We have 4x² + y² + z² – 4xy – 2yz + 4xz = (2x)² + (–y)² + (z)² + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z)
= [2x + (–y) + z]² (Using Identity V)
= (2x – y + z)² = (2x – y + z)(2x – y + z)
So far, we have dealt with identities involving second degree terms. Now let us extend Identity I to compute (x + y)³. We have:
(x + y)³ = (x + y) (x + y)²
= (x + y)(x² + 2xy + y2)
= x(x² + 2xy + y²) + y(x² + 2xy + y²)
= x³ + 2x²y + xy² + x²y + 2xy² + y³
= x³ + 3x²y + 3xy² + y3
= x³ + y³ + 3xy(x + y)

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).

\(\frac{1}{7}=0.142857142857 \ldots=0 . \overline{142857}\) and \(\frac{2}{7}=0.28571428571428 \ldots=0 . \overline{285714}\)
Here, the two decimal expansions are non-terminating recurring.
Hence, 1/7​ and 2/7​ are two rational numbers.
We know, between any two rational numbers, there are infinitely many irrational numbers.
An irrational number has non-terminating non-recurring decimal expansions.
Then an irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\) is 0.15015001500015.
Similarly, 0.21020020002... is another irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\)

Here, we have
8x³ + y³ + 27z³ – 18xyz
= (2x)³ + y³ + (3z)³ – 3(2x)(y)(3z)
= (2x + y + 3z)[(2x)² + y2 + (3z)² – (2x)(y) – (y)(3z) – (2x)(3z)]
= (2x + y + 3z) (4x² + y2 + 9z² – 2xy – 3yz – 6xz)

The true option is (iii) y = 3x + 5 has infinitely many solution
Reason: For every value of x, there is a corresponding value of y and vice-versa.

Consider the lamp as a point and table as a plane.Choose any two perpendicular edges of the table.Measure the distance of the lamp from the longer edge, suppose it is 25 cm. Again, measure the distance of the lamp from the shorter edge, and suppose it is 30 cm.You can write the position of the lamp as (30,25) or (25, 30), depending on the order you fix.

Let x=\(0.\overline { 47 } \) =0.4777....
Multiplying both sides by 10, we get
10x = 9.9999...
10x = 4.7777...
10x = 4.3+0.47777...
10x = 4.3+x
10x - x = 4.3
9x = 4.3
x = 4.3/9 = 43/90
Thus, \(0.\overline { 47 } \) = 43/90
Here p =4 3
q = 90(\(\neq 0\))

(2,2) and (-3,-3) both satisfy y=x

O(0,0) does not satisfy y=mx+c

x+5=6 ⇒ x=1

\(42+y-y^2=42+7y-6y-y^2\)
\(=7(6+y)-y(6+y)\)
\(=(6+y)(7-y)\)

Definition of quadratic polynomial

(i) We have 2x + 3y = 9.3\(\overline{5}\)
∴ (2)x + (3)y + (-9.3\(\overline{5}\)) = 0
Comparing it with ax + bx + c = 0, we have a = 2, b = 3 and c = -9.3\(\overline{5}\).
(ii) We have x - \(\frac{y}{5}\)- 10 = 0
or x + \(\left(-\frac{1}{5}\right)\) y + (-10) = 0
Comparing with ax + bx + c = 0, we get
a = 1, b = \(-\frac{1}{5}\) and c = -10
Note: Above equation can also be compared by:
Multiplying throughout by 5,
5(x) - (5x\(\frac{y}{5}\)) - (5 x 10) = 0
or 5x - y - 50 = 0
or 5(x) + (-1)y + (-50) = 0
Comparing with ax + by + c = 0, we get a = 5, b = -1 and c = -50.
(iii) We have -2x + 3y = 6
⇒ -2x + 3y - 6 = 0
⇒ (-2)x + (3)y + (-6) = 0
Comparing with ax + bx + c = 0, we get a = -2, b = 3 and c = -6.
(iv) We have x = 3y
⇒ x - 3y = 0
⇒ (1)x + (-3)y + 0 = 0
Comparing with ax + bx + c = 0, we get a = 1, b = -3 and c = 0.
(v) We have 2x = -5y
⇒ 2x + 5y = 0
⇒ (2)x + (5)y + 0 = 0
Comparing with ax + by + c = 0, we get a = 2, b = 5 and c = 0.
(vi) We have 3x + 2 = 0
⇒ 3x + 2 + 0y = 0
⇒ (3)x + (0)y + (2) = 0
Comparing with ax + by + c = 0, we get a = 3, b = 0 and c = 2.
(vii) We have y - 2 = 0
⇒ (0)x + (1)y + (-2) = 0
Comparing with ax + by + c = 0, we have a = 0, b = 1 and c = -2.
(viii) We have 5 = 2x
⇒ 5 - 2x = 0
⇒ -2x + 0y + 5 = 0
⇒ (-2)x + (0)y + (5) = 0
Comparing with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

From the figure, we have
(i) The co-ordinates of B are (-5, 2).
(ii) The co-ordinates of C are (5, -5).
(iii) The point E is identified by the co-ordinates (-3, -5)
(iv) The point G is identified by the co-ordinates (2, -4).
(v) The abscissa of the point D is 6.
(vi) The ordinate of the point H is -3.
(vii) The co-ordinates of the point L are (0, 5).
(viii) The co-ordinates of the point M are (-3, 0).
Plotting a point in the plane when coordinates of the point are given
To plot a point in the coordinate plane we draw the coordinate axes and choose our units such that we can mark equal distances on the x-axis or on the y-axis or on both the axes. We take origin as zero for x-axis as well as on y-axis. The distances marked along OX and OY are taken as positive and those along OX and OY are taken as negative.

\(-\frac { 31 }{ 4 } \)

First, we draw AB = 9.3 units. Now, from B, mark a distance of 1 unit. Let this point be C. Let D be the mid-point of AC. Now, draw a semi-circle with centre D and radius DA. Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.
\(\therefore\)   Distance, BD=\(\sqrt { 9.3 } \)
Draw an arc with centre B and radius BD, which intersects the number line at point E. So point E represents \(\sqrt { 9.3 } \) .

(i) (d) (-4, 3) 
(ii) (b) (-3, -2)
(iii) (b) (8, 4.5)
(iv) (d) IV
(v) (b) 10 units

(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