ML Aggarwal Simultaneous Linear Equations Chapter Test Class 9 ICSE Maths Solutions. Step by step solutions of Simultaneous Linear Equations Ch-Test problems as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Simultaneous Linear Equations Chapter Test Class 9 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-5 | Simultaneous Linear Equations |

Topics | Solution of Ch-Test Questions |

Academic Session | 2024-2025 |

### Solution of Ch-Test Questions

ML Aggarwal Simultaneous Linear Equations Chapter Test Class 9 ICSE Maths Solutions

**Solve the following simultaneous linear equations (1 to 4):**

**Question 1.**

**(i) 2x – (¾)y = 3,**

**5x – 2y = 7**

**(ii) 2(x-4) = 9y+2**

**x – 6y = 2**

**Answer :**

**(i) 2x-(3y/4)=3**

(8x-3y)/4=3

8x-3y = 12 …(i)

5x-2y = 7 …(ii)

Multiply (i) by 5 and (ii) by 8, we get

40x-15y = 60 (iii)

40x -16y = 56 (iv)

Subtract (iv) from (iii), we get

y = 4

Substitute y in (i)

8x-3×4 = 12

8x = 12+12

8x = 24

x = 24/8

x = 3

Hence x = 3 and y = 4.

**(ii) 2(x-4) = 9y+2**

**x – 6y = 2**

2(x-4) = 9y+2

2x-8 = 9y+2

2x-9y = 2+8

2x-9y = 10 ….(i)

x-6y = 2 ….(ii)

Multiply (ii) by 2, we get

2x -12y = 4 ….(iii)

Subtract (iii) from (i), we get

2x-9y = 10

-2x +12y = -4

0+3y = 6

3y = 6

y = 6/3

y = 2

Substitute the value of y in (i)

2x-9×2 = 10

2x-18 = 10

2x = 10+18

2x = 28

x = 28/2

x = 14

Hence x = 14 and y = 2.

**Question 2. **

**(i) 97x+53y = 177**

**53x+97y = 573**

**(ii) x+y = 5.5**

**x-y = 0.9**

**Answer :**

Given equations are as follows.

**(i) 97x+53y = 177 …(i)**

53x+97y = 573 …(ii)

Multiply (i) by 53 and (ii) by 97

53(97x+53y) = 53×177

5141x+2809y = 9381 …..(iii)

97(53x+97y) = 97×573

5141x+9409y = 55581 …..(iv)

Subtract (iv) from (iii)

5141x+2809y = 9381 …..(iii)

5141x+9409y = 55581 …..(iv)

——————————

0x -6600y = -46200

-6600y = -46200

y = -46200/-6600

y = 7

Substitute the value of y in (i)

97x+53×7 = 177

97x+371 = 177

97x = 177-371

97x = – 194

x = -194/97

x = -2

Hence x = -2 and y = 7.

**(ii) x+y = 5.5**

**x-y = 0.9**

x+y = 5.5 …(i)

x-y = 0.9 …(ii)

——————–

Adding (i) and (ii), we get

2x = 5.5+0.9

2x = 6.4

x = 6.4/2

x = 3.2

Substitute value of x in (i)

3.2+y = 5.5

y = 5.5-3.2

y = 2.3

Hence x = 3.2 and y = 2.3.

**Question 3. **

**(i) x+y = 7xy**

**2x-3y+xy = 0**

**(ii) 30/(x-y) …….**

**Answer :**

**(i) x+y = 7xy …(i)**

**2x-3y+xy = 0 …..(ii)**

Divide (i) by xy, we get

(x/xy)+(y/xy)=7xy/xy

(1/y)+(1/x)=7……(iii)

Divide (ii) by xy, we get

(2x/xy) -(3y/xy)+(xy/xy)=0

(2/y)-(3/x)+1=0

(-3/x)+(2/y)=-1…….(iv)

Multiplying (iii) by 3, we get

(3/x)+(3/y)=3×7

(3/x)+(3/y)=21……..(v)

Adding (v) and (iv), we get

5/y=20

y=5/20

y=1/4

Substitute value of y in (iv)

(-3/x)+2×4=-1

(-3/x)+8=-1

(-3/x)=-1-8

-3/x=-9

x=3/9

x=1/3

Hence x = 1/3 and y = 1/4.

**(ii)**

Substitute (v) in (i), we get

Now solve for (v) and (vi)

x+y = 11

x-y = 5

Add (v) and (vi)

2x = 16

x = 16/2 = 8

Substitute x in (v)

8+y = 11

y = 11-8

y = 3

Hence x = 8 and y = 3.

