SChand CLASS 9 Chapter 5 Simultaneous Linear Equations Exercise 5(A)

 Exercise 5(A)

Question 1

If (5, k) is a solution of the equation 2x + y – 7 = 0 find the value of k.

Sol :

∴(5,k) is a solution of the equations :-

⇒2x+k-7=0

⇒2×5+k-7

⇒10+k-7=0

⇒3+k=0

k=-3

Question 2

x + y = 5, x – y = 3

Sol :

⇒x+y=5...(i)

⇒x-y=3...(ii)

From (ii) x=3+y

Substituting the value of x in (i)

⇒3+y+y=5

⇒2y=5-3=

⇒$y=\frac{2}{2}=1$

∴x=3+y=3+1=4

Hence x=4 , y=1

Question 3

y = 2x – 6, y = 0

Sol :

Let y=2x-6

y=0

From (i) and (ii)

⇒2x-6=0

⇒2x=6

$x=\frac{6}{2}$

⇒x=3

So, x=3 , y=0

Question 4

p = 2q – 1, q = 5 – 3p

Sol :

Let P=2q-1....(i)

⇒q=5-3p...(ii)

Substituting the value of p in (ii)

⇒q=5-3(2q-1)

⇒q+6q=8

⇒7q=8

⇒$q=\frac{8}{7}$

From equation (ii)

$p=2 \times \frac{8}{7}-1$

$p=\frac{16}{7}-1$

$=\frac{16-7}{7}=\frac{9}{7}$

∴$p=\frac{9}{7}, q=\frac{8}{7}$

Question 5

9x + 4y = 5, 4x – 5y = 9

Sol :

9x+4y=5...(i)

4x-5y=9...(ii)

Multiplying both (i) by 5 and (ii) by 4 , we get

4x+20y=-25..(iii)

16x-20y=36...(iv)

On adding (iii) and (iv)

61x=61

$\Rightarrow x=\frac{61}{61}=1$

From (i)

⇒9×1+4y=5

⇒4y=5-9=-4

⇒$y=\frac{-4}{-4}=-1$

∴x=1 ,y=-1

Question 6

x + 3y = 5, 3x – y = 5

Sol :

x+3y=5...(i)

3x-y=5...(ii)

From (i) x=5-3y

So, we substituting the value of x in (ii)

⇒3(5-3y)-y=5

⇒15-9y-y=5

⇒-10y=5-15=-10

$y=\frac{-10}{-10}=1$

∴x=5-3y=5-3×1=5-3=2

Hence , x=2  y=1

Question 7

3x – 7y + 10 = 0, y – 2x – 3 = 0

Sol :

3x-7y+10=0 or 

3x-7y=-10...(i)

y-2x-3=0...(ii)

From (ii) y=3+2x

Substituting the value of y in (i)

⇒3x-7(3+2x)=-10

⇒3x-21-14x=-10

⇒-11x=-10+21=11

⇒$x=\frac{11}{-11}$=-1

∴y=3+2x=3+2(-1)

=3-2=1

So, x=-1 , y=1

Question 8

20u – 30v = 13, 10v – 10u = – 5

Sol :

20u-30v=13...(i)

10v-10u=-5 or 

10u-10v=5....(ii)

Multiplying (i) by 1 and (ii) by 3,  we get

$\begin{aligned}20 u-30 v&=13 \\20 u-20 v&=10 \\-\phantom{20u}+\phantom{20v}&\phantom{=}- \\ \hline-10 v&= 3 \end{aligned}$

$v=\frac{-3}{10}$

From (i) $10u-10 \times\left[\frac{-3}{10}\right]=5$

10u+3=5

10u=5-3

$\therefore u=\frac{2}{10}=\frac{1}{5}$

Hence $u=\frac{1}{5}, v=\frac{-3}{10}$

Question 9

2x – 3y = 1.3, y – x = 0.5

Sol :

2x-3y=1.3...(i)

y-x=0.5....(ii)

From (ii) y=0.5+x

Substituting the value of y in (i)

⇒2x-3(0.5+x)=1.3

⇒2x-1.5-3x=1.3

⇒-x=1.3+1.5

⇒-x=2.8

⇒x=-2.8

∴y=0.5+(-2.8)=-2.3

Hence , x=-2.8 and y=-2.3

Question 10

11x + 15y + 23 = 0, 7x – 2y – 20 = 0

Sol :

