Question 3

We compare the ratios of coefficients to determine consistency.

Step 1 — Analyze Part (i)

Let's write the equations in standard form. The first equation is $3x + 2y - 5 = 0$. The second equation is $2x - 3y - 7 = 0$. We identify the coefficients. $a_1 = \mathbf{3}$, $b_1 = \mathbf{2}$, $c_1 = \mathbf{-5}$. $a_2 = \mathbf{2}$, $b_2 = \mathbf{-3}$, $c_2 = \mathbf{-7}$.

Now, we calculate the ratios. $\frac{a_1}{a_2} = \frac{3}{2}$

$\frac{b_1}{b_2} = \frac{2}{-3}$

We compare these two ratios. $\frac{3}{2} \neq \frac{2}{-3}$ This means $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. The lines intersect at one point.

$\boxed{\text{Consistent}}$

Step 2 — Analyze Part (ii)

Let's write the equations in standard form. The first equation is $2x - 3y - 8 = 0$. The second equation is $4x - 6y - 9 = 0$. We identify the coefficients. $a_1 = \mathbf{2}$, $b_1 = \mathbf{-3}$, $c_1 = \mathbf{-8}$. $a_2 = \mathbf{4}$, $b_2 = \mathbf{-6}$, $c_2 = \mathbf{-9}$.

Now, we calculate the ratios. $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$ The first two ratios are equal. So, we calculate the third ratio. $\frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$ Now, we compare all three ratios. $\frac{1}{2} = \frac{1}{2} \neq \frac{8}{9}$ This means $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. The lines are parallel and do not intersect.

$\boxed{\text{Inconsistent}}$

Step 3 — Analyze Part (iii)

Let's write the equations in standard form. The first equation is $\frac{3}{2}x + \frac{5}{3}y - 7 = 0$. The second equation is $9x - 10y - 14 = 0$. We identify the coefficients. $a_1 = \mathbf{\frac{3}{2}}$, $b_1 = \mathbf{\frac{5}{3}}$, $c_1 = \mathbf{-7}$. $a_2 = \mathbf{9}$, $b_2 = \mathbf{-10}$, $c_2 = \mathbf{-14}$.

Now, we calculate the ratios. $\frac{a_1}{a_2} = \frac{3/2}{9} = \frac{3}{18} = \frac{1}{6}$

$\frac{b_1}{b_2} = \frac{5/3}{-10} = \frac{5}{-30} = -\frac{1}{6}$

We compare these two ratios. $\frac{1}{6} \neq -\frac{1}{6}$ This means $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. The lines intersect at one point.

$\boxed{\text{Consistent}}$

Step 4 — Analyze Part (iv)

Let's write the equations in standard form. The first equation is $5x - 3y - 11 = 0$. The second equation is $-10x + 6y + 22 = 0$. We identify the coefficients. $a_1 = \mathbf{5}$, $b_1 = \mathbf{-3}$, $c_1 = \mathbf{-11}$. $a_2 = \mathbf{-10}$, $b_2 = \mathbf{6}$, $c_2 = \mathbf{22}$.

Now, we calculate the ratios. $\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$ The first two ratios are equal. So, we calculate the third ratio. $\frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$ Now, we compare all three ratios. $-\frac{1}{2} = -\frac{1}{2} = -\frac{1}{2}$ This means $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. The lines are coincident and have many solutions.

$\boxed{\text{Consistent}}$

Step 5 — Analyze Part (v)

Let's write the equations in standard form. The first equation is $\frac{4}{3}x + 2y - 8 = 0$. The second equation is $2x + 3y - 12 = 0$. We identify the coefficients. $a_1 = \mathbf{\frac{4}{3}}$, $b_1 = \mathbf{2}$, $c_1 = \mathbf{-8}$. $a_2 = \mathbf{2}$, $b_2 = \mathbf{3}$, $c_2 = \mathbf{-12}$.

Now, we calculate the ratios. $\frac{a_1}{a_2} = \frac{4/3}{2} = \frac{4}{6} = \frac{2}{3}$

$\frac{b_1}{b_2} = \frac{2}{3}$ The first two ratios are equal. So, we calculate the third ratio. $\frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}$ Now, we compare all three ratios. $\frac{2}{3} = \frac{2}{3} = \frac{2}{3}$ This means $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. The lines are coincident and have many solutions.

$\boxed{\text{Consistent}}$

Answer

(i) Consistent (ii) Inconsistent (iii) Consistent (iv) Consistent (v) Consistent