Exercise 3.3

1.Solve the following pair of linear equations by the substitution method.

(i) x + y = 14

x-y=4

the given pair of linear equations is

x+y=14………(1)

x-y=4…………(2)

from equation(1)

y=14-x………….(3)

substitute this value of y in equation (2)

we get

x-(14-x)=4

x-14+x=4

2x-14=4

2x=4+14

2x=18

X=\frac{18}{2}=9

Substituting this value of x in equation (3), e get

Y=14-9=5

Therefore the solution is

X=9,   y=5

Remark

Verification (for our confidence that our solution is correct)

Substituting x = 9 and y = 5,

we find that both the equations (1) and (2) are satisfied as shown below:

x + y = 9 + 5 = 14

x y = 9 - 5 = 4

This verifies the solution.

(ii)s -- t = 3

$$\frac{s}{3}+\frac{t}{2}=6$$

The given pair of linear equations is

s - t = 3……….(1)

$$\frac{s}{3}+\frac{t}{2}=6$$..............(2)

From equation (1),

s =t + 3 ...(3)……………(3)

Substitute this value of s in equation (2), we get

$$\frac{t+3}{3}+\frac{t}{2}=6$$

$$\frac{2(t+3)+3t}{6}=6$$

2(t + 3) + 3t = 36

2t + 6 + 3t = 36

5t + 6 = 36

5t = 36 — 6

5t = 30

$$t=\frac{30}{5}=6$$

substituting this value of t in equation (3), we get

s= s = 6 + 3 = 9

Therefore the solution is

s = 9, t = 6

(iii) 3x y = 3

9x — 3y = 9

The given pair of linear equations is

3x-y=3................(1)

9x-3t=9...............(2)

From equation (1), 3x - 3 = y

i.e.,       y = 3x - 3          ...(3)

Substitute this value of y in equation (2), we get

9x - 3(3x - 3) = 9

9x - 9x + 9 = 9

9 = 9

which is true. Therefore, equations (1) and (2) have infinitely many solutions.

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

The given system of linear equations is

0.2x + 0.3y = 1.3          ...(1)

0.4x + 0.5y = 2.3          ...(2)

Multiplying each of the two equations (1) and (2) by 10 to get rid of decimals, we have

2x+ 3y= 13      .......... ...(3)

And      4x + 5y = 23     ...(4)

From equation (3),

3y = 13 — 2x

$$Y=\frac {13-2x}{3}$$.................(4)

Substituting this value of y in equation (4), we get

Multiplying by LCM = 3,

12x + 5 (13 — 2x) = 69 12x + 65 — 10x = 69

12x — 10x = 69 — 65

2x= 4

$$X=\frac{4}{2}=2$$

Substituting this value of x in equation (5), we get

$$Y=\frac{13-4}{3}=\frac{9}{3}=3.$$

Therefore, the solution is x = 2, y = 3.

v)$$\sqrt{2x}+\sqrt{3y}=0$$

$$\sqrt{3x}-\sqrt{8y}=0$$

The given pair of linear equation is

$$\sqrt{2x}+\sqrt{3y}=0$$………(1)

$$\sqrt{3x}-\sqrt{8y}=0$$…………(2)

From equation (2)

$$\sqrt{3x}=\sqrt{8y}$$

$$X=\frac{\sqrt{8}}{\sqrt{3}}y$$…………(3)

Substituting this value of x in equation (1), we get

$$\sqrt{2}.\frac{\sqrt{8}}{\sqrt{3}}y+\sqrt{3}y=0$$

$$\frac{4}{\sqrt{3}}y+\sqrt{3}y=0$$

($$\because  \sqrt{2}\sqrt{8}=\sqrt{2x8}=\sqrt{16}=4$$)

$$\Rightarrow \:\:  (\frac{4}{\sqrt{3}}+\sqrt{3})y=0$$

$$\Rightarrow \:\:    y=0  \:\:  \because  (\frac{4}{\sqrt{3}}+\sqrt{3})≠0$$

Substituting this value of y in equation (3), we get

$$X=\frac{\sqrt{8}}{\sqrt{3}}(0)=0$$

Therefore, the solution is

X=0,   y=0.

Remark . the reader can also do this question by the method we have employed in

(i).iii). and iv) part.

vi)$$\frac{3x}{2}-\frac{5y}{3}=-2$$

$$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$$

The given system of linear equation is

$$\frac{3x}{2}-\frac{5y}{3}=-2$$…………(1)

$$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$$………….(2)

Multiplying equation (1)by 6

(LCM of denominators 2 and 3),

$$\Rightarrow$$      9x-10y=-12…………(3)

Multiplying equation (2)by 6.

                 (LCM of denominators 2,3 and 6)

2x+3y=13………….(4)

From equation (3)

9x=10y-12

$$X=\frac{10y-12}{9}$$……….(5)

Substituting the value of x in equation (4), we get

$$2(\frac{10y-12}{9})+3y=13$$

20y-24+27y=117

47y=117+24

$$Y=\frac{141}{47}$$

Y=3

Substituting the value of y in equation (5), we get

$$X=\frac{10(3)-12}{9}=\frac{30-12}{9}$$

=$$\frac{18}{2}=2$$

Therefore, the solution is

X=2, y=3

Remark. The reader can also do it by the method we have applied in (i), (ii), (iii) and (iv) part.

