Let the contributions ofYarnini and Fatima be Rs.x and Rs.y respectively.
Then according to the question
x + y = 100
This is the linear equation which the given data satisfies.
Now, x + y = 100
⇒ y = 100-x
Table of solution

We plot the points (0, 100) and (50, 50) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 100.

(i) The point (-2,4) lies in the II quadrant.
(ii) The point (3, -1) lies in the IV quadrant.
(iii) The point (- 1,0) lies on the negative x-axis.
(iv) The point (1, 2) lies in the I quadrant.
(v) The point (- 3, - 5) lies in the Ill quadrant.

This expression is a polynomial in one variable x because in the expression there is only one variable (x) and all the indices of x are whole numbers.

Yes! zero is a rational number.We can write zero in the form \(\frac { p }{ q } \),where p and q are integers and \(q\neq 0\)as follows:
  \(0=\frac { 0 }{ 1 } =\frac { 0 }{ 2 } =\frac { 0 }{ 3 } \)etc.

Here (1, 2) is a solution of a linear equation you are looking for. So, you are looking for any line passing through the point (1, 2). One example of such a linear equation is x + y = 3. Others are y – x = 1, y = 2x, since they are also satisfied by the coordinates of the point (1, 2). In fact, there are infinitely many linear equations which are satisfied by the coordinates of the point (1, 2).

Let p(x) = x + 2.
Then p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0
Therefore, –2 is a zero of the polynomial x + 2, but 2 is not.

We want to write \(\frac { 1 }{ \sqrt { 2 } } \)as an equivalent expression in which the denominator is a rational number. We know that \(\sqrt{2} \cdot \sqrt{2} \) is rational. We also know that multiplying \(\frac{1}{\sqrt{2}} \text { by } \frac{\sqrt{2}}{\sqrt{2}}\) will give us an equivalent expression, since \(\frac{\sqrt{2}}{\sqrt{2}}=1\) . So, we put these two facts together to get 
\(\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
In this form, it is easy to locate \(\frac{1}{\sqrt{2}}\) on the number line. It is half way between 0 and \(\sqrt{2} \text {. }\)

\((2 \sqrt{2}+5 \sqrt{3})+(\sqrt{2}-3 \sqrt{3})=(2 \sqrt{2}+\sqrt{2})+(5 \sqrt{3}-3 \sqrt{3})\)
\(=(2+1) \sqrt{2}+(5-3) \sqrt{3}=3 \sqrt{2}+2 \sqrt{3}\)

2x+0y-5=0

Convention

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
= (2x + 3y - 4z)² [Using Identity V]
= (2x + 3y - 4z)(2x + 3y - 4z)
(ii) 2x² + y²+ 8z² - 2 \(\sqrt{2}\)xy + 4\(\sqrt{2}\) yz - 8xz
= (-2 \(\sqrt{2}\)x)² + (y)² + (2\(\sqrt{2}\)z)² + 2 (-\(\sqrt{2}\)x)(y) + 2 (2\(\sqrt{2}\)z)(y) + 2 (2\(\sqrt{2}\)z) (-\(\sqrt{2}\)x)
= (-\(\sqrt{2}\)x + y + 2 \(\sqrt{2}\))²
= (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)( -\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)

(i) 2x + y = 7
When x = 0, 2(0) + y = 7
⇒ 0+ y = 7
⇒ y = 7
∴ Solution is (0, 7).
When x = 1, 2(1) + y = 7
⇒ y = 7 - 2
⇒ y = 5
∴ Solution is (1, 5).
When x = 2, 2(2) + y = 7
⇒ y = 7 - 4
⇒ y = 3
∴ Solution is (2, 3).
When x = 3, 2(3) + y = 7
⇒ y = 7 - 6
⇒ y = 1
∴ Solution is (3, 1).
(ii) πx + y = 9
When x = 0, π(0) + y = 9
⇒ y = 9 - 0 
⇒ y = 9
∴ Solution is (0, 9).
When x = 1, π(1) + y = 9
⇒ y =9-π
∴ Solution is {1, (9 - π)}.
When x = 2, π(2) + y = 9
⇒ y = 9 - 2π
∴ Solution is {2, (9 - 2π)}.
When x = -1, π(-1) + y = 9
⇒ -π+ y = 9
⇒ y=9+π
∴ Solution is {-1, (9 + π)}.
(iii) x = 4y
When x = 0, 4y = 0
⇒ y = 0
∴ Solution is (0, 0).
When x = 1, 4y = 1
⇒ y= \(\frac{1}{4}\)
∴ Solution is (1,\(\frac{1}{4}\)).
When x = 4, 4y = 4
⇒ y = \(\frac{4}{4}\) = 1
∴ Solution is (4, 1).
When x =-4, 4y =-4
y = \(\frac{-4}{4}\) =-1
∴ Solution is (-4, -1).

Let EW and NS be two main roads such that the road EW is along the East-West direction and road NS is along the North-South direction. Then, the angle between the two roads is 90°, i.e. the roads EW and NS are perpendicular to each other.
Let us consider EW along X-axis and NS along Y-axis and let the center of ciry is 0
Here, the distance be in two consecutive streets in same direction is 200 m and all the streets are parallel to the main

The street plan is shown in the above figure:
(i) Because of the two reference lines that we have used for locating them, there is only one cross-street which can be referred to as (4, 3).
(ii) Similarly, there is only one cross-street which can be referred to as (3, 4).

(i) We have, \(\frac { 1 }{ \sqrt { 7 } } \)
On multiplying both numerator and denominator by \(\sqrt { 7 } \) , we get
\(\frac { 1 }{ \sqrt { 7 } } \times \frac { \sqrt { 7 } }{ \sqrt { 7 } } =\frac { \sqrt { 7 } }{ 7 } \)
(ii) We have,\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \)
On multiplying both numerator and denominator by \(\sqrt { 7 } +\sqrt { 6 } ,\) we get
\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \times \frac { \left( \sqrt { 7 } +\sqrt { 6 } \right) }{ \left( \sqrt { 7 } +\sqrt { 6 } \right) } \)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ { \left( \sqrt { 7 } \right) }^{ 2 }-{ \left( \sqrt { 6 } \right) }^{ 2 } } [\because (a-b)(a+b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ 7-6 } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ 1 } \)
\(=\sqrt { 7 } +\sqrt { 6 } \)
(iii) \(\left[ \frac { \sqrt { 5 } -\sqrt { 2 } }{ 3 } \right] \)
(iv) \(\left[ \frac { \sqrt { 7 } +1 }{ 3 } \right] \)