True False[1 Marks ]

Question 11 Mark

The boundaries of the solids are curves.

Answer

False.Solution:
The given statement is false because boundaries of solids are surfaces.

Question 21 Mark

The coordinates of points in the table.

$x$	$0$	$1$	$2$	$3$	$4$
$y$	$2$	$3$	$4$	$-5$	$6$

represent some of the solutions of the equation $x – y + 2 = 0.$

Answer

The coordinates of p0oints are $(0, 2), (1, 3), (2, 4), (3, –5)$ and $(4, 6).$
Given equation is $x0 - y + 2 = 0$
At point $(0, 2), 0 – 2 + 2 = 0 \Rightarrow 0=0,$ it satisfies.
At point $(1, 3), 1 - 3 + 2 = 3-3 = 0 \Rightarrow 0 = 0,$ it satisfies.
At point $(2, 4), 2 - 4 + 2 = 4 - 4 = 0 \Rightarrow 0 = 0,$ it satisfies.
At point $(3, -5), 3 – (- 5) + 2 = 3 + 5 + 2 = 10 \neq 0,$ it does not satisfy.
At point $(4, 6), 4 - 6 + 2 - 6 - 6 = 0 \Rightarrow 0 = 0,$ it satisfies.
Hence, point $(3, – 5)$ does not satisfy the equation.

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Question 31 Mark

Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.

Answer

False Solution: Since, every point on the graph of the linear equation represents a solution.

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Question 41 Mark

The graph of every linear equation in two variables need not be a line.

Answer

False Solution: Since, the graph of a linear equation in two variables always represent a line.

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Question 51 Mark

The graph given below represents the linear equation $x + y = 0.$

Answer

If the given points $(-1, 1)$ and $(- 3, 3)$ lie on the linear equation $x + y = 0,$
then both points will satisfy the equation. So, at point $(-1, 1),$
we put $x = -1,$ and $y = 1$ in $LHS$ of the given equation,
we get $LHS = x + y = -1 + 1 = 0 = RHS$ Again,
at point $(-3, 3)$ put $x = -3$ and $y = 3$ in $LHS$ of the given equation,
we get, $LHS = x + y= -3 + 3 = 0 = RHS$
Hence, $(-1, 1)$ and $(-3, 3)$ both satisfy the given linear equation $x + y = 0.$

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Question 61 Mark

The graph of the linear equation $x + 2y = 7$ passes through the point $(0, 7).$

Answer

If we put $x = 0$ and $y = 7$ in $LHS$ of the given equation, we get, $LHS = (0) + 2 (7)= 0 + 14 = 14 ≠ 7 = RHS$ Hence, $(0, 7)$ does not lie on the line $x + 2y = 7.$

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Question 71 Mark

The point $(0, 3)$ lies on the graph of the linear equation $3x + 4y = 12.$

Answer

If we put $x = 0$ and $y = 3$ in $LHS$ of the given equation, we find, $LHS = 3 × 0 + 4 × 3 = 0 + 12 = 12 = RHS$ Hence, $(0, 3)$ lies on the linear equation $3x + 4y = 12.$

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