Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5B Goyal Brothers ICSE Maths Solutions

Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5B Goyal Brothers ICSE Maths Solutions Ch-5. In this article you will learn how to Solve by Criss Cross Method easily. Visit official Website CISCE for detail information about ICSE Board Class-9.

Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5B Goyal Brothers ICSE Maths Solutions Ch-5

Board	ICSE
Subject	Maths
Class	9th
Chapter-5	Simultaneous Linear Equations
Writer	RS Aggrawal
Topics	Solving Simultaneous Linear Equations by Criss Cross Method
Academic Session	2024-2025

Solving Simultaneous Linear Equations by Criss Cross Method

The equations are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

Page- 76

Exercise- 5B

(Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5B Goyal Brothers ICSE Maths Solutions Ch-5)

Solve each of the following systems of equations by using the method of cross multiplication :

Que-1: 2x-5y+8 = 0, x-4y+7 = 0

Sol: 2x – 5y + 8 = 0………..(i)
x – 4y + 7 = 0………….(ii)
Multiplying by 2 to (ii)
2x – 8y = -14…….(iii)
subtracting (ii) from (i)
-5y + 8y = -8 + 14
3y = 6
y = 2 and
2x – 5 × 2 + 8 = 0
2x – 10 + 8 = 0
2x – 2 = 0
x = 1

Que-2: 5x-4y+2 = 0, 2x+3y = 13

Sol: 5x-4y+2 = 0 …….(i)
2x+3y = 13 ……….(ii)
Multiply the first equation by 3 and the second equation by 4 to make the coefficients of y equal in magnitude:
3(5x−4y+2) = 0×3
4(2x+3y) = 13×4
This gives:
15x−12y+6 = 0
8x+12y = 52
Now, add the two equations together to eliminate y :
(15x−12y+6) + (8x+12y) = 0+52
15x+8x+6 = 52
23x+6 = 52
23x = 46
x = 2
Put the value of x in eq(ii), we get
2x+3y = 13
2(2)+3y = 13
4+3y = 13
3y = 13-4
y = 9/3
y = 3.

Que-3: 3x-5y=19, 7x-3y=1

Sol: The given equations are.
3x–5y–19 = 0 ————(1)
−7x+3y+1 = 0 ————(2)
Multiplying (1) by 3 and (2) by 5, we get,
9x–15y = 57——–(3)
−35x+15y = −5 ———(4)
Adding (3) and (4), we get
−26x = 52 ⇒ x = −2
Substituting x = −2 in (1), we get
3(−2)–5y = 19
⇒ −6–5y = 19
−5y = 19+6
⇒ −5y = 25
⇒ y = −5
x = −2 and y = −5

Que-4: 2x+3y=17, 3x-2y=6

Sol: Given equations: 2x + 3y = 17 and 3x – 2y = 6
a1 = 2, b1 = 3, c1 = −17 and a2 = 3, b2 = −2, c2 = −6
By cross multiplication method
x = (b1c2−b2c1)/(a1b2−a2b1) &
y = (c1a2−c2a1)/(a1b2−a2b1)
Where a1x + b1y + c1 = 0 & a2x + b2y + c2 = 0
are 2 lines
⇒ x = [3(−6)−(−2)(−17)]/[2(−2)−(3×3)]
= −52/−13 = 4
y = [(−17)3−(−6)2]/[2(−2)−(3×3)]
= −39/−13 = 3
⇒ x = 4 & y = 3

Que-5: x+2y+1=0, 2x-3y=12

Sol: x+2y+1 = 0……….(i)
2x−3y−12 = 0……….. (ii)
Here a1 = 1, b1 = 2, c1 = 1
a2 = 2, b2 = −3, c2 = −12
By cross multiplication method,
[x/(−24+3)] = [−y/(−12−2)] = [1/(−3−4)]
(x/−21) = (−y/−14) = (1/−7)
Now,
(x/−21) = (1/−7)
= x = 3
And,
(−y/−14) = (1/−7)
= y = −2

Que-6: 2x+5y=1, 2x+3y=3

Sol: 2x+5y = 1—-(1)
2x+3y = 3—-(2)
No need for cross multiply because coefficients of x are already same
(1) – (2)
2x−2x+5y−3y = 1−3
0x+2y = −2
y = −1
so
2x+5(−1) = 1 (from first equation)
2x−5 = 1
2x = 6
x = 3.

Que-7: 8x-3y=12, 5x=2y+7

Sol: 8x – 3y – 12 = 0 → (1)
5x – 2y – 7 = 0 → (2)
Use the coefficients for cross multiplication
= [x/{21-(+24)}] = [y/{-60-(-56)}] = [1/{-16-(-15)}]
= [x/(21-24)] = [y/(-60+56)] = [1/(-16+15)]
= (x/-3) = (y/-4) = (1/-1) = -1
= (x/-3) = -1
x = 3.
(y/-4) = -1
y = 4.

