Conditions for solvability

Compare the coefficients of the two equations. The second equation has coefficients 8 and 18, which are twice 4 and 9, the corresponding coefficients in the first equation. However, twice the first right-hand side is \\(2\\times2=4\\), not 7. Thus, the left sides are proportional but the constants are not.

The equations therefore have the same slope but different positions. Their graphs are distinct parallel lines, so they never meet and no ordered pair satisfies both equations simultaneously. This is an inconsistent pair, giving option B. A unique solution would require different slopes, while infinitely many solutions would require the constant ratio to match as well.

Here, \(\frac{a_1}{a_2}=\frac{13}{26}=\frac{1}{2}\), \(\frac{b_1}{b_2}=\frac{4}{8}=\frac{1}{2}\), and \(\frac{c_1}{c_2}=\frac{6}{12}=\frac{1}{2}\). Therefore, both equations represent the same line, so the pair has infinitely many solutions. Option B would be correct only if the first two ratios were equal but the third ratio were different. Exam tip: when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the pair has infinitely many solutions.

Equating coefficient ratios gives (v-2=3). Thus (v=5), and the different constant ratio gives no solution.

A pair of linear equations has a unique solution when the determinant of its coefficient matrix is non-zero. Here, the determinant is \(w(-24)-16(-6)=96-24w=24(4-w)\). It is non-zero when \(w\ne4\). At \(w=4\), the coefficients of \(x\) and \(y\) are in the same ratio, but the constants are not, so the lines are parallel and there is no solution. Exam tip: for a unique solution, check that \(a_1b_2-a_2b_1\ne0\).

The coefficient ratio is (\frac{1}{6}). For infinitely many solutions, (\frac{\lambda}{48}=\frac{1}{6}), so (\lambda=8).

(\frac{14}{21}=\frac{2}{3}). Equating coefficient ratios, (\frac{\mu}{6}=\frac{2}{3}) gives (\mu=4).

For infinitely many solutions, (\frac{9}{27}=\frac{\beta+4}{18}=\frac{15}{45}) must hold. This gives (\beta=2).

Equating coefficient ratios, (\frac{5}{25}=\frac{\gamma}{15}) gives (\gamma=3). The constant ratio is different.

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.

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 \(a_1b_2-a_2b_1\neq0\).

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

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.

The direct answer is option B: \(g=5\). Infinitely many solutions occur when the two equations are the same line, so the ratios of corresponding coefficients and constants must be equal: \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). In this problem, \(\frac{10}{20}=\frac{15}{30}=\frac12\). Therefore \(\frac{g-1}{8}\) must also equal \(\frac12\). Solve step by step: \(\frac{g-1}{8}=\frac12\), so \(g-1=4\), and hence \(g=5\). Option B is correct. Option A, \(g=4\), gives \((g-1)/8=3/8\), not \(1/2\); the equations would not be identical. Option B gives \(4/8=1/2\), matching both the y-coefficient ratio and the constant ratio. Option C, \(g=6\), gives \(5/8\), so it fails the condition. Option D, \(g=7\), gives \(6/8=3/4\), also unequal to \(1/2\). It is important not to compare only one pair of coefficients: all three ratios must agree for infinitely many solutions. Geometrically, identical lines have every point of the line in common, so there are infinitely many solutions. Memory cue: first simplify the known ratios; here both are one-half, then make the unknown ratio one-half too.

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 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.

Two linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) have a unique solution when \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, \(\frac{r}{9}\neq\frac{2}{3}\) is required. Since \(\frac{2}{3}=\frac{6}{9}\), we get \(r\neq6\). If \(r=6\), the coefficients on the left-hand sides become proportional, so the solution cannot be unique. Exam tip: compare the ratios of the coefficients of \(x\) and \(y\) first.

Equating coefficient ratios, (\frac{s}{21}=\frac{4}{12}) gives (s=7). The constant ratio is not equal.