I = 24 cm b = 12 cm h = 8 cm
Volume of the brick = I x b x h = (24 \(\times\) 12 \(\times\) 8) cm³= 2304 cm³
Wall dimensions are:
L = 20 m = 2000 cm
B = 48 cm
H = 6 m = 600 cm
No. of brieks required to build the wall = \(\frac { Volume\ of\ the\ wall }{ Volume\ of\ a\ bricks } \)

No. of brieks required = 25000

Total number of families S = 1500
n(S) = 1569
Let P be the event of selecting a family having part time maids and F be the event of selecting a family having full time maids.
(i) Both types of maids Let P \(\cap\) F be the event of selecting a family having both types of maids.
Let (P \(\cap\) F) = 259
\(\mathrm{P}(\mathrm{P} \cap \mathrm{F})=\frac{\mathrm{n}(\mathrm{P} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{250}{1500}=\frac{1}{6}\)
(ii) Part time maids Part time maids = only part time maid + both
= 860 + 250 = 1110 '
n(P) = 1110
\(\mathrm{p}(\mathrm{P})=\frac{\mathrm{n}(\mathrm{P})}{\mathrm{n}(\mathrm{S})}=\frac{1110}{1500}=\frac{111}{150}\)
(iii) No maids Let (P \(\cup\) P)'be the event of choosing a family not having maids and (P u F) be the event of choosing a family having part time or fuli time or both maids.
n(P \(\cup\) F ) = only n(P) + only n(F) + n(p \(\cap\) p) = 850 + 370 + 250
n(P\(\cup\)F) = 1480
n(P\(\cup\)F)' = n(S)-n(P\(\cup\)F)
= 1500 - 1480
n(P \(\cup\) F)' = 20
\(\mathrm{P}(\mathrm{P} \cup \mathrm{F})^{\prime}=\frac{\mathrm{n}(\mathrm{P} \cup \mathrm{F})^{\prime}}{\mathrm{n}(\mathrm{S})}=\frac{20}{1500}=\frac{1}{75}\)

(i) side ofa cube = 8 m
TSA of the cube = 6a² = 6 \(\times\) 64 = 384 m²
LSA of the cube = 4a² = 4 \(\times\) 64 = 256 m²
(ii) side a = 21 cm
TSA= 6a² = 6 \(\times\) 21 \(\times\) 21 = 2646 cm².
LSA = 4a² = 4 \(\times\) 21 \(\times\) 21 = 1764 cm².
(iii) side a = 7.5 cm
TSA = 6a² = 6 \(\times\) 7.5 \(\times\) 7.5 cm² = 337.5 cm²
LSA = 4a² = 4 \(\times\) 7.5 \(\times\) 7.5 cm² = 225 cm².

Area of the quadrilateral
= Area of Δ ABD + Area of Δ BCD
Δ ABD is right angled triangle
∴ Area = \(\frac{1}{2}\)b h
=
In Δ ABD, BD²= AD² + AB²
= 15²+ 20²= 225 + 400 = 625 m²
BD =\(\sqrt { 625 } \) = 25 m
∴ In ΔBCD, s = \(\frac { 25+25+17 }{ 2 } \)
= \(\frac { 68 }{ 2 } \) = 4
Area of Δ BCD = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 34(34-25)(34-26)(34-17) } \)
= \(\\ \sqrt { 34\times 9\times 8\times 17 } =\sqrt { 41616 } \)
∴ Area of the quadrilateral
= (150 + 204) m² = 354 m².

Let the age of the father be x .
Let the sum of the age of his sons be y
At present x = 2y \(\Rightarrow\) x - 2y = 0
After 20 years x + 20 = y + 40
x-y = 40 -20
x-y = 20             ...(1)
(1) \(\Rightarrow\) x-2y = 0
(2) \(\Rightarrow\) \(\cfrac { x-y=20 }{ -y=-20 } \)
y = 20
Substitute y = 20 in (1)
x-2(20) = 0
x-40 = 0
x = 40
\(\therefore\) The age of the father is 40 years

