Explore topic-wise InterviewSolutions in Current Affairs.

(a) Firstly, to calculate the total rain in India, we can get an estimate of it and then knowing the weight of water we can estimate the weight of clouds.

(b) The pressure generated by the wind can give us the estimation of its speed.

It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ∼ 10^–15 s) are used to measure time intervals in several physical and chemical processes.
X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.
The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ∼ 10^–15 s) are used to measure time intervals in several physical and chemical processes.

X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing. 

The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

(a) Beyond 2F 

(b) At 2F 

(c) Between F and 2F 

(d) Between F and optical centre

For maximum work angle must be zero degree and for minimum work force and displacement must be perpendicular to each other.

(i) When force is in direction of displacement, then work done , W = F × S

(ii) When force is at an angle θ to the direction of displacement, then work done,

W = F S cos^θ

जब एक दण्ड चुम्बक मुक्त रूप से एक कुण्डली के बीच से होकर गिरता है तो कुण्डली में वैद्युत धारा प्रेरित हो जाती है, जो सदैव उस कारण का विरोध करती है, जिससे वह उत्पन्न होती है। अत: चुम्बक का त्वरण (a), गुरुत्वीय त्वरण (g) से कम होगा।

729 x 8000 = 3√729 x 8000

= 3√(9 x 9 x 9) x (20 x 20 x 20)

= 9 x 20 = 180

In previous question there are three numbers which are not perfect cubes. 

(i) 130 

Apply subtraction method, 

130 – 1 = 129 

129 – 7 = 122 

122 – 19 = 103 

103 – 37 = 66 

66 – 61 = 5 

∵ Next number to be subtracted is 91, which is greter than 5 

Hence, 130 is not a perfect cube. So, to make it perfect cube we subtract 5 from it. 

130 – 5 = 125 (which is a perfect cube of 5) 

(ii) 345 

Apply subtraction method, 

345 – 1 = 344 

344 – 7 = 337 

337 – 19 = 318 

318 – 37 = 281 

281 – 61 = 220 

220 – 91 = 129 

129 – 127 = 2 

∵ Next number to be subtracted is 169, which is greter than 2 

Hence, 345 is not a perfect cube. So, to make it a perfect cube we subtract 2 from it. 

345 – 2 = 343 (which is a perfect cube of 7) 

(iii) 792 

Apply subtraction method, 

792 – 1 = 791 

791 – 7 = 784 

784 – 19 = 765 

765 – 37 = 728 

728 – 61 = 667 

667 – 91 = 576 

576 – 127 = 449 

449 – 169 = 280 

280 – 217 = 63 

∵ Next number to be subtracted is 271, which is greter than 63 

Hence, 792 is not a perfect cube. So, to make it a perfect cube we subtract 63 from it. 

792 – 63 = 729 (which is a perfect cube of 9)

The correct option is: (D) 163.86 cm²

Explanation:

Total surface area of the block = Total surface area of cube + Curved surface area of hemisphere - Area of base of hemisphere

= 6 x (5)² + 2π x (2.1)² - π x (2.1)2 

= 150 + π x (2.1)² = 150 + 13.86 = 163.86 cm²

The correct option is:

(B) If two sphere are melted to form a cylinder, then surface area of cylinder is the sum of surface area of two sphere. 

(B) The volumes of the two cylinders thus formed is 210c.c.

Let one number be ‘x’. 

another number be ‘y’. 

Their difference is 26. 

∴ x – y = 26 ………. (i) 

One number is three times the other, 

∴ x = 3y …………. (ii) 

Substituting x = 3y in eqn. 

(i), x – y = 26 

3y – y = 26 

2y = 26

∴ y = \( \frac{26}{2}\)

∴ y = 13 

Substituting the value of ‘y’ in eqn. (ii), 

x = 3y

∴ x = 3 × 13 

∴ x = 39, y = 13

The given equations are

x + 2y = – 1 ..(i)

and 2x – 3y = 12 …(ii)

From (i) x = – 1 – 2y …(iii)

Substituting this value of x in

(ii), we get

2( – 1 – 2y) – 3y = 2

⇒ – 2 – 4y – 3y = 12

⇒ – 7y = 14

⇒ y = -2

Putting y = – 2 in (iii), we get

x = -1-2×(-2) = -1+4 = 3

Hence the required solution is x = 3 and y = – 2.

