- Solve the following system of equations by the method of substitution:
(i) x + y = 10
x – y = 12
(ii)3x – 7y = 7
11x + 5y = 87
(iii) 3x – 4y = 10
4x + 3y = 5
(iv) 3a – 2b = 12
4a – 5b = 16
(v) 5x + 4y – 4 = 0
x – 20 = 12y
(vi) 7x + 5y = 10
3x + y = 2
(vii) 2p + 3q = -5
3p – 2q = 12
(viii) 2x + 3y = 5
3x + 4y = 6
Solution
(i) x + y = 10…………….(1)
x – y = 12 ……………..(2)
from (1), we have, x = 10 – y …………..(3)
Substitute (3) in (2), we get,
10 – y – y = 12
10 – 2y = 12
-2y = 12 – 10 = 2
y = –2/2 = -1
Since x = 10 – y , we have y = – 1,
Therefore, x = 10 – (-1) = 10 + 1 = 11
Hence, x = 11 and y = -1
(ii)3x – 7y = 7 ………………….(1)
11x + 5y = 87……………………(2)
From (1), x = 7+7y/3……………….(3)
Substitute (3) in (2), we get,
11(7+7y/3) + 5y = 87
3[11(7+7y/3) + 5y = 87]
11(7 + 7y) + 15y = 261
77 + 77y + 15y = 261
+92y = 261 – 77 = 184
92 y = 184
y = 2.
Since, x = 7+7y/3 , we have y = 2, then,
x = 7+7(2)/3 = 7+14/3 = 21/3 = 9
Thus, x = 9 and y = 2.
(iii) 3x – 4y = 10……………(1)
4x + 3y = 5…………….(2)
From (1), x = 10+4y/3 …………..(3)
Substitute (3) in (2), we get
4(10+4y/3) + 3y = 5
3[4(10+4y/3) + 3y = 5]
4(10 + 4y) + 9y = 15
40 + 16y + 9y = 15
40 + 25y = 15
25y = 15 – 40 = -25
y = -1
Since, x = 10+4y/3 and we have y = -1, then
x = 10+4(-1)/3= 10-4/3 = 6/3 = 2
Therefore, x = 2 and y = -1
(iv) 3a – 2b = 12…………….(1)
4a – 5b = 16………………..(2)
From (1), we have a = 12 + 2b/3 ………..(3)
Substitute (3) in (2), we get
4(12 + 2b/3) – 5b = 16
3[4(12 + 2b/3) – 5b = 16]
4(12+2b) – 15b = 48
48 + 8b – 15b = 48
-7b = 0
b= 0
Since a = 12 + 2b/3 and b = 0
Then, a = 12+0/3 = 12/3 = 4
Therefore , a = 4 and b = 0
(v) 5x + 4y – 4 = 0⇒5x + 4y = 4….……..(1)
x – 20 = 12y⇒x – 12y = 20………..(2)
From (1), we have, x = 4-4y/5 …………..(3)
Substitute (3) in (2)
4-4y/5 – 20 = 12y
(4-4y/5 – 20 = 12y)x5
4 – 4y – 100 = 60y
-96 = 60y + 4y = 64y
y = -96/64 = -3/2
Since, y = –3/2 then x = 4-4y/5 = [4 – 4(-3/2)]/5 = 4 + 6/5 = 10/5 = 2
Therefore, x = 2 and y = -3/2
(vi) 7x + 5y = 10 …………..(1)
3x + y = 2………………….(2)
From (1)
x = 10-5y/7 …………………..(3)
Substitute (3) in (2)
3(10-5y/7 ) + y = 2
[3(10-5y/7 ) + y = 2]x7
= 3(10 – 5y) + 7y = 14
30 – 15y + 7y = 14
-8y = 14 – 30 = – 16
8y = 16
y = 2
Since y = 2 then x = 10-5(2)/7 = 10-10/7 = 0
(vii) 2p + 3q = -5……………..(1)
3p – 2q = 12………………….(2)
From (1), p = -5-3q/2 …………(3)
Substitute (3) in (2)
3(-5-3q/2 ) – 2q = 12
[3(-5-3q/2 ) – 2q = 12]x2
3(-5-3q) – 4q = 24
-15 – 9q – 4q = 24
-13q = 24 + 15 = 39
q = 39/-13 = -3
Since q = -3 we have, p = -5-3(-3)/2 = -5+9/2 = 4/2 = 2
Therefore q = -3 and p = 2
(viii) 2x + 3y = 5 ……………….(1)
3x + 4y = 6 ……………………(2)
From (1), x = 5-3y/2 ……………(3)
Substitute (3) in (2), we get,
3(5-3y/2 ) + 4y = 6
[3(5-3y/2 ) + 4y = 6]x2
3(5 – 3y) + 8y = 12
15 – 9y + 8y = 12
-y = 12 – 15 = -3
y = 3
