Practice set 5.1 9th Standard Linear Equations in Two Variables 9th Standard

Table Of Contents

Question (1) By using variables x and y form any five linear equations in two variables.
Question (2) Write five solutions of the equation x+y=7
Question (3) Solve the following sets of simultaneous equations.

Question (1) By using variables x and y form any five linear equations in two variables.

x + y = 1
2x + 2y = 1
3x + 3y = 1
4x + 4y = 1
5x + 5y = 1

Like this you can write any number ( excluding zero ) this will become a Linear Equation in two variables.

Question (2) Write five solutions of the equation x+y=7

x+ y = 7 …(i)
1st solution : Let x = 1 then y = ?
Substituting x= 1 in equation number (i) we get,
x + y = 7
1 + y = 7
y = 7 – 1 …( here, +1 become -1 )
y= 6
hence 1st solution x = 1, y = 6.

x+ y = 7 …(i)
2nd solution : Let x = 2 then y= ?
Substituting x= 2 in equation number (i) we get,
x + y = 7
2 + y = 7
y = 7 – 2 … ( here, +2 become -2 )
y= 5
hence 2nd solution x = 2, y = -1

x+ y = 7 …(i)
3rd solution : Let x = 3 then y= ?
Substituting x= 3 in equation number (i) we get,
x + y = 7
3 + y = 7
y = 7 – 3 …( here, +3 become -3 )
y= 4
hence 3rd solution x = 3, y = 4

x+ y = 7 …(i)
4th solution : Let y = 1 then x = ?
Substituting y= 1 in equation number (i) we get,
x + y = 7
x + 1 = 7
x = 7 – 1 …( here, +1 become -1 )
x = 6
hence 4th solution x = 6, y = 1

x+ y = 7 …(i)
5th solution : Let y = 2 then x = ?
Substituting y= 2 in equation number (i) we get,
x + y = 7
x + 2 = 7
x = 7 – 2 …( here, +2 become -2 )
x = 5
hence 5th solution x = 5, y = 2.

Question (3) Solve the following sets of simultaneous equations.

(i) x + y = 4 ; 2x – 5y = 1

Solution: x + y = 4 … (i)
2x – 5y = 1 … (ii)
Now, multiplying equation number (i) by 5 we get,
∴ 5×x + 5×y = 4×5
∴ 5x + 5y = 20 … (iii)
Now, Adding equation (ii) and equation (iii) we get,
∴ 2x – 5y = 1
+ 5x + 5y = 20
————————
7x = 21 ( explanation -5y +5y = 0 )
x = 21÷7
∴. x = 3
∴ Substituting x= 3 in equation ( i ) we get,
x + y = 4
3 + y = 4
y = 4 – 3 …( +3 become -3 )
y = 1
Answer: x=3 and y = 1 or ( x, y ) = (3,1) is the solution of the given Equations.

(ii) 2x + y = 5; 3x-y = 5

Solution: 2x +y = 5 ….(i)
3x – y = 5 …(ii)
Adding equation number (i) and (ii), we get,
2x + y = 5 …(i)
3x – y = 5 …(ii)
———————————
5x. = 10 … ( explanation 2x + 3x = 5x )
∴x =10÷5
∴ x = 2 ….( explanation 5 two ja 10 )
Substituting x=2 in equation number (i) , we get
2x + y = 5
∴ 2 ×2 + y = 5
∴4 + y = 5
∴ y = 5-4
∴ y= 1
Answer : x = 2 and y = 1 or (x, y) = (2, 1) is the solution of the given Equations.

(iii) 3x – 5y =16; x – 3y = 8

Solution: 3x – 5y=16 …(i)
x – 3y = 8 …(ii)
multiplying equation (ii) by 3, we get.
3×x – 3×3y = 8×3
3x – 9y = 24 …(iii)
Subtracting equation (iii) from equation (i),
3x – 5y = 16 …. ( i)
3x – 9y = 24. …(ii)
———————
4y = -8 …. ( Explanation +3x and -3x = 0 )
∴ y= -8÷4
∴ y = -2 ….(Explanation 4 two ja 8 )
Substituting y = -2 in equation number (ii), we get
∴x – 3y = 8
∴x – 3 × (-2) = 8
∴x + 6 = 8 ( explanation 3 two ja 6 ) ( – multiple by – become + )
∴ x = 8 – 6
∴ x = 2
Answer : x = 2 and y= -2 or (x,y) = ( 2, -2) is the solution of the given Equations.

(iv) 2y – x = 0 ; 10x + 15y = 105

Solution: 2y – X = 0 …. (i)
10x + 15y = 105. … ( ii)
in equation number ( ii ), Dividing both the sides by 5, we get
∴ 2x + 3y = 21… (iii) … ( explanation 105 ÷ 5 = 21 )
Multiplying equation number (i) by 2
∴ 4y – 2x = 0
∴ i.e. -2x + 4y = 0 ….(iv)
Adding equation number ( iii) and ( iv )
2x + 3y = 21 … (iii)
-2x + 4y = 0 ….(iv)
————————————
7y = 21 ……. ( explanation: 3y + 4y = 7y )
∴ y = 21 ÷ 7
∴ y = 3
Substitution y = 3 in equation number (i)
2y – X = 0 …. (i)
2× 3 – X = 0
∴ 6 – X = 0
∴6 = x
∴ x = 6
x= 6 and y = 3 or ( x,y) = ( 6,3 ) is the solution of the given Equations.

(v) 2x + 3y + 4 = 0 ; x – 5y = 11

2x + 3y + 4 = 0 i.e
2x + 3y = –4 …. (i)
x – 5y = 11 … (ii)
Multiply equation number (ii) by 2, we get
∴2×x – 2×5y = 11×2
∴2x – 10y = 22 …(iii)
Subtracting equation number (iii) by eq (i)
2x – 10y = 22 …(iii)
2x + 3y = –4 …. (i)
– – +
—————————————
0 – 13y = 26
∴Y = 26 ÷ ( – 13 )
∴Y= –2
Substituting y = -2 in equation number (ii)
x – 5y = 11
x – 5×2 = 11
∴x – 10 = 11
∴X = 11 – 10
∴x= 1
Answer : x = 1 and y = -2 OR (x,y) = (1, -2) is the solution of the given equations.

(vi) 2x – 7y = 7 ; 3x + y = 22

Solution: 2x – 7y = 7 …… (i)
3x + y = 22 … (ii)
Multiplying equation number (ii) by 7, we get
7×3x + 7×y = 7×22
21x + 7y = 154 …. (iii )
Adding equation number (i) and equation number (iii)
2x – 7y = 7 …… (i)
21x + 7y = 154 …. (iii )
—————————————
23x = 161 ( explanation +7y -7y = 0 )
X = 161 ÷ 23
X = 7 ( explanation 23 seven ja 161 )
Substitute X = 7 in equation number (i)
2x – 7y = 7 …… (i)
2×7 – 7y = 7
14 – 7y = 7
-7y = 7 – 14
-7y = – 7
7y = 7
y= 7 ÷ 7
Y = 1
Answer: x = 7 and y = 1 or (x,y) = ( 7,1) is the solution of the given equations.