10x+15y+23=0...(i)

7x-2y-20=0....(ii)

Multiplying (i) by 2 and (ii) by 15 , then adding  we get 

$\begin{aligned}22 x+30 y+46=&0\\105 x-30 y-300=&0 \\ \hline 127 x \quad \quad-254=&0\end{aligned}$

⇒127x=254

⇒$x=\frac{254}{127}=2$

From (ii)

⇒7(2)-2y-20=0

⇒14-7y-20=0

⇒-2y=20-14=6

$y=\frac{6}{-2}=-3$

∴x=2 , y=-3

Question 11

3 – (x – 5) = y + 2

2(x + y) = 4 – 3y

Sol :

⇒3-(x-5)=y+2

⇒3-x+5=y+2

⇒-x-y=2-3-5

⇒-x-y=-6

⇒x+y=6...(i)

⇒2(x+y)=4-3y

⇒2x+2y=4-3y

⇒2x+2y+3y=4

⇒2x+5y=4...(ii)

From (i) x=6-y

Substituting the value of x in (ii)

⇒2(6-y)+5y=4

⇒12-2y+5y=4

⇒3y=4-12=-8

⇒$y=\frac{-8}{3}$

∴$x=6-y+\frac{8}{3}=\frac{18+8}{3}=\frac{26}{3}$

Hence , $x=\frac{26}{3}, y=\frac{-8}{3}$

Question 12

(a) $\frac{x}{2}+\frac{y}{4}=6 \text{ , } \frac{x}{5}-\frac{y}{2}$ 

(b) $\frac{x}{4}-3=\frac{y}{6}\text{ , } \frac{1}{2}x-y=-2$

Sol :

(i)

$\frac{x}{2}+\frac{y}{4}=6$....(i)

$\frac{x}{5}-\frac{y}{2}=0$...(ii)

Multiplying (i) by 1 and (ii) by $\frac{1}{2}$ ,  we get

⇒$\frac{x}{2}+\frac{y}{4}=6$

⇒$\frac{x}{10}-\frac{y}{4}=0$

Adding we get

⇒$\frac{x}{2}+\frac{x}{10}=6$

⇒$\frac{5 x+x=60}{10}$

⇒6x=60

⇒$x=\frac{60}{6}=10$

From (ii)

⇒$\frac{10}{5}-\frac{y}{2}=0$

⇒$2-\frac{y}{2}=0$

⇒$\frac{y}{2}=2$

⇒y=4

∴x=10 , y=4

(ii) 

⇒$\frac{x}{4}-\frac{y}{6}=3$...(i)

⇒$\frac{1}{2} x-y=-2$...(ii)

Multiplying (i) 1 and (ii) by $\frac{1}{6}$ , we get

⇒$\frac{x}{4}-\frac{y}{6}=3$

⇒$\frac{1}{12} x-\frac{y}{6}=\frac{-2}{6}=\frac{-1}{3}$

On Subtracting, we get

⇒$\frac{x}{4}-\frac{x}{12}=3+\frac{1}{3}=\frac{10}{3}$

⇒$\frac{3 x-x=40}{12}$

⇒2x=40

⇒$x=\frac{40}{2}$=20

From (ii)

⇒$\frac{20}{2}-y$

⇒-2=10-y=-2

⇒y=10+2=12

∴x=20 , y=12

Question 13

(a) $\frac{7}{x}+\frac{8}{y}=2 \text{ , } \frac{2}{x}+\frac{12}{y}=20$

(b) $\frac{1}{7x}+\frac{1}{6y}=3 \text{ , }\frac{1}{2x}-\frac{1}{3y}=5$

Sol :

(i)

$\frac{7}{x}+\frac{8}{y}=2$...(i)

$\frac{2}{x}+\frac{12}{4}=20$....(ii)

Multiplying (i) by 3 and (ii) by 2,  we get

$\frac{21}{x}+\frac{24}{y}=6$

$\frac{4}{x}+\frac{24}{y}=40$

Subtracting we get

⇒$\frac{17}{x}=-34$

⇒$x=\frac{17}{-34}=\frac{-1}{2}$

From (i)

