Maths M.C.Q (1 Marks)

The given systems of equations are
$2x + 3y = 7$
$(a + b)x + (2a - b)y = 21$
For the equations to have infinite number of solutions, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Here, $a_1=2, a_2=(a+b), b_1=3, b_2=(2 a-b), c_1=7, c_2=21$
$\frac{2}{\text{a}+\text{b}}=\frac{3}{2\text{a}-\text{b}}=\frac{7}{21}$
Let us take $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
$\frac{2}{\text{a}+\text{b}}=\frac{3}{2\text{a}-\text{b}}$
By cross multiplication we get,
$2(2a - b) = 3(a + b)$
$4a - 2b = 3a + 3b$
$4a - 3a = 3b + 2b$
$a = 5b ......(i)$
Now take $\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\frac{3}{2\text{a}-\text{b}}=\frac{7}{21}$
$\frac{3}{2\text{a}-\text{b}}=\frac{1}{3}$
By cross multiplication we get,
$3 \times 3 = 1 \times 2a - b$
$9 = 2a - b .....(ii)$
Substitute $a = 5b$ in the above equation
$9 = 2 × 5b - b$
$9 = 10b - b$
$9 = 9b$
$\frac{9}{9}=\text{b}$
$1 = b$
Substitute $b = 1$ in equation $(i)$ we get $a = 5b$
$a = 5 \times 1$
$a = 5$
Therefore $a = 5$ and $b = 1$
Hence, the correct choice is $b$.

Has a unique solution.
Given $\text{am}\neq\text{bl},$ the system of equations has
$ax + by = c$
$lx + my = n$
We know that intersecting lines have unique solution $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\text{a}_1\times\text{b}_2\neq\text{a}_2\times\text{b}_1$
Here $a_1=a, a_2=l, b_1=b, b_2=m$
$\frac{\text{a}}{\text{l}}\neq\frac{\text{b}}{\text{m}}$
$\text{a}\times\text{m}\neq\text{l}\times\text{b}$
Therefore intersecting lines, have unique solution.
Hence, the correct choice is $a$.

If a pair of linear equations in two variables is consistent, then its solution exists.
$\therefore$ The lines represented by the equations are either intersecting or coincident.
Hence, correct choice is d.

Given
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\ .....(\text{i})$
Eq. (i) cut x-axis and y-axis at a and b respectively.
Area of $\triangle=\frac{1}{2}\times\text{base}\times\text{height}$
$=\frac{1}{2}\times\text{a}\times\text{b}$
$=\frac{1}{2}\text{ab}$

The given equation are,
$kx - 5y = 2$
$6x + 2y = 7$
If $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
Here $a_1=k, a_2=6, b_1=-5, b_2=2$
$\frac{\text{k}}{6}=\frac{-5}{2}$
$2\text{k}=-30$
$\text{k}=-15$
Hence, the correct choice is $d$.