Perimeter of an equilateral triangle = 180 cm
∴ one side (a) = \(\\ \frac { 180 }{ 3 } \) = 60 m.
Area of an equilateral triangle =\(\frac { \sqrt { 3 } }{ 4 } \) a² sq.units

= 900\(\sqrt { 3 } \) m²
= 900 \(\times\) 1.732 = 1558.8 m²

Let the time taken by the larger pipe be x hours
and Set the time taken by the smaller pipe be y hours.
\(\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 1 }{ 24 } \)
In 1.hour the larger pipe can fill it = \(\cfrac { 1 }{ x } \)
In 1hour the smaller pipe can fill it  = \(\cfrac { 1 }{ y } \)
\(\cfrac { 8 }{ x } +\cfrac { 18 }{ y } =\cfrac { 1 }{ 2 } \)
Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
\(a+b\ =\cfrac { 1 }{ 24 } \)
24a + 24b = 1
24a + 24b - 1 = 0    ...(1)
8a + 18b = \(\cfrac { 1 }{ 2 } \)
16a+36b = 1
16a + 36b - 1 = 0    ..(2)
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ -24+36 } =\cfrac { b }{ -16+24 } =\cfrac { 1 }{ 864-384 } \)
\(\cfrac { a }{ 12 } =\cfrac { b }{ 8 } =\cfrac { 1 }{ 480 } \)
\(\cfrac { a }{ 12 } =\cfrac { 1 }{ 480 } \ \)  \( \cfrac { b }{ 8 } =\cfrac { 1 }{ 480 } \\ \)

\(\therefore\) x = 40, y = 60

"The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle."
\(\therefore\) The mid points of the sides of the triangle are given as
  x₁y_{1 ,} x₂y, x₃y₃
\(\left( \frac { 3 }{ 2 } ,5 \right) \left( 7,\frac { -9 }{ 2 } \right) \left( \frac { 13 }{ 2 } ,\frac { -13 }{ 2 } \right) \)
\(\therefore\)Centroid \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 3 }{ 2 } +\frac { 7 }{ 1 } +\frac { 13 }{ 2 } }{ 3 } ,\frac { 5+\left( \frac { -9 }{ 2 } \right) +\frac { (-13) }{ 2 } }{ 3 } \right) =\left( \frac { \frac { 3+14+13 }{ 2 } }{ 3 } ,\frac { \frac { 10-9-13 }{ 2 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 30 }{ 2 } }{ 3 } ,\frac { \frac { -12 }{ 2 } }{ 3 } \right) =\left( \frac { 15 }{ 3 } ,\frac { -6 }{ 3 } \right) =(5,-2)\)

Centroid G (x,y) = (4, -2)
two vertices (x₁, y₁) = (3, -2)
(x₂, y₂) = (5, 2), (x₃, y₃) =?

\(\frac { 8+{ x }_{ 3 } }{ 3 } =4\ \ \frac { { y }_{ 3 } }{ 3 } =-2\)
8 + x₃ = 12  y₃ = -6
x₃ = 4
∴ The third vertex (x₃, y₃) = (4, -6)

In the figure,
\(\triangle\)OPM, \(\triangle\)OQN are similar triangles. In similar triangles the sides are in the same proportional.
\(\cfrac { QN }{ PM } =\cfrac { QO }{ PO } \)
\(\cfrac { h }{ 5 } =\cfrac { 35 }{ 25 } \)
\(h=\cfrac { 5\times 35 }{ 25 } \)

h = 7m

\(sinA=\cfrac { 4 }{ 5 } \)
\(tanA=\cfrac { 4 }{ 3 } \)
\(\therefore \cfrac { sinA-cosA }{ 2tanA } =\cfrac { \frac { 4 }{ 5 } -\frac { 3 }{ 5 } }{ 2\times \frac { 4 }{ 3 } } =\cfrac { \frac { 1 }{ 5 } }{ 2\times \frac { 4 }{ 3 } } =\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 5 } \times \cfrac { 3 }{ 4 } =\cfrac { 3 }{ 40 } \)