⇒$\frac{-7 \times 2}{1}+\frac{8}{y}=2$

⇒$-14+\frac{8}{y}=2$

⇒$\frac{8}{y}=2+14=16$

⇒$y=\frac{8}{16}=\frac{1}{2}$

∴$x=\frac{-1}{2}, y=\frac{1}{2}$

(ii) 

$\frac{1}{7 x}+\frac{1}{6 y}=3$...(i)

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

Multiplying (i) by 1 and (ii) by $\frac{1}{2}$ , we get

$\frac{1}{7 x}+\frac{1}{6 y}=3$

$\frac{1}{4 x}-\frac{1}{6y}=\frac{5}{2}$

On adding we get

⇒$\frac{1}{7 x}+\frac{1}{4 x}=3+\frac{5}{2}$

⇒$\frac{4x +7x}{7 x \times 4x}=\frac{6+5}{2}$

⇒$\frac{11x}{28x^2}=\frac{6+5}{2}$

⇒$\frac{11}{28x}=\frac{11}{2}$

⇒$\frac{1}{14x}=\frac{1}{1}$

⇒14x=1

⇒$x=\frac{1}{14}$

From (i)

$\frac{1}{7 \times \frac{1}{4}}+\frac{1}{6 y}=3$

$\frac{14}{7}+\frac{1}{6 y}=3$

$=2+\frac{1}{6 y}=3$

$=\frac{1}{6 y}$=3-2=1=6y=1

$y=\frac{1}{6}$

∴$x=\frac{1}{14}, y=\frac{1}{6}$

Question 14

(a) 65x – 33p = 97, 33x – 65y = 1

(b) 23x + 31y = 77, 31x + 23y = 85

Sol :

(i) 

65x-33y=97...(i)

33x-65y=1...(ii)

So, we adding and get

98x-98y=98 [dividing both side by 98]

x-y=1...(iii)

Subtracting (2) from (1)

32x+32y=96 [dividing both sides by 32]

x+y=3...(iv)

Adding (iii) and (iv)

2x=4

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

and subtracting (iii) from (iv)

2y=2

$y=\frac{2}{2}=1$

(ii)

23x+31y=77...(i)

31x+23y=85...(ii)

Adding  we get

54x+54y=162...(iii)

x+y=3  [Dividing by 54]

and subtracting (i) from (ii)

8x-8y=8 [Dividing both sides by 8]

x-y=1...(iv)

Now adding (iii) and (iv) , we get

2x=4

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

and subtracting (iv) and (iii)

2y=2

$y=\frac{2}{2}=1$

∴x=2 ,  y=1

Question 15

(a) $4 x+\frac{6}{y}=15$ , $6 x-\frac{8}{y}=14$ ; y≠0

Find p, if y = px – 2

(b) $\frac{4}{x}+5y=7$ , $\frac{3}{x}+4y=5$ ; x≠0

Sol :

(i)

$4 x+\frac{6}{y}=15$....(i)

$6 x-\frac{8}{y}=14$...(ii)

Multiplying (i) by 4 and (2) by 3 , we get

$16 x+\frac{24}{y}=60$

$18 x-\frac{24}{y}=42$

Adding we get

34x=102

$x=\frac{102}{34}=3$

From (i)

$4 \times 3+\frac{6}{y}=15$

$=12+\frac{6}{y}=17$

$\frac{6}{y}=15-12$=3

3y=6

$y=\frac{6}{3}=2$

Hence , x=3 , y=2

Now, y=px-2

2=3p-2

3p=2+2=4

$p=\frac{4}{3}$

(ii) 

$\frac{4}{x}+5 y=7$...(i)

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

First we will multiplying (i) by 4 and (ii) by 5 , we get

$\frac{16}{x}+20 y=28$

$\frac{15}{x}+20 y=25$

On subtracting we get

$\frac{16}{x}-\frac{15}{x}=28-25$

$\frac{16-15}{x}=3$

$x=\frac{1}{3}$

From (i) 

$\frac{4}{\frac{1}{3}}+5 y=7$

$=\frac{4 \times 3}{1}+5 y=7$

=12+5y=7

=5y=7-12=-5

$=y=\frac{-5}{5}=-1$

$\quad 50 x=\frac{1}{3}$, y=-1

Question 16

(a) $\frac{6}{x+y}=\frac{7}{x-y}+3$ , $\frac{1}{2(x+y)}=\frac{1}{3(x-y)}$ , where x + y ≠ 0, x – y ≠ 0.

