1. Home
2. /
3. Blog
4. /
5. Uncategorised
6. /
7. Practice set 1.1, Algebra,...

# Practice set 1.1, Algebra, ssc,

21 Dec 2020 6:51 pm

Practice set 1.1

Complete the following activity to solve the simultaneous equations.

1. 5x + 3y = 9 —–(I)
2x + 3y = 12 —(II)

#### SOLUTION

Disclaimer: There is error in the Q. In (II) there should have been 2x – 3y = 12
5x + 3y = 9 —–(I)
2x – 3y = 12 —– (II)
7x = 21
x = 3
Putting the value of x = 3 in (I) we get
5(3)+3y=9
⇒15+3y=9
⇒3y=9−15=−6
⇒y=−2
Thus, (x, y) = (3, -2)

SOLVE THE FOLLOWING SIMULTANIOUS EQUATIONS

1. 3a + 5b = 26; a + 5b = 22

Solution

3a + 5b = 26                                  …..(I)
a + 5b = 22                                    …..(II)
Subtracting (II) from (I)
2a = 4

⇒  a = 2
Putting the value of a = 2 in (II)
5b = 22 – 2 = 20

⇒ b = 205=4
Thus, a = 2 and b = 4.

2. x + 7y = 10; 3x – 2y = 7

#### SOLUTION

x + 7y = 10;                             …..(I)
3x – 2y = 7                               ….(II)
Multiplying (I) with 3
3x + 21y = 30;                         …..(III)
3x – 2y = 7                               ….(IV)
Subtracting (IV) from (III), we get,
23y = 23
y = 1
Putting the value of y in (IV) we get
3x – 2 = 7
⇒ 3x = 7 + 2 = 9
⇒ 3x = 9
⇒ x = 3
Thus, (x, y) = (3, 1)

3. 2x – 3y = 9                         …..(I)
2x + y = 13                       …..(II)

Subtracting (II) from (I) we get
– 3y − y = 9 − 13
⇒−4y=−4
⇒y=1
Putting this value in (I) we get
2x−3(1)=9
⇒2x=9+3=12
⇒x=122=6
Thus, (x, y) = (6, 1)

4. 5m – 3n = 19; m – 6n = –7

Solution

5m – 3n = 19               …..(I)
m – 6n = –7                      …..(II)
Multiplying (I) with 2 we get
10m – 6n = 38               …..(III)
Subtracting (II) from (III) we get
10m−m−6n−(−6n)=38−(−7)
⇒9m=45
⇒m=459=5
Putting the value of m = 5 in (II) we get
5−6n=−7
⇒−6n=−7−5
⇒−6n=−12
⇒n=−12−6=2
Thus, (m, n) = (5, 2).

5. 5x + 2y = –3; x + 5y = 4

Solution

5x + 2y = –3                      …..(I)
x + 5y = 4                                …..(II)
Multiply (II) with 5 we get
5x + 25y = 20                          …..(III)
Subtracting (III) from (I) we get
5x−5x+2y−25y=−3−20
⇒−23y=−23
⇒y=−23−23=1
Putting the value of y = 1 in (II) we get
x+5(1)=4
⇒x+5=4
⇒x=4−5=−1
Thus, (x, y) = (−1, 1)

6.

Solution

7. 99x + 101y = 499; 101x + 99y = 501

Solution

99x + 101y = 499                       …..(I)

101x + 99y = 501                       …..(II)

200x + 200y = 1000

or, x + y = 5                              ……(III)

Subtracting (1) from (2), we get,

2x – 2y = 2

Or, x – y = 1                             …..(IV)

Adding (3) and (4), we get,

2x = 6

⇒ x = 3

Putting the value of x in (3), we get,

3 − y = 1

y = 2

∴ y = 2, x = 3.

8. 49x – 57y = 172; 57x – 49y = 252

Solution

49x – 57y = 172                         …..(I)
57x – 49y = 252                              …..(II)
49x+57x−57y−49y=172+252
⇒106x−106y=424
⇒x−y=4…..(III)
Subtracting (II) from (I) we have
49x−57y−57y−(−49y)=252−172
⇒−8x−8y=−80
⇒−x−y=−10
⇒x+y=10…..(IV)
x−y=4
x+y=10
⇒2x=14
⇒x=7
Putting the value of x = 7 in (IV) we get
7+y=10
⇒y=10−7
⇒y=3
Thus, (x, y) = (7, 3)

#### VISITORS COUNT

Users Today : 148
Total Users : 364229
Views Today : 400
Total views : 1257972