By the Pythagoras theorem
x=\(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } \)
= \(\sqrt { 25-9 } \)
= \(\sqrt { 16 } =4\)

sin B = \(\frac { 9 }{ 41 } \);
cos B = \(\frac { 40 }{ 41 } \);
tan B =\(\frac { 9 }{ 40 } \) ;
cosec B = \(\frac { 1 }{ sinB } \) = \(\frac { 41 }{ 9 } \);
sec B =\(\frac { 1 }{ cotB } \) = \(\frac { 41 }{ 40 } \);
cot B =\(\frac { 1 }{ tanB } \) = \(\frac { 40 }{ 9 } \)

\(M(x,y)=\left( \frac { 3+p }{ 2 } ,\frac { 4+7 }{ 2 } \right) \)
\(\frac { 3+p }{ 2 } =x\quad \frac { 4+7 }{ 2 } =y\)
\(y=\frac { 11 }{ 2 } \)
substitute the values of (x, y) in 2x + 2y +1 = 0

3 + p + 11 + 1 = 0
P + 15 = 0
p = -15

(i) \(\cfrac { 2x+2 }{ 3 } =\cfrac { 3x-6 }{ 5 } \)
5(2x+2) = 3(3x - 6) [By cross multiplication]
10x + 10 = 9x - 18
10x = 9x - 18 - 10
10x = 9x-28
10x- 9x = -28
x = -28
(ii)

2 = 4 (x + 1)-x
2 = 4x+4-x
3x + 4 = 2'
3x+4-4 = 2-4
3x =-2

\(3\left( \cfrac { 1 }{ 4 } +1 \right) =3\left( 5-\cfrac { 1 }{ 4 } \right) -2\left( 5+\cfrac { 1 }{ 4 } \right) \)

\(3\left( \cfrac { 5 }{ 4 } \right) =3\left( \cfrac { 20-1 }{ 4 } \right) -2\left( \cfrac { 20+1 }{ 4 } \right) \)
\(\cfrac { 15 }{ 4 } =\cfrac { 57 }{ 4 } -\cfrac { 42 }{ 4 } \)
\(\cfrac { 15 }{ 4 } =\cfrac { 15 }{ 4 } \)
Yes \(\cfrac { 1 }{ 4 } \) is a solution of the given equation

The side of the cube (a) = 5 cm
Total Surface Area = 6a² = 6(5²) = 150 sq. cm
Lateral Surface Area = 4a² = 4(5²) = 100 sq. cm

The given system of equations are 3x + 5y − 21 = 0; −7x − 6y + 49 = 0
Now using the coefficients for cross multiplication, we get,

\(\Rightarrow \frac { x }{ (5)(49)-(-6)(-21) } =\frac { y }{ (-21)(-7)-(49)(3) } =\frac { 1 }{ (3)(-6)-(-7)(5) } \)
\(\frac { x }{ 119 } =\frac { y }{ 0 } =\frac { 1 }{ 17 } \)
\(\Rightarrow \frac { x }{ 119 } =\frac { 1 }{ 17 } ,\frac { y }{ 0 } =\frac { 1 }{ 17 } \)
\(\Rightarrow x=\frac { 119 }{ 17 } ,y=\frac { 0 }{ 17 } \)
\(\Rightarrow\) x = 7, y = 0
Verification :
3x + 5y = 21 ...(1)
3(7) + 5(0) = 21
21 + 0 = 21
21 = 21 True
-7x - 6y = -49 ...(2)
-7(7) - 6(0) = - 49
-49 = -49
-49 = -49 True

(i) y = 2x

 Put x = -1, y = 2 \(\times\) - 1 = -2
When x = 0, y = 2 \(\times\) 0 = 0
When x = 1, y = 2 \(\times\)- 1 = 2

The points (x,y) to be plotted.
(ii) y = 4x - 1

When x = -1\(\Rightarrow\) y = 4(-1)-1
y = -4 - 1 = -5
x = 0 \(\Rightarrow\) y = 4 \(\times\)0 -1 = -1
x = 1 \(\Rightarrow\) y = 4\(\times\)1 -1 = 3

The points (x, y) to be plotted:
(-1, -5), (0, -1), (1, 3)
(iii) \(y=\left( \frac { 3 }{ 2 } \right) x+3\)