(b) $\frac{44}{x+y}+\frac{30}{x-y}=10$ , $\frac{55}{x+y}+\frac{40}{x-y}=13$ , where x + y ≠ 0, x – y ≠ 0.

Sol :

(a) $\frac{6}{x+y}=\frac{7}{x-y}+3$

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

Let x+y=a and x-y=b then

$\frac{6}{a}=\frac{7}{b}+3$

$\frac{6}{a}-\frac{7}{b}=3$...(i)

$\frac{1}{2 a}=\frac{1}{3 b}$

2a=3b...(ii)

$a=\frac{3}{2} b$

Substituting the value of a in (i)

$\frac{6}{\frac{3}{2} b}-\frac{7}{b}=3$

$\frac{12}{3 b}-\frac{7}{b}=3$

$\frac{4}{b}-\frac{7}{b}=3$

$\frac{-3}{b}=3$

$b=\frac{-3}{3}=-1$

∴$a=\frac{3}{2}$ $b=\frac{3}{2}(-1)=\frac{-3}{2}$

Now $x+y=\frac{-3}{2}$

x-y=-1

Adding we get

$2 x=\frac{-5}{2}$

$x=\frac{-5}{4}$

and subtracting 

$2 y=\frac{-1}{2}$

$y=\frac{-1}{2 \times 2}=\frac{-1}{4}$

So, $x=\frac{-5}{4}, y=\frac{-1}{4}$

(ii)

$\frac{44}{x+4}+\frac{30}{x-y}=10$

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

Hence x+y=a , x-y=b  then

$\frac{44}{a}+\frac{30}{b}=10$...(i)

$\frac{55}{a}+\frac{40}{b}=13$...(ii)

Multiplying (i) by 4 and (ii) by 3 , we get

$\frac{126}{a}+\frac{120}{b}=40$

$\frac{165}{a}+\frac{120}{b}=39$

On Subtracting 

$\frac{11}{a}=1$

a=11

From (i)

$\frac{44}{11}+\frac{30}{b}=10$

$4+\frac{30}{6}=10$

$\frac{30}{b}=10-4=\frac{6}{1}$

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

Now, x+y=11

x-y=5

Adding we get

2x=16

$x=\frac{16}{2}=8$

and subtracting 

2y=6

$y=\frac{6}{2}=3$

So , x=8 , y=3

Question 17

(i) 2 (3u – v) = 5uv, 2 (w + 3v) = 5uv

(ii) 3 (a + 3b) = 11ab, 3 (2a + b) = 7ab

Sol :

(i) 2 (3u – v) = 5uv

Let 6u-2v=5uv

Dividing by uv

$\frac{6}{v}-\frac{2}{v}=5$...(i)

and 2(u+3v)=5uv

2u+6v=5uv  (Dividing by uv both side)

$\frac{2}{v}+\frac{c}{u}=5$....(ii)

Multiplying both (i) by 3 and (ii) by 1

$\frac{18}{v}-\frac{6}{u}=15$

$\frac{2}{v}+\frac{6}{v}=5$

Adding both sides we get

$\frac{20}{v}=20$

$v=\frac{20}{20}=1$

From (i) 

$\frac{6}{1}-\frac{2}{u}=5$

$\frac{-2}{v}=5-6=-1$

u=2

Hence , u=2 , v=1

(ii) 3a+9b=11ab

$\frac{3}{b}+\frac{9}{a}=11$...(i)  (Dividing both ab)

3(2a+b)=7ab

6a+3b=7ab

$\frac{6}{b}+\frac{3}{a}=7$...(ii)  (Dividing both ab)