Given equations are

3x – y = 3 …(i)

9x – 3y = 9 …(ii)

From eqⁿ (i), y = 3x – 3 …(iii)

On substituting y = 3x – 3 in eqⁿ (ii), we get

⇒ 9x – 3(3x – 3) = 9

⇒ 9x – 9x + 9 = 9

⇒ 9 = 9

This equality is true for all values of x, therefore given pair of equations have infinitely many solutions.

Given equations are

0.5x + 0.8y = 3.4 …(i)

0.6x – 0.3y = 0.3 …(ii)

From eqⁿ (ii), 2x – y = 1

y = 2x – 1 …(iii)

On substituting y = 0.2x – 1 in eqⁿ (i), we get

⇒ 0.5x + 0.8(2x – 1) = 3.4

⇒ 0.5x + 1.6x – 0.8 = 3.4

⇒ 2.1x = 3.4 + 0.8

⇒ 2.1x = 4.2

⇒ x = 4.2/2.1 = 2

Now, on putting x = 2 in eqⁿ (iii), we get

⇒ y = 2(2) – 1

⇒ y = 4 – 1

⇒ y = 3

Thus, x = 2 and y = 3 is the required solution.

Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the given condition,

 the sum of a two digit number and the number 

obtained by interchanging its digits is 99. 

∴ 10y + x + 10x +y = 99 

∴ 11x + 11y = 99 

Dividing both sides by 11,

 x + y = 9 

if y = 1, then x = 8

If y = 2, then x = 7 

If y = 3, then x = 6 and so on. 

∴ The number can be 18, 27, 36, … etc

Five solutions of x + y = 5 are given below:

 (1,4), (2, 3), (3, 2), (4,1), (0, 5) 

The above solutions also satisfy the equation 2x + 2y = 10. 

∴ x + y = 5 …[Dividing both sides by 2] 

∴ If the two equations are the same, then the two 

equations in two variables have all solutions common.

Correct option is (C) 42, 18

Let present age of son be x years and

present age of father be y years.

Then according to given conditions, we get

y - 6 = 3 (x - 6)

\(\Rightarrow\) y = 3x - 12       ____________(1)

and y + 6 = 2 (x+6)

\(\Rightarrow\) 2x+12 = y+6

\(\Rightarrow\) 2x - y + 6 = 0   ____________(2)

From (1) & (2), we get

2x - (3x - 12) + 6 = 0

\(\Rightarrow\) 2x - 3x + 12 + 6 = 0

\(\Rightarrow\) x = 18

\(\therefore\) y = 3x - 12

= 54 - 12 = 45       (From (1))

Hence, their present ages are 18 & 42 years.

Correct option is C) 42, 18

8 – y = 2 

∴ y = 8 – 2 

∴ y = 6 

∴ 8 – 6 = 2

Correct option is (C) 12

Let the two-digit number be ab whose unit digit is b and ten's digit is a.

\(\therefore\) Required number is ab = 10a + b   _________(1)

Reversed number is ba = 10b + a     _________(2)

Sum of digits = a + b

According to given conditions, we have

10a + b = 4 (a+b)

and 10a + b + 9 = 10b + a

\(\Rightarrow\) 6a - 3b = 0

and 9a - 9b + 9 = 0

\(\Rightarrow\) 2a - b = 0          _________(3)

and a - b + 1 = 0   _________(4)

Subtract equation (4) from (3), we get

(2a - b) - (a - b + 1) = 0 - 0

\(\Rightarrow\) a - 1 = 0

\(\Rightarrow\) a = 1

\(\therefore\) b = 2a = 2    (From (3))

\(\therefore\) Required number is ab = 12.   (From (1))

Correct option is C) 12

Correct option is (A) x = 4, y = 3

Take \(\frac1x=X\;\&\;\frac1y=Y\) then given equations are convert into

5Y - 2X \(=\frac76\)         __________(1)

and 36X - 24Y = 1    __________(2)

Multiply equation (1) by 18, we get

90Y - 36X = 21    __________(3)

By adding equations (2) & (3), we get

(36X - 24Y) + (90Y - 36X) = 1+21

\(\Rightarrow\) 90Y - 24Y = 22

\(\Rightarrow\) 66Y = 22

\(\Rightarrow\) Y = \(\frac{22}{66}=\frac13\)

\(\therefore y=\frac1Y=3\)    \((\because Y=\frac1y)\)

Then from (1),

2X = 5Y - \(\frac76\)

\(=\frac{5}{3}-\frac{7}{6}=\frac{10-7}{6}\)

\(=\frac36=\frac12\)

\(\therefore X=\frac14\)

\(\Rightarrow\) x = 4   \((\because X=\frac1x)\)

Hence, x = 4 & y = 3 is the solution of given system of equations.