Que-8: 7x-2y=20, 11x+15y+23-0

Sol: Consider the given equations.
7x−2y = 20 ………(1)
11x+15y = −23 ……..(2)
From equation (1), we get
⇒ y = (7x−20)/2
On putting value of y in equation (2), we get
⇒ 11x+15[(7x−20)/2] = −23
⇒ 11x+(105x/2)−150 = −23
⇒ (127x/2) = 127
⇒ x = 2
Therefore,
⇒ y = [7×2−20]/2
⇒ y = −3
Hence, the value of x and y is 2,−3.

Que-9: ax+by=(a-b), bx-ay=(a+b)

Sol: ax+by = a−b ………. (1)
bx−ay = a+b ………. (2)
Multiply equation (1) with a and equation (2) with b ,
a²x+aby = a²−ab ……….(3)
b²x−aby = ab+b² ……… (4)
Adding (3) and (4), we get,
(a²+b²)x = a²+b²
⇒ x = 1
Now, from (1),
ax+by = a−b
a+by = a−b
by = −b
y = −1

Que-10: 3x+2y+25=0, 2x+y+10=0

Sol: Given equations are,
3x+2y+25 = 0…………….(i)
2x+y+10 = 0……………….(ii)
From (ii), we can get
y = −2x−10
Substitute y value in equation (i) , we get
3x+2(−2x−10)+25 = 0
⇒ 3x−4x−20+25 = 0
⇒ −x+5 = 0
∴ x = 5
Hence,
y = −2x−10
= −2(5)−10
= −10−10
y = −20
Solution is (x,y) = (5,−20)

Que-11: (5/x)-(4/y)+2=0, (2/x)+(3/y)=13 (x≠0, y≠0)

Sol: (5/x)-(4/y)+2 = 0
(2/x)+(3/y) = 13
Let 1/x be ′a′ and 1/y be ′b′
We get our equations as
2a+3b = 13−−−−−−(i)
5a−4b = −2−−−−(ii)
Multiplying the (i) by 4 and the (ii) by 3, we get
8a+12b = 52−−−−(iii)
15a−12b = −6−−−(iv)
Now adding the two equations, we get
23a = 46
⇒ a = 2
Substituting a = 2 in the (i), we get
2 × 2 + 3b = 13
3b = 9
⇒b = 3
Earlier we have assumed that a = 1/x & b = 1/y
Therefore, x = 1/2 and y = 1/3.

Que-12: (1/x)+(1/y)=7, (2/x)+(3/y)=17 (x≠0, y≠0)

Sol: 1/x+1/y = 7 —–(1)
2/x+3/y = 17 ——(2)
let 1/x = u and 1/y = v
so equation are written as
u+v−7 = 0…..(3)
2u+3v−17 = 0….(4)
by cross multiplication
[u/(−17+21) = [v/(−14+17) = 1/(3−2)
u/4 = v/3 = 1/1
u/4 = 1/1 ⇒ u = 4
and
v/3 = 1/1 ⇒ v = 3
But 1/x = u
1/x = 4
so x = 1/4
and
1/y = v
1/y = 3
so y = 1/3
x = 1/4 and y = 1/3

Que-13: [10/(x+y)]+[2/(x-y)]=4, [15/(x+y)]-[5/(x-y)]+2=0, where x≠y and x≠-y

Sol: [10/(x+y)] + [2/(x-y)] = 4
[15/(x+y)] – [5/(x-y)] + 2 = 0
Let [1/(x+y)] = p and [1/(x-y)] = q
The given equations reduce to:
10p + 2q = 4
=> 10p + 2q – 4 = 0 ….(1)
15p – 5q = -2
=> 15p – 5q + 2 = 0 …(2)
Using cross-multiplication method, we obtain:
[p/(4-20)] = [q/(-60-20)] = [1/(-50-30)]
p/-16 = q/-80 = 1/-50
p = 1/5 and q =1
p = 1/(x + y) = 1/5 and q = [1/(x-y)] = 1
x + y = 5 …..(3)
x – y = 1 ….(4)
Adding equation (3) and (4), we obtain:
2x = 6
x = 3
Substituting the value of x in equation (3), we obtain:
y = 2
∴ x = 3, y = 2

Que-14: [5/(x+1)]-[2/(y-1)]=1/2, [10/(x+1)]+[2/(y-1)]=5/2, where x≠-1 and x≠1

Sol: [5/(x+1)]-[2/(y-1)] = 1/2
[10/(x+1)]+[2/(y-1)] = 5/2
Let [1/(x+1)] = u and [1/(y-1)] = v
Then, the given system of equations becomes
5u – 2v = 1/2 ….(i)
10u + 2y = 5/2 ……(ii)
Adding equation (i) equation (ii), we get
5u + 10u = (5/2) + (1/2)
15u = 6/2
u = 6/30 = 1/5
Putting u = 1/5 in equation (i) we get
5(1/5)-2v = 1/2
1-2v = 1/2
1-(1/2) = 2v
1/2 = 2v
v = 1/4
Now, u = [1/(x+1)]
[1/(x+1)] = 1/5
x+1 = 5
x = 4
And v = [1/(y-1)]
[1/(y-1)] = 1/4
y-1 = 4
y = 5.