Question

Choose correct alternative for each of the following questions
(1) To draw graph of 4x +5y = 19, Find y when x = 1.

A) 4

(B) 3

(C) 2

(D) –3

(2) For simultaneous equations in variables x and y, D_x = 49, D_y = –63, D = 7 then what is x ?

A) 7

(B) –7

(C) $\frac{1}{7}$

(D) $\frac{-1}{7}$

(3) Find the value of $\left|\begin{array}{cc}5& 3\\ -7& -4\end{array}\right|$

A) –1

(B) –41

(C) 41

(D) 1

(4) To solve x + y = 3 ; 3x – 2y – 4 = 0 by determinant method find D.

A) 5

(B) 1

(C) –5

(D) –1

(5) ax + by = c and mx + ny = d and an ≠ bm then these simultaneous equations have -

(A)	Only one common solution.	(A)	No solution.
(C)	Infinite number of solutions.	(D)	Only two solution.

Answer 1

(1) 4x +5y = 19
When x = 1, then y will be
$$\begin{gathered} 4\left(1\right)+5y=19 \\ \Rightarrow 4+5y=19 \\ \Rightarrow 5y=19-4=15 \\ \Rightarrow 5y=15 \\ \Rightarrow y=\frac{15}{5}=3 \end{gathered}$$
Hence, the correct answer is option (B).

(2) $x=\frac{D_x}{D}=\frac{49}{7}=7$
Hence, the correct answer is option (A).

(3) $\left|\begin{array}{cc}5& 3\\ -7& -4\end{array}\right|=5\times \left(-4\right)-3\times \left(-7\right)=-20+21=1$
Hence, the correct answer is option (D).

(4) x + y = 3 ; 3x – 2y – 4 = 0
$D=\left|\begin{array}{cc}1& 1\\ 3& -2\end{array}\right|=1\times \left(-2\right)-1\times 3=-2-3=-5$
Hence, the correct answer is option (C).

(5) ax + by = c and mx + ny = d
$D=\left|\begin{array}{cc}a& b\\ m& n\end{array}\right|=an-bm$
an ≠ bm
So, D ≠ 0.
So, the given equations have a unique solution or only one common solution.
Hence, the correct answer is option A.