Correct option is A) x = 4, y = 3

Let 1/x = u and 1/y = v, 

So the given equations becomes, 

u/2 + v/3 = 2 ………………(i) 

u/3 + v/2 = 13/6 ……………(ii) 

From (i), we get 

u/2 + v/3 = 2 

⇒ 3u + 2v = 12 

⇒ u = \(\frac{(12 – 2v)}{3}\) ………….(iii)

Using (iii) in (ii) 

[(12 – 2v)/3]/3 + \(\frac{v}{2}\) = \(\frac{13}{6}\) 

⇒ \(\frac{(12 – 2v)}{9 }\)+ v/2 = \(\frac{13}{6}\) 

⇒ 24 – 4v + 9v = (\(\frac{13}{6}\)) x 18 [after taking LCM] 

⇒ 24 + 5v = 39 

⇒ 5v = 15 

⇒ v = 3 

Substituting v in (iii) 

u = \(\frac{(12 – 2(3))}{3}\)

⇒ u = 2 

Thus, x = \(\frac{1}{u}\)

⇒ x = \(\frac{1}{2}\) and y = \(\frac{1}{v}\) 

⇒ y = \(\frac{1}{3}\) 

The solution for the given pair of equations is x = \(\frac{1}{2}\) and y = \(\frac{1}{3}\) respectively.

x + 5 = 14 

∴ x = 14 – 5 

x = 9 

∴ 9 + 5 = 14

2x + y = 5 …(i) 

3x – y = 5 …(ii) 

Adding equations (i) and (ii),

 2x + y = 5 + 3x – y = 5 

5x = 10 

∴ x = \(\frac{10}{5}\)

∴ x = 2

 Substituting x = 2 in equation (i),

 2(2) + y = 5 

4 + y = 5 

∴ y = 5 – 4 = 1

∴ (2, 1) is the solution of the given equations.

Let 1/x = u and 1/y = v 

So, the given equations becomes 

15x + 2y = 17 ……………(i) 

x + y = \(\frac{36}{5}\)………………(ii) 

From equation (i) we get, 

2y = 17 – 15x 

y = \(\frac{(17 − 15x)}{ 2}\) …………(iii) 

Substituting (iii) in equation (ii) we get, 

x + \(\frac{(17 − 15x)}{2}\) = \(\frac{36}{5}\)

2x + 17 – 15x = (36 x 2)/ 5 [after taking LCM] 

-13x = \(\frac{72}{5}\) – 17

-13x = -\(\frac{13}{5}\) 

⇒ x = \(\frac{1}{5}\) 

⇒ u = \(\frac{1}{x}\) = 5 

Putting x = \(\frac{1}{5}\) in equation (ii) , we get 

\(\frac{1}{5}\) + y = \(\frac{36}{5}\) 

⇒ y = 7 

⇒ v = \(\frac{1}{y}\) = \(\frac{1}{7}\) 

The solution of the pair of equations given are u = 5 and v = \(\frac{1}{7}\) respectively.

2y – x = 0 

∴ x = 2y …(i) 

10x + 15y = 105 …(ii) 

Substituting x = 2y in equation (ii), 

10(2y) + 15y = 105

∴ 20y + 15y = 105

∴ 35y = 105 

∴ y = \(\frac{105}{35}\)

∴ y = 3 

Substituting y = 3 in equation (i), 

x = 2y

∴ x = 2(3) = 6 

∴ (6, 3) is the solution of the given equations.

i. m + 3 = 5

 m = 5 – 3 

∴m = 2

ii. 3y + 8 = 22

∴ 3y = 22 – 8 

∴ 3y = 14 

∴ y = 14/9

iii. x/3 = 2 

∴ x = 2 × 3 

∴ x = 6

iv. 2p = p + 4/9

∴ 2p – p = 4/9

∴ p = 4/9

Let 1/x = u and 1/y = v 

So, the given equations becomes 

3u – v = -9…………(i) 

