(i) E,F,G,H,M,N
(ii) \(\overline{EG}\),\(\overline{FH}\),\(\overline{EF}\),\(\overline{GH}\),\(\overline{MN}\)
(iii) M,E,G,B
(iv) \(\overleftrightarrow { AB } ,\overleftrightarrow { CD } ,\overleftrightarrow { PQ } ,\overleftrightarrow { RS } \)

Some of Euclid's axioms are
(i) Things which are equal to the same things, are equal to one another.
(ii) If equals are added to equals, then wholes are equal.
(iii) Things which are double of the same things, are equal to one another.
Thus, given three axioms are Euclid's axioms. So, here we cannot deduce any statement from these axioms which contradicts any axiom. So, given system of axioms is consistent.

We have
AB = BC
\(\Rightarrow \) AB - BX = BC - BX
|If equals are subtracted from equals, the remainders are equal (Euclid's Axiom (iii))
AB - BX = BC - BY  \(|\)\( \because\)  BX = BY
AB - BX coincides with AX;
BC - BY coincides with CY
[Things which coincide with one another are equal to one another (Euclid's Axiom (iv))]

According to the question, the linear
equation for the given problem is
=> y = 500 + 20x ... (1)
Table of solution

We plot the points (0, 500) and (5, 600) on a graph paper and join the same by a ruler to get the line which is the graph of the equation (1).

Let the present ages of father and son be x years and y year respectively.
Then, Age of father 5 years ago =(x-5)years
Age of his son 5 years ago(y-5) years
According to the question,
x-5=7(y-5)+2
x-y=7y-35+2
x-7y+28=0
which is the required linear equation in two variables.

(i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0.
(ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0.
(iii) 2x = 3 can be written as 2x + 0.y – 3 = 0.
(iv) 5y = 2 can be written as 0.x + 5y – 2 = 0.

Given, ㄥ2 = ㄥ1 and ㄥ3 = ㄥ4    ......(i)
Now,  ㄥ2 = ㄥ1
ㄥ2 + ㄥ3 = ㄥ1 + ㄥ3    [if equals are added to equals, then the wholes are equal ]
⇒ ㄥ2 + ㄥ3 = ㄥ2 + ㄥ3      [from Eq. (i)]
⇒ ㄥ2 + ㄥ3 + 30° = ㄥ2 + ㄥ3 + 30° [if equals are added to equals, then the wholes are equal]
⇒ ㄥ2 + ㄥ3 + 30° =ㄥ2 +ㄥ3 + θ ⇒ 30° = θ
[since, if equals are subtracted from equals, then the remainders are equal]
ஃ  θ = 30°

In the given figure, AC coincides with AB + BC.
Also, Euclid's axiom 4 says that things which coincide with one another, are equal to one another. So, it can be deduced that
AB + BC = AC

(i)1
(ii)11

2x+0y-3=0

O(0,0) satisfies y=x

x+y+0=0

2x+0y-5=0

Given equation is x + y = 4.
To draw the graph, we need at least two solutions of the equation.
Given equation can be written as y = 4 - x.
When x = 0, then y = 4
When x = 4, then y = 4 - 4 = 0
Thus, we have the following table
So, plot the points A(0,4) and B(4,0) on thegraph paper and join them by a line.

Hence, line AB represents the required graph of given linear equation.

(ii) x - y = 2
⇒ y = x-2
If we have x = 0, then y = 0 -2 = -2
x = 1, then y = 1 - 2 = - 1
x = 2, then y = 2 - 2 = 0
∴ We have the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown below:
Thus, the line PQ is required graph of x - y = 2.

(iii) y = 3x
If x = 0, then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) ⇒ y = 3
x = -1, then y = 3(-1) ⇒ y = -3
We get the following table:

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get the straight line LM.
Thus, LM is the required graph of y = 3x.

Note: The graph of the equation of the form y = kx is a straight line which always passes through the origin.
(iv) 3 = 2x + y ⇒ y = 3 - 2x
∴ If x = 0, then y = 3 - 2(0) ⇒ y = 3
If x = 1, then y = 3 - 2(1) ⇒ y = 1
If x = 2, then y = 3 - (2) ⇒ y = -1

Plot the ordered pairs (0, 3), (1, 1) and (2, -1) on the graph paper. Joining these points, we get a line CD.

