Answer: Option C. -> passes through the origin
:
C
If x = 0,
0 + 2y = 0
$\Rightarrow$ y = 0
$\therefore$ The line x + 2y = 0 passes through the origin.
We know that for x + 2y = 0 ,
y = 0 when x = 0.
Therefore, (0,3) does not lie on the line.

Answer: Option A. -> 0x + 2y - 3 = 0
:
A
The standard form is given by ax + by + c = 0
So,2y = 3 can be written in the form0x + 2y - 3 = 0.

Answer: Option A. -> True
:
A
Let us consider the equation x + y = 3
when x = 0 then y = 3
similarly, when y = 0 then x = 3
Let us consider the equation x + y = 5
when x = 0 then y = 5
similarly, when y = 0 then x = 5
By plotting the graphs ofboth of the equations, we can easily observethat the line x + y = 3 lies closer to the origin.

Answer: Option B. -> False
:
B
At the y-axis, value of x= 0
Putting x = 0 in the given equation,
3 x 0 - 2y = 6
$\Rightarrow$ y= -3.
Hence, the line 3x - 2y = 6 cuts y-axis at the point(0,-3).

Answer: Option C. -> 34
:
C
Let the number be $x$.
'Subract¼ from the number'is mathematically written as 'x - ¼'.
Multiplying the above obtained result by½ is mathematically shown as:
'(x - ¼) × ½'.
According to the question,
$(x-\frac{1}{4})\times \frac{1}{2}=\frac{1}{4}$
$\Rightarrow x-\frac{1}{4}=2\times \frac{1}{4}$
(by multiplying both the sides by 2)
$\Rightarrow x-\frac{1}{4}=\frac{1}{2}$
$\Rightarrow x=\frac{1}{2}+\frac{1}{4}$
(by transposing¼ from LHS to the RHS)
$\Rightarrow x=\frac{3}{4}$
So, the required number is $\frac{3}{4}$.

Answer: Option A. -> True
:
A
Linear dependency signifies that "a","b"and "c" are linearly related, which means that if one of them can be found, the other two can be found.
Hence, the given equation can be reduced to a linear equation if the exact relation between "a","b"and "c" is given.
For a better understanding, let us express the dependencies as equations.
Since b is linearly dependent on a, we can write b = ka + p, where k and p are some constants. Note that in this equation, b and a are the only variables, whereas k and p are known.
Similarly, a can be written as a = gc+ u, where g and u are constants (known values).
Substituting these values of a and b in the given equation, we get,
3a + 2b + 4c = 5
3(gc + u) + 2(ka + p) + 4c = 5
3gc + 3u + 2p + 2k(gc + u) + 4c = 5
(3g + 2kg + 4)c + (3u + 2p + 2ku - 5) = 0.
Note that in the final equation, the only c is unknown, the others are all constants and hence is a linear equation in one variable.

Answer: Option A. -> True
:
A
$\frac{x+9}{4}+\frac{2x+3}{5}=10$ is the given equation.
We can check if $x=11$ is a solution by substituting this value in the LHS to check whether itis equal to RHS or not
$\frac{(11+9)}{4}+\frac{(22+3)}{5}$
= 5 + 5
= 10
Thus we see that$x=11$ makes LHS = RHS andis the solution of the given equation.

Answer: Option A. -> True
:
A
$\frac{5}{3x}$ =$\frac{25}{27}$
By cross multiplication, we get
$5\times 27=3x\times 25$
$\Rightarrow 3x=\frac{5\times 27}{25}$
$\Rightarrow x=\frac{5\times 27}{3\times 25}$
$\Rightarrow x=\frac{9}{5}$
$=1.8$, which lies between 1 and 2.

:
$\frac{x+5}{5x+8}=\frac{11}{38}$
38x + 190 = 55x + 88
55x -38x = 190 - 88
17x = 102
$x=\frac{102}{17}=6$

Answer: Option A. -> 1
:
A
$\frac{3}{x+1}=\frac{6}{5x-1}$
On cross multiplication, we get
$\Rightarrow 3(5x-1)=6(x+1)\Rightarrow 15x-3=6x+6\Rightarrow 9x=9\Rightarrow x=1$