2u + 3v = 5 ……….(ii)

Multiplying equation (i) x 3 and (ii) x 1 we get, 

9u – 3v = -27 …….(iii)

2u + 3v = 5 ………(iv) 

Adding equation (iii) and (iv) we get , 

9u + 2u – 3v + 3v = -27 + 5 

⇒ 11u = -22 

⇒ u = -2 

Now putting u = -2 in equation (iv) we get, 

2(-2) + 3v = 5 

⇒ 3v = 9 

⇒ v = 3 

Hence, x = 1/u = −1/2 and, y = 1/v = 1/3.

i. x = 1, y = 6

ii. x = -1, y = 8 

iii. x = 5, y = 2 

iv. x = 0, y = 7 

v. x = 10, y = -3

2x – 7y = 7 …(i)

 3x + y = 22

∴ y = 22 – 3x ……(ii) 

Substituting y = 22 – 3x in equation (i), 

2x – 7(22 – 3x) = 7

 ∴ 2x – 154 + 21x = 7 

∴ 23x = 7 + 154 

∴ 23x = 161 

∴ x = \(\frac{161}{23}\)

∴ x = 7 

Substituting x = 7 in equation (ii),

 y = 22 – 3x 

∴ y = 22 – 3(7) 

∴ 7 = 22 -21= 1 

∴ (7, 1) is the solution of the given equations

Let 1/x = u and 1/y = v 

So, the given equations becomes 

2u + 5v = 1…………(i)

60u + 40v = 19 …………(ii)

Multiplying equation (i) x 8 and (ii) x 1 we get, 

16u + 40v = 8 …………(iii) 

60u + 40v = 19 ………(iv) 

Subtracting equation (iii) from (iv) we get, 

60u – 16u + 40v – 40v = 19 – 8 

⇒ 44u = 11 

⇒ u = 1/4 

Now putting u = 1/4 in equation (iv) we get, 

60(1/4) + 40v = 19 

⇒ 15 + 40v = 19 

⇒ v = 4/ 40 = 1/10 

Hence, x = 1/u = 4 and, y = 1/v = 10.

D. 10x - 14y + 4 = 0

Condition for dependent linear equations -

a₁ /a₂ = b₁/b₂ = c₁/c₂ …(i)

Given equation of line is, - 5x + 7y - 2 = 0;

Comparing with ax+ by +c = 0;

Here, a₁ = - 5, b₁ = 7, c₁ = - 2;

For second equation, let’s assume a₂x + b₂y + c₂ = 0;

From Eq. (i), -5/a₂ = 7/b₂ = -2/c₂ = 1/k

Where, k is any arbitrary constant.

Putting k = - 1/2 then

a₂ = 10, b₂ = - 14, c₂ = 4;

∴ The required equation of line becomes

a₂x + b₂y + c₂ = 0;

10x - 14y + 4 = 0;

Substitution Method:

x + y = 4

∴ x = 4 – y …(i)

 2x – 5y = 1 ……(ii) 

Substituting x = 4 – y in equation (ii), 

2(4 – y) – 5y = 1 

∴ 8 – 2y – 5y = 1

∴ 8 – 7y = 1 

∴ 8 – 1 = 7y 

∴ 7 = 7y 

∴ y = 7/7

∴ y = 1 

Substituting y = 1 in equation (i)

, x = 4 – 1 = 3

∴ (3,1) is the solution of the given equations.

Alternate method: 

Elimination Method:

x + y = 4 …(i)

 2x – 5y = 1 ……(ii)

 Multiplying equation (i) by 5,

 5x + 5y = 20 … (iii)

 Adding equations (ii) and (iii)

, 2x – 5y = 1 + 5x + 5y = 20 

7 = 21 

∴ x = 21/7

∴ x = 3 

Substituting x = 3 in equation (i),

 3 + y = 4 

∴ y = 4 – 3 = 1 

(3,1) is the solution of the given equations.