\(x=-2\Rightarrow y=\left( \cfrac { 3 }{ 2 } \right) \left( 2 \right) +3\)
= -3 + 3 = 0
\(x=0\Rightarrow y=\left( \cfrac { 3 }{ 2 } \right) \left( 0 \right) +3\)
y = 0 + 3 = 3
\(x=2\Rightarrow y=\left( \cfrac { 3 }{ 2 } \right) \left( 2 \right) +3\)
= 3 + 3
y = 6

The points to be plotted
(-2, 0), (0, 3),(2, 6)
(iv) 3x + 2y = 14

2y = -3x + 14
\(y=\cfrac { -3x+14 }{ 2 } \)
\(\Rightarrow y=\left( \cfrac { -3 }{ 2 } \right) \left( 0 \right) +7=7\)
When x = -2, \(y=\left( \cfrac { -3 }{ 2 } \right) \left( -2 \right) +7=3+7=10\)
When x = 0 , \(y=\left( \cfrac { -3 }{ 2 } \right) \left( 0 \right) +7=7\)
When x = 2, y = -3 +7 = 4

The points to be plotted (-2, 10), (0, 7), (2, 4)

Let A(x, 3), B(6, y), C(8, 2) and D(9, 4) be the vertices of the parallelogram ABCD. By definition, diagonals AC and BD bisect each other.

Mid-point of AC = Mid-point of BD
\(\left( \frac { x+8 }{ 2 } ,\frac { 3+2 }{ 2 } \right) =\left( \frac { 6+9 }{ 2 } ,\frac { y+4 }{ 2 } \right) \)
equating the coordinates on both sides, we get
\(\frac { x+8 }{ 2 } =\frac { 15 }{ 2 } \)
x + 8 = 15
x = 7
\(\frac { 5 }{ 2 } =\frac { y+4 }{ 2 } \)
5 = y + 4
y = 1
Hence, x = 7 and y = 1.

By Pythagoras theorem,
\(AB=\sqrt { { BC }^{ 2 }{ AC }^{ 2 } } \)
= \(\sqrt { { \left( 25 \right) }^{ 2 }-{ 7 }^{ 2 } } \)
= \(\sqrt { 625-49 } =\sqrt { 576 } \) = 24
The six trignometric ratios are
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { 7 }{ 25 } \)
\(tan\theta =\frac { oppositeside }{ adjacent\ side } =\frac { 7 }{ 24 } \)
\(sec\theta =\frac { hypotenuse }{ adjacent\ side } =\frac { 25 }{ 24 } \)
\(cos\theta =\frac { adjacentside }{ hypotenuse } =\frac { 24 }{ 25 } \)
\(cosec\theta =\frac { hypotenuse }{ oppositeside } =\frac { 25 }{ 7 } \)
\(cot \theta\ \frac { adjacentside }{ oppositeside } =\frac { 24 }{ 7 } \)

In the quadrilateral ABCD, join one of the diagonals, say AC.
Area of \(\triangle\)ABC = \(\frac{1}{2}\)\(\times\) base \(\times\) height
=\(\frac{1}{2}\)\(\times\)8\(\times\)15\(\times\) 60 cm²
By Pythagoras theorem, in right angled triangle ABC,
AC² = AB² + BC²
= 8² +15² = 64 + 225 = 289 cm
Therefore, AC =\(\sqrt{289}\) =17cm
Now, for\(\triangle\)ACD, let us consider a = 17 cm, b =12 cm, c =25 cm
then, s = \(\frac{a+b+c}{2}=\frac{17+12+25}{2}=\frac{54}{2}\) = 27cm

Area of \(\triangle\)ACD =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt{27(27-17)(27-12)(27-25)}\)
=\(\sqrt{27\times10\times15\times2}\)
=\(\sqrt{3\times3\times3\times2\times5\times5\times3\times2}\)
= 3 × 3 × 2 × 5 = 90cm²
Therefore, Area of quadrilateral ABCD
=Area of \(\triangle\)ABC + Area of \(\triangle\)ACD
= 60 + 90 = 150 cm²

The given system of equation can be written as
2x + 7y − 5 = 0
−3x + 8y +11 = 0
For the cross multiplication method, we write the coefficients as