**Question 4. **

**(i) ax+by = a-b**

**bx-ay = a+b**

**(ii)** **3x + 2y = 2xy**

**(1/x)+(2/y)=7/6**

**Answer :**

**(i) ax+by = a-b …(i)**

bx-ay = a+b …(ii)

multiplying (i) by a and (ii) by b, we get

a(ax+by) = a(a-b)

a^{2}x +aby = a^{2}-ab …(iii)

b(bx-ay) = b(a+b)

b^{2}x -aby = ab+b^{2} …(iv)

Adding (iii) and (iv)

a^{2}x +aby = a^{2}-ab

b^{2}x -aby = ab+b^{2}

———————–

(a^{2}+b^{2})x = (a^{2}+b^{2})

x = (a^{2}+b^{2})/ (a^{2}+b^{2})

x = 1

Substitute the value of x in (i), we get

a×1+by = a-b

a+by = a-b

by = -b

y = -b/b

y = -1

Hence x = 1 and y = -1.

**(ii)** **3x + 2y = 2xy**

**(1/x)+(2/y)=7/6**

3x + 2y = 2xy …(i)

(1/x)+(2/y)=7/6….(ii)

Divide (i) by xy

(3x/xy)+(2y/xy)=2xy/xy

(3/y)+(2/x)=2……..(iii)

Multiply (ii) by 2, we get

(2/x)+(4/y)=7/3……(iv)

Subtract (iii) from (iv)

Substitute y in (iii)

(3/3) + (2/x) = 2

1+(2/x) = 2

(2/x) = 1

x = 2

Hence x = 2 and y = 3.

**Question 5. Solve 2x -(3/y) =9, ****3x + (7/y) = 2. ****Hence find the value of k if x = ky + 5.**

**Answer :**

2x -(3/y) = 9 …(i)

3x + (7/y) = 2 …(ii)

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

6x-(9/y) = 27 ..(iii)

6x+(14/y) = 4 …(iv)

Subtracting (iv) from (iii), we get

-23/y = 23

y = 23/-23

y = -1

Substitute y in (i)

2x-(3/-1) = 9

2x+3 = 9

2x = 9-3

2x = 6

x = 6/2

x = 3

Hence x = 3 and y = -1.

Given x = ky+5

Substitute x and y in above eqn

3 = k×-1+5

3 = -k+5

k = 5-3

k = 2

Hence the value of k is 2.

**Question 6.** **Solve, ****1/(x+y)-1/2x=1/30, ****5/(x+y)+1/3=4/3 ****Hence find the value of 2x**^{2}-y^{2}.

^{2}-y

^{2}.

**Answer :**

**1/(x+y)-1/2x=1/30….(i)**

**5/(x+y)+1/3=4/3………(ii)**

Let (x+y) = a

(1/a)-(1/2x)

1/a-1/2x=1/30……(iii)

5/a+1/x=4/3

Multiply (iii) by 5

5/a – 5/2x=1/6………(iv)

5/a + 1/x=4/3

Subtracting (ii) from (iv)

(-5/2x) -1/x=1/6 – 4/3

(-5-2)/2x=(1-8)/6

-7/2x =-7/6

2x =6

x=3

Substitute x in (iii)

(1/a) -1/(2×3) = 1/30

(1/a) – (1/6) = 1/30

1/a = (1/30)+(1/6)

1/a = (1+5)/30

1/a = 6/30

a = 30/6

a = 5

Substitute a in x+y = a

3+y = 5

y = 5-3

y = 2

Hence x = 3, y = 2.

2x^{2}-y^{2} = 2×3^{2}-2^{2}

= 2×9-4

= 18-4

= 14

Hence the value of 2x^{2}-y^{2} is 14.

**Question 7. Can x, y be found to satisfy the following equations simultaneously ?**

**(2/y)+(5/x)=19**

**(5/y)-(3/x)=1**

**3x+8y=5**

**If so, find them.**

**Answer :**

(2/y)+(5/x)=19…….(i)

(5/y)-(3/x)=1………..(ii)

3x+8y=5……..(iii)

Multiply (i) by 5 and (ii) by 2, we get

(10/y) +(25/x)=95……(iv)

(10/y)-6/x =2…….(v)

Subtract (v) from (iv)

31/x = 95-2

31/x = 93

x = 31/93

x = 1/3

Substitute x in (i)

(2/y)+5÷(1/3) = 19

(2/y)+5×3 = 19

(2/y) = 19-15

(2/y) = 4

y = 2/4

y = 1/2

Substitute x and y in (iii)

3×(1/3) + 8×(1/2) = 5

1+4 = 5

The value of x and y satisfies (iii).

Hence the given equations are simultaneous.

— : End of ML Aggarwal Simultaneous Linear Equations Chapter Test Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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