2x + 3y + 4 = 0 …(i) 

x – 5y = 11 

∴x = 11 + 5y …(ii)

 Substituting x = 11 + 5y in equation (i), 

2(11 +5y) + 3y + 4 = 0

∴ 22 + 10y + 3y + 4 = 0 

∴ 13y + 26 = 0 

∴ 13y = -26 

∴y = \(\frac{-26}{13}\)

∴ y = -2 

Substituting y = -2 in equation (ii)

 x = 11 + 5y

∴ x = 11 + 5(-2) 

∴ x = 11 – 10 = 1 

∴ (1, -2) is the solution of the given equations.

3x – 5y = 16 …(i) 

x – 3y = 8

∴x = 8 + 3y …..(ii)

 Substituting x = 8 + 3y in equation (i),

 3(8 + 3y) – 5y = 16

 24 + 9y- 5y = 16 

∴4y= 16 – 24 

∴ 4y = -8 

∴ y = -8/4

 y = -2

Substituting y = -2 in equation (ii), 

x = 8 + 3 (-2) 

∴ x = 8 – 6 = 2 

∴ (2, -2) is the solution of the given equations.

Let 1/x = u and 1/y = v 

So, the given equations becomes 

u/5 + v/6 = 12………(i) 

u/3 – 3v/7 = 8………(ii)

Taking LCM for both equations, we get 

6u + 5v = 360………(iii) 

7u – 9v = 168……….(iv) 

Subtracting (iii) from (iv) 

7u – 9v – (6u + 5v) = 168 – 360 

⇒ u – 14v = -192 

⇒ u = (14v – 192)………. (v) 

Using (v) in equation (iii), we get 

6(14v – 192) + 5v = 360 

⇒ 84v -1152 + 5v = 360 

⇒ 89v = 1512 

⇒ v = 1512/89 

⇒ y = 1/v = \(\frac{89}{1512}\) 

Now, substituting v in equation (v), we find u 

u = 14 x (1512/89) – 192

 ⇒ u = \(\frac{4080}{89}\) 

⇒ x = 1/u = \(\frac{89}{4080}\)

Hence, the solution for the given system of equations is x = \(\frac{89}{4080}\) and y = \(\frac{89}{1512}\).

 \(\frac{4}{x}\)+ 3y = 14

\(\frac{3}{x}\)+ 4y = 23

Multiply eq1 by 4 and eq2 by 3 and adding

⇒ 25/x = 125

⇒ x = 1/5

Thus,

20 + 3y = 14

⇒ y = - 2

Let 1/√x = u and 1/√y = v, 

So, the given equations becomes 

2u + 3v = 2…………(i) 

4u – 9v = -1 ……….(ii) 

Multiplying (ii) by 3 and Adding equation (i) and (ii) x 3 we get,

6u + 9v + 4u – 9v = 6 – 1 

⇒ 10u = 5 

⇒ u = \(\frac{1}{2}\) 

Substituting u = \(\frac{1}{2}\) in (i), we find v 

2(\(\frac{1}{2}\)) + 3v = 2 

⇒ 3v = 2 – 1 

⇒ v = \(\frac{1}{3}\) 

Since, 1/√x = u we get x = 1/u² 

⇒ x = 1/(\(\frac{1}{2}\))² = 4 

And, 1/√y = v y = 1/v² 

⇒ y = 1/(\(\frac{1}{3}\))² = 9 

Hence, the solution is x = 4 and y = 9.

 \(\frac{22}{x+y}\)+\(\frac{15}{x-y}\) = 5

\(\frac{55}{x+y}\)+\(\frac{45}{x-y}\) = 14

Multiplying eq1 by 3 and subtracting from eq2

⇒ - 11/(x + y) = - 1

⇒ x + y = 11 ---------- (3)

Multiplying eq1 by 5 and eq2 by 2 and subtracting

⇒ 15/(x – y) = 3

⇒ x – y = 5 ------ (4)

(3) + (4)

⇒ 2x = 16

⇒ x = 8

Thus,

y = 3

Taking 1/x = u, the given equation becomes 

4u + 3y = 14…………(i) 

3u – 4y = 23…………(ii) 

Adding (i) and (ii), we get 

4u + 3y + 3u – 4y = 14 + 23 

⇒ 7u – y = 37 

⇒ y = 7u – 37…………(iii) 

Using (iii) in (i), 

4u + 3(7u – 37) = 14 

⇒ 4u + 21u – 111 = 14 

⇒ 25u = 125 

⇒ u = 5 

⇒ x = 1/u = \(\frac{1}{5}\) 

Putting u= 5 in (iii), we find y 

y = 7(5) – 37 

⇒ y = -2 

Hence, the solution for the given system of equations is x = \(\frac{1}{5}\) and y = -2.

