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 :
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 :
⇒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 :
⇒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 :
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 :
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-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 :
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 :
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 :
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 :
⇒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)
$\frac{x}{5}-\frac{y}{2}=0$...(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 :
$\frac{2}{x}+\frac{12}{4}=20$....(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 :
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 :
$6 x-\frac{8}{y}=14$...(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) $\frac{6}{x+y}=\frac{7}{x-y}+3$
$\frac{1}{2 a}=\frac{1}{3 b}$
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 :
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
$\frac{3 x-6}{4}+\frac{3}{1}-\frac{5 y-4}{2}=\frac{5 y}{2}$
Sol :
$=\frac{3 x-6+12-10 y+8=10y}{4}$ [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 :
$\frac{11}{2(x+y)}=1$
$x+y=\frac{11}{2}$
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
Sol :
$\frac{x}{2}+y=\frac{8}{10}$
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 :
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
∴$\frac{x}{y}=\frac{15}{5}=3$
Sol :
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 :
$x=\frac{a^{2}+b^{2}}{a^{2}+b^{2}}=1$
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 :
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?
Sol :
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 :
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
No comments:
Post a Comment