Conditions for solvability

A unique solution occurs when the lines intersect at one point. Its ratio form is (\frac{a_1}{a_2}\neq\frac{b_1}{b_2}).

When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the two linear equations represent the same line; that is, the lines are coincident. Hence, every point on the line satisfies both equations, giving infinitely many solutions. No solution occurs in the close but different case \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Exam tip: If all three ratios are equal, identify coincident lines and infinitely many solutions immediately.

The slope-intercept form of a line is y = mx + c, where m is the slope and c is the y-intercept. If two lines have the same m and the same c, both equations describe exactly the same line, not two different lines. Every point lying on that common line satisfies both equations. Consequently, the pair has infinitely many common solutions. Option C is correct. Equal slopes with different intercepts would instead give parallel lines and no solution, while different slopes would produce one intersection and therefore a unique solution. The origin is not necessarily on the line.

For distinct parallel lines, coefficient ratios are equal and the constant ratio is different. This is the condition for no solution.

For equations written as \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\), the expression \(a_1b_2-a_2b_1\) is the determinant of the coefficient matrix. If it is non-zero, the coefficients are not proportional. Consequently, the two lines have different slopes and cannot be parallel or identical; they must intersect at one point.

The given condition says this determinant is not zero. Therefore the system is solvable in exactly one way: there is one ordered pair \((x,y)\) satisfying both equations. This is called a unique solution, so choice C is correct. A zero determinant would require further comparison of the constant terms, because it could describe either infinitely many solutions or no solution. A non-zero determinant avoids both of those cases.

When all three determinants are zero, the equations may be dependent. In Class (10), link this with infinitely many solutions.

A pair of linear equations does not have a unique solution when the determinant of the coefficient matrix is zero: \((z+2)(z-1)-3\times6=0\). Thus, \(z^2+z-20=0\,\), or \((z+5)(z-4)=0\). Therefore, \(z=4\) or \(z=-5\), so option C is correct. Exam tip: to test uniqueness, first set \(a_1b_2-a_2b_1=0\).

A pair of linear equations has a unique solution when the determinant of its coefficient matrix is non-zero. Here, the determinant is \(a\cdot a-4\cdot2=a^2-8\). Therefore, for a unique solution, \(a^2-8\neq0\), i.e. \(a^2\neq8\). The condition \(a\neq2\) is not sufficient because the determinant is also zero at \(a=-2\sqrt2\). Exam tip: for a unique solution, check that \(a_1b_2-a_2b_1\neq0\).

For infinitely many solutions, the equations must be proportional, so 2/b = b/8 = 6/24. The constant ratio is 6/24 = 1/4. Thus 2/b = 1/4, which gives b = 8, but the more direct simultaneous coefficient condition is 2/b = b/8, so b² = 16 and b = ±4. Checking the listed choices, b = 4 satisfies all ratios: 2/4 = 4/8 = 6/24 = 1/4. The negative value is not among the options and would not satisfy the constant ratio. Therefore option C is correct.

Equating coefficient ratios, (\frac{c}{9}=\frac{6}{18}) gives (c=3). The constant ratio (\frac{5}{10}) is different.

For infinitely many solutions, the two linear equations must represent the same line. Hence the ratios of corresponding coefficients must be equal: 4/12 = d/15 = 7/21. Both 4/12 and 7/21 simplify to 1/3. Therefore, d/15 = 1/3, giving d = 5. Substitution verifies the result: 4/12 = 5/15 = 7/21 = 1/3. Thus option C is correct. The other options make the y-coefficient ratio different, so the equations would not be coincident and could not have infinitely many common solutions.

A pair of linear equations has a unique solution when the ratios of the coefficients are unequal. Here, \(\frac{e}{18}\ne\frac{-4}{-12}=\frac13\), which gives \(e\ne6\). If \(e=6\), all three ratios become \(\frac13\), so the pair has infinitely many solutions rather than a unique solution. Exam tip: for a unique solution, check \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\).

To make the coefficient ratio (\frac{1}{4}), (f=5) is required. The different constant ratio gives no solution.

For infinitely many solutions, (\frac{g-1}{8}=\frac{10}{20}=\frac{15}{30}) must hold. This gives (g=5).

For two linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\), a unique solution exists when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{h}{10}\ne\frac{12}{15}=\frac45\), which gives \(h\ne 8\). Thus, option B is correct. If \(h=8\), the coefficients on the left-hand sides become proportional, so the pair cannot have a unique solution. Exam tip: for a unique solution, compare only \(a_1/a_2\) and \(b_1/b_2\).

Equating coefficient ratios, (\frac{6}{18}=\frac{i}{27}) gives (i=9). The constant ratio is not equal.

For infinitely many solutions, the two linear equations must represent the same line; hence \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here, \(\frac{5}{7}=\frac{10}{14}\), so \(\frac{j}{14}=\frac{5}{7}\). Therefore, \(j=10\). For example, if \(j=8\), then \(\frac{j}{14}\neq\frac{5}{7}\), so the equations cannot have infinitely many solutions. Exam tip: for infinitely many solutions, verify that all three ratios are equal.

A pair of linear equations has no solution when the coefficient ratios are equal but the constant ratio is different: a1/a2 = b1/b2 ≠ c1/c2. Here, 3/6 = 1/2. Therefore, 2/k must also equal 1/2, which gives k = 4. However, the constant ratio is 4/11, and 4/11 is not equal to 1/2. Thus the lines have equal slopes but different intercepts; they are parallel and never intersect. Therefore, there is no solution and option C is correct. Other listed values do not make the lines parallel.

A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, \(\frac{m}{16}\neq\frac{7}{14}=\frac12\) is required. Therefore, the pair has a unique solution for \(m\neq8\). If \(m=8\), then \(\frac{m}{16}=\frac{7}{14}=\frac{13}{26}=\frac12\), so both equations represent the same line and have infinitely many solutions. Exam tip: for a unique solution, the ratios of the corresponding coefficients must be unequal.

Infinitely many solutions require the equations to be proportional. Therefore, 3/5 = n/10 = 18/30. The first and third ratios both equal 3/5. Hence n/10 = 3/5, and multiplying by 10 gives n = 6. Verification confirms that 3/5 = 6/10 = 18/30. Thus option C is correct. If n were 4, 5, or 8, the y-coefficient ratio would not equal the x-coefficient and constant ratios, so the two equations would not represent the same line and could not have infinitely many common solutions.