2 Marks Questions

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 $4.$
$\therefore (\sqrt2,4\sqrt2)$ 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
$x – 2y =40 – 2(2) = –4$, which is not $4.$
$\therefore (0, 2)$ 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 $y=9-\pi\left(\frac{9}{\pi}\right)=9-9=0$
$\therefore$ 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

$2x + y = 7$
$\Rightarrow y = 7 – 2x$
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 $y=7-2(3)=7-6=1$
$\therefore$ 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 $A B$ 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 72 Marks

Solve the equation $2x +1 = x - 3$, and represent the solution(s) on the number line.

Answer

we have the equation $2x + 1 = x - 3$ for representing on number line and cartesian system. first we need to solve for $x2x +1 = x - 3$
$\Rightarrow 2x - x = -3 -1$ [ by transposing the values ] we get
$\Rightarrow x = - 4$
Representation on number line
$x = - 4$

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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 92 Marks

The graph is given in Fig. the select which equation for the 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 $x+2 y=6$

Answer

We have By inspection, $x=2, y=2$ is a solution because for $x=2, y=2 x+2 y=2+4=6$
Now, let us choose $x=0$. With this value of $x$, the given equation reduces to $2 y=6$ which has the unique solution $y=3$. So $x=0, y=3$ is also a solution of $x+2 y=6$.
Similarly, taking $y=0$, the given equation reduces to $x=6$. So, $x=6, y=0$ is a solution of $x+2 y=6$ 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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