\(\frac { x }{ (7)(11)-(8)(-5) } =\frac { y }{ (-5)(-3)-(11)(2) } =\frac { 1 }{ (2)(8)-(-3)(7) } \)
\(\frac { x }{ 77+40 } =\frac { y }{ 15-22 } =\frac { 1 }{ 16+21 } \)
\(\frac { x }{ 117 } =\frac { y }{ -7 } =\frac { 1 }{ 37 } \)
\(\frac { x }{ 117 } =\frac { 1 }{ 37 } ,\frac { y }{ -7 } =\frac { 1 }{ 37 } \)
Hence the solution is \(\left( \frac { 117 }{ 37 } ,\frac { -7 }{ 37 } \right) \)
Verification:
2x + 7y -5 = 0 .....(1)
\(2\left( \frac { 117 }{ 37 } \right) +7\left( \frac { -7 }{ 37 } \right) -5=0\)
\(\frac { 234 }{ 57 } -\frac { 49 }{ 37 } -5=0\)
\(\frac { 185 }{ 37 } -5=0\)
5 - 5 =0 True
3x+8y+11 = 0 ...(2)
\(3\left( \frac { 117 }{ 37 } \right) +8\left( \frac { -7 }{ 37 } \right) +11=0\)
\(\frac { -351 }{ 37 } -\frac { 56 }{ 37 } +11=0\)
\(\frac { -407 }{ 37 } +11=0\)
-11 + 11 = 0 True

The given system of equations are 8x − 3y = 5xy ...(1)
6x − 5y = −2xy ...(2)
Observe that the given system is not linear because of the occurrence of xy term. Also note that if x =0, then y =0 and vice versa. So, (0,0) is a solution for the system and any other solution would have both x \(\neq \) 0 and y \(\neq \) 0
Let us take up the case where x \(\neq \) 0 and y \(\neq \) 0
Dividing both sides of each equation by xy,
\(\frac { 8x }{ xy } -\frac { 3y }{ xy } =\frac { 5xy }{ xy } \)
\(\frac { 6x }{ xy } -\frac { 5y }{ xy } =\frac { -2xy }{ xy } \)
Let \(a=\frac { 1 }{ x } ,b=\frac { 1 }{ y } \)
We get, \(\frac { 8 }{ y } -\frac { 3 }{ x } =5\)  .....(3)
\(\frac { 6 }{ y } -\frac { 5 }{ x } =-2\)  .......(4)
(3)&(4) respectively become, 8b − 3a = 5 ...(5)
b − 5a = −2 ...(6)
which are linear equations in a and b.
To eliminate a, we have,  (5) \(\times\) 5\(\Rightarrow\) 40b −15a = 25 .....(7)
(6) × 3\(\Rightarrow\) 18b −15a = −6 .....(8)
Now proceed as in the previous example to get the solution\(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \).
Thus, the system have two solutions \(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \) and (0,0).

(i) (cos0⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ - cos 45⁰ + cos 60⁰)
= \(\left[ 1+\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \left[ 1-\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \)
= \(\left[ \frac { 2\sqrt { 2 } +2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] \left[ \frac { 2\sqrt { 2 } -2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] =\left[ \frac { 3\sqrt { 2 } +2 }{ 2\sqrt { 2 } } \right] \left[ \frac { 3\sqrt { 2 } -2 }{ 2\sqrt { 2 } } \right] \)
= \(\frac { 18-4 }{ 4\left( \sqrt { 2 } \right) ^{ 2 } } =\frac { 14 }{ 4\times 2 } =\frac { 7 }{ 4 } \) 
(ii) tan²60⁰ - 2tan²45⁰ - cot²30⁰ +2sin²30⁰ + \(\frac { 3 }{ 4 } \) cosec²45⁰
= \(\left( \sqrt { 3 } \right) ^{ 2 }-2(1)^{ 2 }-\left( \sqrt { 3 } \right) ^{ 2 }+2\left( \frac { 1 }{ 2 } \right) ^{ 2 }+\frac { 3 }{ 4 } \left( \sqrt { 2 } \right) ^{ 2 }\)
= \(3-2-3+\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \)
= -2 + \(\frac { 4 }{ 2 } \) = -2 + 2 = 0