Multiplying both (i) by 1 and (ii) by 3

$\frac{3}{b}+\frac{9}{a}=11$

$\frac{18}{b}+\frac{9}{a}=21$

Subtracting , we get

$\frac{-15}{b}=-10$-10b=-15

$b=\frac{-15}{-10}=\frac{3}{2}$

From (i) equation $\frac{3}{3}+\frac{9}{a}=11$

$=\frac{3 \times 2}{3}+\frac{9}{a}=11$

$\Rightarrow 2+\frac{9}{a}=11$

$=\frac{9}{a}=11-2=9$

$=a=\frac{9}{9}=1$

∴we get a=1 , $b=\cdot \frac{3}{2}$

Question 18

$\frac{3 x-6}{4}+\frac{3}{1}-\frac{5 y-4}{2}=\frac{5 y}{2}$

$\frac{y-x}{4}+\frac{x}{8}-\frac{7x-5y}{3}=y-2x$

Sol :

$\frac{3 x-6}{4}+\frac{3}{1}-\frac{5 y-4}{2}=\frac{5 y}{2}$

$=\frac{3 x-6+12-10 y+8=10y}{4}$ [we take LCM of 24=4]

then

3x-10y-10=-12-8+6

3x-20y=-14...(i)

$=\frac{y-x}{4}+\frac{x}{8}-\frac{7 x-5 y}{3}=y-2 x$

$=\frac{6 y-6 x+3 x-56 x+40 y=24 y-48 x}{24}$

[We take LCM of 8,3=24]

=6y-6x+3x-56x+40y-24y+48x=0

=-6x+3x-56x+48x+6y+40y-24y=0

-62x+51x+46y+24y=0

-11x+22y=0

22y=11x

$x=\frac{22}{11}$ 

x=2y

From (i)

3x-20y=-14

3×2y-20y=-114

6y-20y=-14

-14y=-14

$y=\frac{-14}{-14}=1$

∴x=2y=2×1=2

Hence x=2  ,y=1

Question 19

x+y=0.9 , $\frac{11}{2(x+y)}=1$

Sol :

x+y=0.9$=\frac{9}{10}$

$\frac{11}{2(x+y)}=1$

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

By adding we get

$2 x=\frac{9}{10}+\frac{11}{2}$

$=\frac{9 \times 55}{10}=\frac{64}{10}$

$x=\frac{64}{10 \times 2}=\frac{32}{10}=3.2$

From (i)

x-y=0.9

3.2-y=0.9

-y=0.9-3.2

-y=-2.3

y=2.3

Hence x=3.2 , y=2.3

Question 20

$\frac{x}{2}+y=0.8$ , $\dfrac{7}{x+\frac{y}{2}}=10$

Sol :

$\frac{x}{2}+y=0.8$

$\frac{x}{2}+y=\frac{8}{10}$

5x+10y=8....(i)

$\frac{7}{x+\frac{y}{2}}=10$

$7=10\left[x+\frac{y}{2}\right]$

10x+5y=7

Multiplying (i) by 1 , (ii) by 2 and then subtracting

$\begin{aligned}5x+10 y=8\\20 x+10 y=14\\ \hline-15 x \quad \quad=-6 \end{aligned}$

$x=\frac{-6}{-15}=\frac{2}{5}$

x=0.4

From (i)

5×0.4+10y=8

2.0+10y=8

10y=8-2=6

$y=\frac{6}{10}=\frac{3}{5}$

Hence ,$x=\frac{2}{5}, y=\frac{3}{5}$

Question 21

If 2x +y = 35, 3x + 4y = 65, find the value of $\frac{x}{y}$

Sol :

2x+y=35...(i)

3x+4y=65...(ii)

Multiplying (i) by 4 and (ii) by 1 then subtracting

8x+4y=140

3x+4y=65

we get 5x=75

$x=\frac{75}{5}=15$

From (i) y=35-2x

=35-2×15

=35-30=5

∴$\frac{x}{y}=\frac{15}{5}=3$

(a) x + y = a – b

ax – by = a² + b²

Sol :

x+y=a-b

ax-by=a²+b²

Multiplying (i) by b , (ii) by 1

$b x+b y=a b-b^{2}$

$a x-b y=a^{2}+b^{2}$

On adding we get 

(a+b)x=a(a+b)

$x=\frac{a(a+b)}{(a+b)}=a$

From (i)

a+y=a-b

y=a-b-a=-b

Hence x=a , y=-b

 

(b) ax + by = a – b … (i)

bx – ay = a + b … (ii)