Let 1/x = u and 1/y = v 

So, the given equations becomes 

2u + 3v = 13…………(i) 

5u – 4v = -2 …………(ii) 

Adding equation (i) and (ii) we get, 

2u + 3v + 5u – 4v = 13 – 2 

⇒ 7u – v = 11 

⇒ v = 7u – 11…….. (iii)

Using (iii) in (i), we get 

2u + 3(7u – 11) = 13 

⇒ 2u + 21u – 33 = 13 

⇒ 23u = 46 

⇒ u = 2 

Substituting u = 2 in (iii), we find v 

v = 7(2) – 11 

⇒ v = 3 

Hence, x = 1/u = 1/2 and, y = 1/v = 1/3.

 \(\frac{44}{x+y}\)+\(\frac{30}{x-y}\) = 10

\(\frac{55}{x-y}\)+\(\frac{40}{x-y}\) = 13

Multiplying eq1 by 5 and eq2 by 4 and subtracting

⇒ - 10/(x – y) = - 2

⇒ x – y = 5

Multiply eq1 by 4 and eq2 by 3 and subtracting

⇒ 11/(x + y) = 1

⇒ (x + y) = 11

Thus,

2x = 16

⇒ x = 8

∴ y = 8 – 5 = 3

Given:

0.2x + 0.3y = 1.3 

⇒ 2x + 3y = 13 …… (1) 

0.4x + 0.5y = 2.3 

⇒ 4x + 5y = 23 …… (2) 

From equation (1) 

2x = 13 – 3y 

⇒ x = \(\frac{13-3y}{2}\)

Substituting x = \(\frac{13-3y}{2}\) equation (2) we get 

\(\frac{13-3y}{2}\) + 5y = 23

⇒ 26 – 6y + 5y = 23 

⇒ -y + 26 = 23 

⇒ y = 26 — 23 = 3 

Substituting y = 3 in equaion (1) we get 

2x + 3(3) = 13 

⇒ 2x + 9 = 13 

⇒ 2x = 13 – 9 

⇒ 2x = 4 

⇒ x = 4/2 = 2 

∴ The solution is (2, 3)

Given: 

3x – 5y = -1 ……. (1) 

x – y = -1 …….. (2) 

From equation (2), 

x – y = – 1

x = y – 1 

Substituting x = y – 1 in equation (1) we get 

3 (y – 1) – 5y = – 1 

⇒ 3y – 3 – 5y = -1 

⇒ – 2y = – 1 + 3 

⇒ 2y = – 2 

⇒ y = -1 

Substituting y = – 1 in equation (1) we get 

3x – 5 (- 1) = -1 

3x + 5 = – 1 

3x = – 1 – 5 

x = \(\frac{-6}{3}\) = -2 

∴ The solution is (-2, -1)

(C) On the line y = x

Since, the given point (a, a) has same value of x and y-coordinates. Therefore, the point (a, a), must be lie on the line y = x.

 \(\frac{4}{x}\)+ 5y = 7

\(\frac{3}{x}\)+ 4y = 5

Multiply eq1 by 4 and eq2 by 5 and subtracting

⇒ 1/x = 3

⇒ x = 1/3

Thus,

12 + 5y = 7

⇒ y = 1

Taking 1/x = u, the given equation becomes 

4u + 5y = 7………(i)

3u + 4y = 5………(ii)

Subtracting (ii) from (i), we get 

4u + 5y – (3u + 4y) = 7 – 5 

⇒ u + y = 2 

⇒ u = 2 – y………(iii) 

Using (iii) in (i), 

4(2 – y) + 5y = 7 

⇒ 8 – 4y + 5y = 7 

⇒ y = -1 

Putting y = -1 in (iii), we find u 

u = 2 – (-1) 

⇒ u = 3

⇒ x = 1/u = \(\frac{1}{3}\) 

Hence, the solution for the given system of equations is x = \(\frac{1}{3}\) and y = -1