2.Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Sol. The given pair of linear equations

2x+3y=11……(1)

2x-4y=-24…………(2)

From equation (1)

3y=11-2x

$$Y=\frac{11-2x}{3}$$……….(3)

Substituting this value of y in equation (2), we get

$$2x — 4 (\frac{11-2x}{3})=24$$

Multiplying every term by LCM = 3,

6x — 4 (11 — 2x) = 72

6x 44 + 8x = — 72

14x 44 = — 72

14x = 44 — 72

14x=-28

$$x=-\frac{28}{14}=-2$$

Substituting this value of x in equation (3), we get

$$y=\frac{11-2(-2)}{3}=\frac{11+4}{3}=\frac{15}{3}=5$$

Therefore the solution is x = — 2, y = 5.

Putting x = - 2 and y = 5 in y= mx + 3

we have 5 = - 2m + 3     2m = 3 — 5

2m=-2    $$\Rightarrow m=\frac{-2}{2}=-1$$

3.Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Sol. Let the two numbers be x and y (x > y), then according to the question, the pair of linear equations formed is

x —y = 26        ...(1)

and      x = 3v   ......(2)

Substitute the value of x from equation

(2) in equation (1), we get

3y — = 26

2y = 26

$$Y=\frac{26}{2}$$

Y=13

Substituting this value of y in equation

(2), we get

x = 3(13) = 39

Hence, the required numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Sol. (ii) Let the larger and the smaller of two supplementary angles be x° and y° respectively.

Then, according to the question, the pair of linear equations formed is

= y° + 18°         ...(1)

and    x° + y° =18°      ….(2)

  |$$ \because$$ the two angles are supplementary

Substitute the value of x° from equation (1) in equation (2), we get

y° + 18° + y° = 180°

2y° + 18° = 180°

2y° = 180° — 18°

2y° = 162°

$$y^{0}=\frac{162^{0}}{2}=81^{0}$$

Substituting this value of y° in equation (1), we get

$$X^{0}=81^{0}+18^{0}=99^{0}$$

Hence, the larger and the smaller of the two supplementary angles are 99° and 810 respectively.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball.

sol.Let the cost of each bat and each ball be rs x and rs y respectively.

Then, according to the question, the pair of linear equations formed is

7x + 6y = 3800 ...(1)

And  3x + 5y = 1750     ...(2)

From equation (2),

5y = 1750 - 3x

$$Y=\frac{1750-3x}{5}$$ ...... (3)

Substitute this value of y in equation (1), we get

$$7x+6(\frac{1750-3x}{5})=3800$$

Multiplying every term by LCM, we have35x + 6(1750 - 3x) = 19000

35x + 10500 - 18x = 19000

17x + 10500 = 19000

17x = 19000 - 10500

17x = 8500

$$x=\frac{8500}{17}=500$$

Substituting this value of x in equation (3), we get

$$y=\frac{1750-3(500)}{5}$$

=$$\frac{1750-1500}{5}=\frac{250}{5}=50$$

Hence, the cost of each bat and each ball is 500 and 50 respectively.

iv) Let the fixed charges be x and the charge per kilometre be Rs y.

Sol.Then, according to the question, the pair of linear equations formed is

x + l0y = 105    .............(l)

and      x + 15y = 155........(2)

From equation (1),

x 105 — 10y    —(3)

Substitute this value of x in equation (2), we get

105 — lOy + 15y = 155

105 + 5y = 155

5y = 155 — 105

5y = 50

$$Y=\frac{50}{5}=10$$

Substituting this value of y in equation (3), we get

x = 105 - 10(10)

= 105 - 100 = 5

Hence, the fixed charges are 5 and the charge per kilometre is 10.

Now, charges for 25 km

= x + 25y = 5 + 25 x 10

= 5 + 250 = 255

Hence the charges for 25 km journey = =rs 255

(v) Let the required fraction be $$\frac{x}{y}$$

Sol.Then, according to the question, the pair of linear equations formed is

$$\frac{x+2}{y+2}=\frac{9}{11}$$

Cross-multiplying

11(x + 2) = 9(y + 2)

11x+ 22 = 9y+ 18

11x= 9y+ 18-22

11x- 9y + 4= 0 …….(1)

and $$\frac{x+3}{y+3}=\frac{5}{6}$$

Cross-multiplying

6(x + 3) = 5(y + 3)

6x+ 18= 5y+ 15

6x — 5y = — 3    ………(2)

From equation (1), 11x = 9y — 4

$$x=\frac{9y-4}{11}$$…..(3)

Substitute this value of x in equation (2),we get

$$6(\frac{9y-4}{11})-5y=-3$$

Multiplying by LCM =11,

6(9y 4) — 55y = — 33

54y — 24 — 55y = — 33

54y — 55y = — 33 + 24

-y=9 $$\Rightarrow$$ y=9 Substituting this value of y in equation (3), we get

$$X=\frac{9(9)-4}{11}=\frac{81-4}{11}=\frac{77}{11}=7$$

Hence , the required fraction is $$\frac{x}{y}=\frac{7}{9}$$.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Sol.Then, according to the question, the pair

of linear equations formed is

x+ 5 = 3(y + 5) x+ 5= 3y+ 15 x — 3y = 10

and      x — 5 = 7(y — 5)

x — 5 = 7y — 35 x— 7y=— 35 + 5 x — 7y = — 30

From equation (1),

x = 3y + 10       (3)

Substitute this value of x in equation

(2), we get

(3y + 10) — 7y = — 30

3y — 7y = — 30 10

— 4y = — 40

$$Y=\frac{-40}{-4}=10$$

Substituting this value of y in equation

(3), we get

x = 3(10) + 10 = 30 + 10 = 40

Hence, the present ages of Jacob and his son are 40 years and 10 years respectively.

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