Solutions: Linear Equations - Test 3

Q1. (a) no solution

$x=0$ is the y-axis, $x=5$ is a line parallel to y-axis. They never meet.

Q2. (b) 15/4

Parallel: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. $\frac{3}{2} = \frac{2k}{5} \Rightarrow 4k = 15 \Rightarrow k = 15/4$.

Q3. (c) 3

Infinite solutions: $\frac{c}{6} = \frac{-1}{-2} = \frac{2}{4}$. $\frac{c}{6} = \frac{1}{2} \Rightarrow c = 3$.

Q4. (a) Both A and R are true and R is the correct explanation of A.

$\frac{2}{4} = \frac{3}{6} = \frac{9}{18} = \frac{1}{2}$. Ratios are equal, so lines are coincident.

Q5. $\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{-2}{4} = -\frac{1}{2}$.
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of equations has a unique solution.
Consistent [2]

Q6. For unique solution: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
$\frac{1}{5} \neq \frac{2}{k} \Rightarrow k \neq 10$.
All real values except 10 [2]

Q7. For no solution: $\frac{3}{2k-1} = \frac{1}{k-1} \neq \frac{1}{2k+1}$.
$3(k-1) = 2k-1 \Rightarrow 3k-3 = 2k-1 \Rightarrow k = 2$.
Check condition: $\frac{1}{2-1} = 1$, $\frac{1}{2(2)+1} = \frac{1}{5}$. $1 \neq \frac{1}{5}$ (True).
k = 2 [2]

Q8. For infinite solutions: $\frac{2}{a-b} = \frac{3}{a+b} = \frac{7}{3a+b-2}$.
From first two: $2(a+b) = 3(a-b) \Rightarrow 2a+2b = 3a-3b \Rightarrow a = 5b$ ...(i)
From last two: $3(3a+b-2) = 7(a+b) \Rightarrow 9a+3b-6 = 7a+7b \Rightarrow 2a-4b=6 \Rightarrow a-2b=3$.
Substitute (i): $5b-2b=3 \Rightarrow 3b=3 \Rightarrow b=1$.
$a = 5(1) = 5$. a=5, b=1 [3]

Q9. $2x - 3y = 8$ and $4x - 6y = 9$.
$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$. $\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$. $\frac{c_1}{c_2} = \frac{8}{9}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.
Parallel lines, they do not cross. [3]

Q10. $2x + y = 6 \Rightarrow$ Points $(0,6), (3,0)$.
$2x - y = -2 \Rightarrow$ Points $(0,2), (-1,0)$. [2]
Plotting graph: Lines intersect at $(1, 4)$. [1]
Vertices of triangle with x-axis: $(-1, 0), (3, 0)$ and $(1, 4)$.
Shaded region is the triangle formed by these points. [2]

(i) $x + 2y = 4$ and $2x + 4y = 12$.
$\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$. [1]

(ii) Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.
The tracks will never meet. [2]

(iii) If eq is $2x + 4y - 8 = 0$, then $\frac{c_1}{c_2} = \frac{-4}{-8} = \frac{1}{2}$.
Ratios become equal. Lines are coincident (same track). [1]