,45): even, odd → different → no.
- (3,30): odd, even → different → no.
- (5,18): odd, even → different → no.
- (6,15): even, odd → different → no.
- (9,10): odd, even → different → no.
All pairs have different parity. So no integer solutions.

b. In general, \(x^2 - y^2 = n\) has at least one solution if and only if \(n\) is odd or \(n\) is divisible by 4. Because if \(n\) is odd, all factor pairs are odd, same parity. If \(n \equiv 2 \pmod{4}\), then all factor pairs have one even, one odd, different parity. If \(n \equiv 0 \pmod{4}\), then both factors even (since if both odd, product odd), so same parity.

10. Let the weights of the five boxes be \(a, b, c, d, e\), all integers less than 100. The sum of each pair is given: 110, 112, 113, 114, 115, 116, 117, 118, 120, 121. There are \(\binom{5}{2} = 10\) pairs, so these are all the pairwise sums.

Let \(S = a + b + c + d + e\). The sum of all pairwise sums is \( \sum_{i<j} (x_i + x_j) = 4S \), because each weight appears in 4 pairs. Sum of the given sums: \(110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121\). Calculate:
- \(110 + 121 = 231\)
- \(112 + 120 = 232\)
- \(113 + 118 = 231\)
- \(114 + 117 = 231\)
- \(115 + 116 = 231\)
Sum: \(231 + 232 + 231 + 231 + 231 = 231 \times 4 + 232 = 924 + 232 = 1156\). So \(4S = 1156\), so \(S = 289\).

The smallest pairwise sum is 110, so the two smallest weights sum to 110. The largest is 121, so the two largest sum to 121. Let the weights in increasing order be \(w_1 < w_2 < w_3 < w_4 < w_5\). Then \(w_1 + w_2 = 110\), \(w_4 + w_5 = 121\), and \(w_1 + w_2 + w_3 + w_4 + w_5 = 289\), so \(w_3 = 289 - 110 - 121 = 58\).

Now, the next smallest sum is 112. This could be \(w_1 + w_3 = 112\), so \(w_1 + 58 = 112\), so \(w_1 = 54\). Then \(w_2 = 110 - 54 = 56\).

Check: \(w_1 = 54\), \(w_2 = 56\), \(w_3 = 58\), \(w_4\), \(w_5\), with \(w_4 + w_5 = 121\).

Now, \(w_1 + w_4 = 54 + w_4\), \(w_1 + w_5 = 54 + w_5\), \(w_2 + w_4 = 56 + w_4\), \(w_2 + w_5 = 56 + w_5\), \(w_3 + w_4 = 58 + w_4\), \(w_3 + w_5 = 58 + w_5\), and \(w_4 + w_5 = 121\).

The pairwise sums are:
- \(w_1 + w_2 = 110\)
- \(w_1 + w_3 = 112\)
- \(w_1 + w_4 = 54 + w_4\)
- \(w_1 + w_5 = 54 + w_5\)
- \(w_2 + w_3 = 114\)
- \(w_2 + w_4 = 56 + w_4\)
- \(w_2 + w_5 = 56 + w_5\)
- \(w_3 + w_4 = 58 + w_4\)
- \(w_3 + w_5 = 58 + w_5\)
- \(w_4 + w_5 = 121\)

We know the sums are: 110, 112, 113, 114, 115, 116, 117, 118, 120, 121.

We have 110, 112, 114. So missing: 113, 115, 116, 117, 118, 120, 121.

\(w_2 + w_3 = 56 + 58 = 114\), already have.

\(w_1 + w_4 = 54 + w_4\). Since \(w_4 > w_3 = 58\), so \(w_4 \geq 59\), so \(w_1 + w_4 \geq 113\). Similarly, \(w_1 + w_5 \geq 54 + 60 = 114\) (since \(w_5 > w_4 \geq 59\), so \(w_5 \geq 60\)).

The smallest unknown sum is 113. So \(w_1 + w_4 = 113\) → \(54 + w_4 = 113\) → \(w_4 = 59\). Then \(w_5 = 121 - 59 = 62\).

Now check all sums:
- \(w_1 + w_2 = 54 + 56 = 110\)
- \(w_1 + w_3 = 54 + 58 = 112\)
- \(w_1 + w_4 = 54 + 59 = 113\)
- \(w_1 + w_5 = 54 + 62 = 116\)
- \(w_2 + w_3 = 56 + 58 = 114\)
- \(w_2 + w_4 = 56 + 59 = 115\)
- \(w_2 + w_5 = 56 + 62 = 118\)
- \(w_3 + w_4 = 58 + 59 = 117\)
- \(w_3 + w_5 = 58 + 62 = 120\)
- \(w_4 + w_5 = 59 + 62 = 121\)

The sums: 110, 112, 113, 114, 115, 116, 117, 118, 120, 121. All match.

So the weights are 54, 56, 58, 59, 62 kg.