Sol :

ax+by=a-b

bx-ay=a+b

Multiplying (i) by a, (ii) by b

$\begin{aligned}a^{2} x+a b y=&a^{2}-a b \\b^{2} x-a b y=&a b+b^{2} \\ \hline (a^2+b^2)x=&a^2+b^2\end{aligned}$

On adding we get

$x=\frac{a^{2}+b^{2}}{a^{2}+b^{2}}=1$

From (i)

a×1+by=a-b

a+by=a-b

by=a-b-a=-b

$y=\frac{-b}{b}=-1$

Hence , x=1 , y=-1

Question 22

3 – 2(3x + 4y) = x, $\frac{x-3}{4}-\frac{y-4}{5}=2\frac{1}{10}$

Sol :

3-2(3x+4y)=x

3-6x-8y=-3 or

-6x-8y-x=-3

-7x-8y=-3 or 7x+8y=3

and $\frac{x-3}{4}-\frac{y-4}{5}=\frac{21}{20}$

$\frac{5 x-15-4 y+16=42}{20}$

5x-4y=42+15-16

5x-4y=41

Multiplying (i) by 1 (ii) by 12 then adding

$\begin{aligned}7 x+8 y=&3\\10 x-8 y=&82\\ \hline17 x=&85\end{aligned}$

$x=\frac{85}{17}=5$

From (i)

7×5+8y=3

35+8y=3

8y=3-35=-32

$y=\frac{-32}{8}=4$

∴x=5 , y=-4

Question 23

Can the following system of equations hold simultaneously?

$\frac{3}{x}+4y=7$ , $\frac{-2}{x}+7y=5$ , $5x+\frac{4}{y}=9$ . if yes, find x and y.

Sol :

$\frac{3}{x}+4 y=7$...(i)

$\frac{-2}{x}+7 y=5$...(ii)

$5 x+\frac{4}{y}=9$...(iii)

Multiplying (i) by 2 , (ii) by 3

$\frac{6}{x}+8 y=14$

$\frac{-6}{x}+21 y=15$

By adding we get

29y=29

$y=\frac{29}{29}=1$

From (i)

$\frac{3}{x}+4 \times 1=7$

$\frac{7}{x}+4=7$

$\frac{3}{x}=7-4=3$

$x=\frac{3}{3}=1$

∴x=1 , y=1

Substituting the value of x and y in (ii)

$5 \times 1+\frac{4}{1}=9$

5+4=9

9=9 is true

Yes, the system of equations are simultaneously and 

x=1 , y=1

Question 24

If the following three equations hold simultaneously for x and y, find p.

3x – 2y = 6, $\frac{x}{3}-\frac{y}{6}=\frac{1}{2}$ ,  x – py = 6

Sol :

3x-2y=6...(i)

$\frac{x}{3}-\frac{y}{6}=\frac{1}{2}$

2x-y=3....(ii)

Multiplying by 6

Multiplying (i) by 1 ,(ii) by 2 

3x-2y=6

4x-2y=6

On subtracting we get

-x=0

or x=0

From (i)

3×0-2y=6

-2y=6

$y=\frac{6}{-2}=-3$

∵The equations are simutaneousy

∴x=0, y=-3 will satisfy the third equation

∴x-py=6

0-p(-3)=6

3p=6

$p=\frac{6}{3}=2$

Hence p=2

Question 25

The sides of an equilateral triangle are (6x + 33y) cm, (8x + 9y – 5) cm and (10x + 12y – 8) respectively. Find the length of each side.

Sol :

∴Its all the sides are equal because its a equilateral triangle 

6x+3y=8x+9y-5

8x+9y-6x-3y=5

2x+6y=5...(i)

10x+12y-8x-9y=-5+8

2x+3y=3...(ii)

Multiplying (i) by 1 and (ii) by 2 On subtracting

$\begin{aligned}2 x+6 y=5\\4 x+6 y=6\\ \hline -2 x=-1 \end{aligned}$

$x=\frac{-1}{-2}=\frac{1}{2}$

2x+6y=5

$2 \times \frac{1}{2}+6 y=5$

1+6y=5

6y=5-1=4

$y=\frac{4}{6}=\frac{2}{3}$

Now 6x+3y

$=6 \times \frac{1}{2}+3 \times \frac{2}{3}$

=3+2=5

∴Each side of the triangle = 5 units