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MATHS 2 Marks Questions

Linear Equations in Two Variables · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 11 questions.

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2 Marks Questions

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11 questions · timed · auto-graded

Question 12 Marks
Find whether $(\sqrt{2},4\sqrt{2})$ is the solution of the equation x – 2y = 4 or not?
Answer
x-2y=4

Put x = $ \sqrt{2}$ , y = $4 \sqrt{2}$ in given equation, we get
 $\sqrt{2}-2(4\sqrt{2})=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}$

which is not 4.
∴ $(\sqrt2,4\sqrt2)$ is not a solution of given equation.

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Question 22 Marks
Find whether (0, 2)is the solution of the equation x – 2y = 4 or not?
Answer
The given equation is x – 2y = 4
Put x = 0 and y = 2 in given equation, we get
x – 2y =4

0 – 2(2) = –4, which is not 4.
∴ (0, 2) is not a solution of given equation.

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Question 32 Marks
Write four solutions of the equation: x = 4y.
Answer
x = 4y
$\Rightarrow$ y = $\frac{x}{4}$
put x = 0, we get y = $\frac{0}{4}$ = 0
Put x = 4, we get y = $\frac{4}{4}$ = 1
Put x = –4, we get y = $\frac{-4}{4}$ = -1
Put x = 2, we get y = $\frac{2}{4}$ = $\frac{1}{2}$
$\therefore$ Four solutions are (0, 0), (4, 1), (–4, –1) and (2, $\frac{1}{2}$).
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Question 42 Marks
Write four solutions of the equation: $\pi$x + y = 9
Answer
$\pi$x + y = 9
⇒ y = 9 – $\pi$x
Put x = 0, we get y = 9 – $\pi$(0) = 9 – 0 = 9
put x = 1, we get y = 9 – $\pi$(1) = 9 – $\pi$
Put x = –1, we get y = 9 – $\pi$( –1 ) = 9 + $\pi$
Put x = $\frac{9}{\pi }$, we get $y = 9 - \pi \left( {\frac{9}{\pi }} \right) = 9 - 9 = 0$
∴ Four solutions are (0, 9), (1, 9 – $\pi$), (–1, 9 + $\pi$) and $\left( {\frac{9}{\pi },0} \right)$.
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Question 52 Marks
Write four solutions of the equation: 2x + y = 7
Answer
2x + y = 7
⇒ y = 7 – 2x
Put x = 0, we get y = 7 – 2(0) = 7 – 0 = 7
Put x = 1, we get y = 7 – 2(1) = 7 – 2 = 5
Put x = 2, we get y = 7 – 2(2) = 7 – 4 = 3
Put x = 3, we get y = 7 – 2(3) = 7 – 6 = 1
∴ Four solutions are (0, 7), (1, 5), (2, 3) and (3, 1).
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Question 62 Marks
Solve the equation 2x +1 = x - 3, and represent the solution(s) on the Cartesian plane.
Answer
We know that x = – 4 can be written as x + 0.y = – 4
which is a linear equation in the variables x and y. This is represented by a line. Now all the values of y are permissible because 0.y is always 0. However, x must satisfy the equation x = – 4. Hence, two solutions of the given equation are x = – 4, y = 0 and x = – 4, y = 2.
Note that the graph AB is a line parallel to the y-axis and at a distance of 4 units to the left of it (see Fig.)

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Question 82 Marks
The graphs given in Fig. select the equation whose graph it is from the choices given below:

Answer
Self learning
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Question 102 Marks
The graph is given in Fig. select which equation for the graph is from the choices given below:

Answer
Self learning
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Question 112 Marks
Find four different solutions of the equation x + 2y = 6
Answer
We have By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2 x + 2y = 2 + 4 = 6
Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6 which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6.
Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the given equation. So four of the infinitely many solutions of the given equation are: (2, 2), (0, 3), (6, 0) and (4, 1). Hence the required Solutions.
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