Question 4

We classify linear equation pairs as consistent or inconsistent. If consistent, we find the solution by graphing.

Step 1 — Check consistency for (i) Let's write the equations in standard form. The first equation is $x + y - 5 = 0$. The second equation is $2x + 2y - 10 = 0$. We compare coefficients $a_1, b_1, c_1$ with $a_2, b_2, c_2$. For the first equation, $a_1 = \mathbf{1}$, $b_1 = \mathbf{1}$, $c_1 = \mathbf{-5}$. For the second equation, $a_2 = \mathbf{2}$, $b_2 = \mathbf{2}$, $c_2 = \mathbf{-10}$. Let's find the ratios of the coefficients.

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-5}{-10}$

$= \frac{1}{2}$ We see that all ratios are equal. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

$\boxed{\text{Consistent (infinitely many solutions)}}$

Step 2 — Graph for (i) The equations represent coincident lines. We can use the first equation $x + y = 5$. Let's find two points for this line. If $x = \mathbf{0}$, then $y = \mathbf{5}$. So, point A is (0, 5). If $y = \mathbf{0}$, then $x = \mathbf{5}$. So, point B is (5, 0). Let's plot these points and draw the line. The second equation $2x + 2y = 10$ is the same line. Divide by $\mathbf{2}$ to get $x + y = 5$. All points on this line are solutions.

Step 3 — Check consistency for (ii) Let's write the equations in standard form. The first equation is $x - y - 8 = 0$. The second equation is $3x - 3y - 16 = 0$. For the first equation, $a_1 = \mathbf{1}$, $b_1 = \mathbf{-1}$, $c_1 = \mathbf{-8}$. For the second equation, $a_2 = \mathbf{3}$, $b_2 = \mathbf{-3}$, $c_2 = \mathbf{-16}$. Let's find the ratios of the coefficients.

$\frac{a_1}{a_2} = \frac{1}{3}$

$\frac{b_1}{b_2} = \frac{-1}{-3}$

$= \frac{1}{3}$

$\frac{c_1}{c_2} = \frac{-8}{-16}$

$= \frac{1}{2}$ We see that the first two ratios are equal, but not the third. $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$\boxed{\text{Inconsistent (no solution)}}$

Step 4 — Check consistency for (iii) Let's use the given equations. The first equation is $2x + y - 6 = 0$. The second equation is $4x - 2y - 4 = 0$. For the first equation, $a_1 = \mathbf{2}$, $b_1 = \mathbf{1}$, $c_1 = \mathbf{-6}$. For the second equation, $a_2 = \mathbf{4}$, $b_2 = \mathbf{-2}$, $c_2 = \mathbf{-4}$. Let's find the ratios of the coefficients.

$\frac{a_1}{a_2} = \frac{2}{4}$

$= \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{1}{-2}$

$= -\frac{1}{2}$ We see that the first two ratios are not equal. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

$\boxed{\text{Consistent (unique solution)}}$

Step 5 — Graph for (iii) We need to find points for each line. For the first equation $2x + y - 6 = 0 \implies y = 6 - 2x$. If $x = \mathbf{0}$, $y = \mathbf{6}$. Point A: (0, 6). If $x = \mathbf{3}$, $y = \mathbf{0}$. Point B: (3, 0).

For the second equation $4x - 2y - 4 = 0 \implies y = 2x - 2$. If $x = \mathbf{0}$, $y = \mathbf{-2}$. Point C: (0, -2). If $x = \mathbf{1}$, $y = \mathbf{0}$. Point D: (1, 0). Let's plot these points and draw the lines. The lines intersect at point (2, 2). This is our unique solution.

Step 6 — Check consistency for (iv) Let's use the given equations. The first equation is $2x - 2y - 2 = 0$. The second equation is $4x - 4y - 5 = 0$. For the first equation, $a_1 = \mathbf{2}$, $b_1 = \mathbf{-2}$, $c_1 = \mathbf{-2}$. For the second equation, $a_2 = \mathbf{4}$, $b_2 = \mathbf{-4}$, $c_2 = \mathbf{-5}$. Let's find the ratios of the coefficients.

$\frac{a_1}{a_2} = \frac{2}{4}$

$= \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-2}{-4}$

$= \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-2}{-5}$

$= \frac{2}{5}$ We see that the first two ratios are equal, but not the third. $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$\boxed{\text{Inconsistent (no solution)}}$

Answer

(i) Consistent, infinitely many solutions. Graphically, the lines are coincident. (ii) Inconsistent, no solution. (iii) Consistent, unique solution. Graphically, the solution is (2, 2). (iv) Inconsistent, no solution.