Let x be the digit at ten’s place and y be the digit at unit place.
Given that x + y = 5 …… (1)

Given, Original number − reversing number = 27
(10x + y) − (10y + x) = 27
10x − x + y −10y = 27
9x − 9y = 27
\(\Rightarrow\) x − y = 3 ... (2)
Also from (1), y = 5 – x ... (3)
Substitute (3) in (2) to get x − (5 − x) = 3
x − 5 + x = 3
2x = 8
x = 4
Substituting x = 4 in (3), we get y = 5 − x = 5 − 4
y = 1
Thus, 10x + y = 10 × 4 +1 = 40 +1 = 41.
Therefore, the given two-digit number is 41.
Verification :
sum of the digits = 5
x + y = 5
4 + 1 = 5
5 = 5 true
Original number – reversed number = 27
41 - 14 = 27
27 = 27 true

Given points are A(–3, 6) and B(1, –2), P(–2, 4) divide AB internally in the ratio m : n.
By section formula,
\(P(x, y)=P\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)
= P(−2, 4) .......(1)
Here x₁ = −3, y₁ = 6, x₂ = 1, y₂ = −2
\((1)\Rightarrow \left( \frac { m(1)+n(-3) }{ m+n } ,\frac { m(-2)+n(6) }{ m+n } \right) \)
Equating x-coordinates, we get
\(\frac{m-3 n}{m+n}=-2\) or m− 3n = −2m− 2n
3m = n
\(\frac{m}{n}=\frac{1}{3}\)
m : n = 1: 3
Hence P divides AB internally in the ratio 1 : 3.

Let A(−2, −1) and B(4, 8) are the given points.

Let P(a,b) and Q(c,d) be the points of trisection of AB, so that AP = PQ = QB .
By the formula proved above,
P is the point.
\(\left( \frac { { x }_{ 2 }+{ 2x }_{ 1 } }{ 3 } ,\frac { { y }_{ 2 }+{ 2y }_{ 1 } }{ 3 } \right) =\left( \frac { 4+2(-2) }{ 3 } ,\frac { 8+2(-1) }{ 3 } \right) \) = (0, 2)
Q is the point
\(\left( \frac { { 2x }_{ 2 }+{ x }_{ 1 } }{ 3 } ,\frac { {2 y }_{ 2 }+{ y }_{ 1 } }{ 3 } \right) =\left( \frac { 2(4)-2 }{ 3 } ,\frac { 2(8)-1 }{ 3 } \right) \) = (2, 5)

Given x + y = 5 ...(1)
2x – y = 4 ...(2)
To draw the graph (1) is very easy. We can find the x and y intercepts and thus two of the points on the line (1).
When x = 0, (1) gives y = 5.
Thus A(0,5) is a point on the line.
When y = 0, (1) gives x = 5.
Thus B(5,0) is another point on the line.
Plot A and B; join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.
When x = 0, (2) gives y = −4.
Thus P(0,−4) is a point on the line.
When y = 0, (2) gives x = 2.
Thus Q(2,0) is another point on the line.
Plot P and Q; join them to produce the line (2).
The point of intersection (3, 2) of lines (1) and (2) is a solution.
The solution is the point that is common to both the lines. Here we find it to be (3,2). We can give the solution as x = 3 and y = 2.

We have already come across one method: forming a table of values, listing and plotting ordered pairs and joining the points.
But, to fix a line, after all, how many points do we need? Just two! These can easily be obtained when a line is given in the form y = mx + c.
The given line y = 4x – 3
put x = 0 to get y-intercept
y = 4(0)–3
y = –3
point is (0, –3) and y-intercept = – 3
put y = 0 to get x-intercept
0 = 4x – 3
3 = 4x
\(\frac { 3 }{ 4 } =x\)
point is \(\left( \frac { 3 }{ 4 } ,0 \right) \) and x-intercept = \(\frac { 3 }{ 4 } \)
The graph may be drawn through two points (0,−3) and \(\left( \frac { 3 }{ 4 } ,0 \right) \)