Thus, the line CD is the required graph of 3 = 2x + y

Since, the point (3, 4) lies on the graph, then it will satisfy its equation.
On putting x = 3 and y = 4 in given equation, we get
3(4)=a(3)+7
\(\Rightarrow 13+3a+7\)
\(\Rightarrow 12-7=3a\)
\(\Rightarrow 3a=5\)
\(\Rightarrow a=\frac{5}{3}\)

Two distinct intersecting lines cannot be parallel to the same line.

Given, AC = BD       ...(i)
From figure, it is clear that
AC = AB + BC and BD = BC + CD.
On putting these values in Eq. (i), we get
AB + BC = BC + CD
On subtracting BC from both sides, we get
AB + BC - BC = BC + CD - BC
⇒ AB = CD    [by axiom 3]
Hence proved

(a) (iii) x + y = 7
(b) (iv) 2
x + y = 7 ⇒ 5 + y = 7
⇒ y = 7 - 5 = 2
(c) (i) (3, 4)
(d) (iv) (0, 7)
(e) (iii) -1
On putting x = 2 and y = -1 in the equation x + 3y - k = 0, we have
2 + 3(-1) - k = 0
⇒ 2 - 3 = k
⇒ k = -1

(a) (iii) 2x + 3y = 60
Let the number of children be x and the number of adults be y then the linear equation in two variable for the given situation is 
2x + 3y = 60.
(b) (i) 10
2x + 3y =60 ⇒ 2(15) + 3y = 60
⇒ 3y = 60 - 30 = 30 
⇒ y = 10
(c) (i) 12
2x + 3y = 60 ⇒ 2x + 3(12) = 60
⇒ 2x 60 - 36 = 24
⇒ x = 12
(d) (iii) 15
On putting x = 5 and y = 0 in the equation 3x + 5y = b, we have 
3 x 5 + 5 x 0 = b
⇒ 15 + 0 = b
⇒ b = 15
(e) (ii) 1.x - 1.y + 5 = 0
y - x = 5 ⇒ y = x + 5
⇒ x - y + 5 = 0
⇒ 1.x - 1.y + 5 = 0

(i) (d) 5x + 2y = 120
Here, the cost of one notebook be Rs. x and that of pen be Rs. y.
According to the statement, we have 5x + 2y = 120
(ii) (b) x = 20, y = 10
5x + 2y = 5(10) + 2(20) = 50 + 40 = 90 ≠ 120
5x + 2y = 5(20) + 2(10) = 100 + 20 = 120
5x + 2y = 5(15) + 2(15) = 75 + 30 = 105 ≠ 120
(c) (ii) Rs. 95
5x + 2y = 5(15) + 2(10) = 75 + 20 = 95
(d) (b) Rs. 10
Here, x = 2y
5(2y) + 2y
= 10y + 2y = 12y = 120
⇒ y = 10
(e) (iv) 0.x + 1.y – 4 = 0
Since, y = 4 ⇒ y – 4 = 0
Thus, standard form of y = 4 is 0.x + 1.y – 4 = 0

(i) (a) 3x + 2y = 80 and 4x + 3y = 110
Here, the cost of one notebook be Rs. x and that of pen be Rs. y.
According to the statement, we have
3x + 2y = 80 and
4x + 3y = 110
(ii) (b) x = 20, y = 10
3x + 2y = 3(20) + 2(10) = 60 + 20 = 80
4x + 3y = 4(20) + 3(10) = 80 + 30 = 110
(b) x = 20, y = 10
(iii) (b) Rs. 10
Cost of 1 pen = Rs. 10
(b) Rs. 10
(iv) (d) Rs. 420
Total cost = Rs. 15 x 20 + Rs. 12 x 10
= 300 + 120
= Rs. 420
(v) (a) Deepak
Ram said that price of each notebook could be Rs. 25.
Ajay felt that Rs. 2.50 for one pen was too little. It should be at least Rs. 16
Deepak guess the cost of one pen is Rs. 10 and
Lohith guess the cost of one notebook is Rs. 30
Therefore, estimation of Deepak is correct

(i) (b) 10x, 5y
Area of one bedroom = 5x sq.m
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
(ii) (d) 54 m
Length of outer boundary = 12 + 15 + 12 + 15 = 54 m
(iii) (d) None of the above
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
So, 10x + 5y = 95  2x + y = 19
Also, x + 2 + y = 15 x + y = 13
(iv) (c) x = 6, y = 7
x + y = 6 + 7 = 13
2x + y = 2(6) + 7 = 19
x = 6, y = 7
x + y = 6 + 7 = 13
2x + y = 2(6) + 7 = 19
 x = 6, y = 7
(v) (a) 30 sq. m
Area of living room = (15 x 7) – 30
= 105 – 30 =75 sq. m