Since y = 3, we have x = 5-3y/2 = 5-3(3)/2 = 5-9/2 = –4/2 = -2
Therefore, y = 3 and x = -2
- Use elimination method for solving the following equations:
(i) 3x + 2y = 0
x + 3y = 7
Solution:
(3x + 2y = 0)x1⇒ 3x + 2y = 0
(x + 3y = 7)x3 ⇒ 3x + 9y = 27
(-) (-) (-)
—————————-
-7y = -27
y = 3
Substituting y = 3 in equation (1)
3x + 2×3 = 0
3x + 6 = 0
3x = -6
x = -2
Therefore, x = -2 and y = 3
(ii) x + y = 3
3x – y = 5
Solution:
x + y = 3……………..(1)
3x – y = 5……………(2)
(1)x3 ⇒ 3x + 3y = 9
(2)x1 ⇒ 3x – y = 5
(-) (+) (-)
—————————-
4y = 4
y = 1
Substitute the value of y in eq(1)
x + y = 3
x = 3 – 1 = 2
Therefore, x = 2 and y = 1
(iii) x – y = 7
3x + y = 10
Solution:
x – y = 7 …………………(1)
3x + y = 10……………….(2)
1)x3 ⇒ 3x – 3y = 21
(2)x1 ⇒ 3x + y = 10
(-) (-) (-)
——————————–
-4y = 11
y = –11/4
Substitute the value of y in (1)
x – y = 7
x – (-11/4) = 7
x = 7 – 11/4 = 28-11/4 = 17/4
Therefore, x = 17/4 and y = -11/4
(iii) 3x + y = 10
x–y = 2
Solution:
3x + y = 10 ………………….(1)
x–y = 2…………………………(2)
1)x1 ⇒ 3x + y = 10
(2)x3 ⇒ 3x – 3y = 6
(-) (+) (-)
——————————–
4y = 4
y =1
Substitute the value of y in (2)
x–1 = 2
x = 2 + 1 = 3
Therefore x = 3 and y = 1
(v) 2x + y = 0
3x – y = -5
Solution:
2x + y = 0 ………………..(1)
3x – y = -5 ………………..(2)
(1)x3 ⇒ 6x + 3y = 0
(2)x2 ⇒ 6x – 2y = – 10
(-) (+) (+)
———————————–
5y = 10
y =2
Substitute the value of y in (1)
2x + y = 0
2x + 2 = 0
2x = – 2
x = -1
Therefore, x = -1 and y = 2
(vi) 2x + y = 7
2x – 3y = 3
Solution:
2x + y = 7 ……………(1)
2x – 3y = 3……………(2)
(-) (+) (-)
———————–
4y = 4
y = 1
Substitute the value of y in (1)
2x + y = 7
2x + 1 = 7
2x = 7 – 1 = 6
x = 3
Therefore, x = 3 and y = 1
(vii) 100x + 200y = 700
200x + 100y = 800
Solution:
100x + 200y = 700………….(1)
200x + 100y = 800………….(2)
(1)x2 ⇒ 200x + 400y = 1400
(2)x1 ⇒ 200x + 100y = 800
(-) (-) (-)
————————————-
300y = 600
y = 600/300 = 2
Substitute the value of y in (1)
100x + 200(2) = 700
100x = 700 – 400 = 300
x = 300/100 = 3
Therefore, x = 3 and y = 2
(viii) 41x + 53y = 135
53x + 41y = 147
Solution:
41x + 53y = 135……………(1)
53x + 41y = 147……………(2)
Adding (1) and (2), we get
94x + 94y = 282 ⇒ x + y = 3 ………….(3)
Subtracting (1) and (2), we get
41x + 53y = 135
53x + 41y = 147
(-) (-) (-)
———————-
-12x +12y = -12 ⇒ -x + y = -1 …………………(4)
From (3) and (4)
x + y = 3
-x + y = -1
—————–
2y = 2
y = 1
Substitite the value of y in (3) i.e,
x + 1= 3
x = 3 – 1 = 2
Therefore, x = 2 and y = 1
- Solve the following system by evaluating proportional value of the variables:
(i) 5x – 4y = -14
3x + 2y = -4
Solution:
(5x – 4y = -14) x 4 ⇒ 20x – 16y = -56
(3x + 2y = -4) x 14 ⇒42x + 28y = -56
(-) (-) (+)
————————————————–
-22x – 44y = 0
-x – 2y=0
x = -2y
Therefore,
x/-2 = y/1 = k
i.e., x = -2k and y = k
substituting in (1)
3(-2k) + 2k = -4
-6k + 2k = -4
-4k = -4
k = -4/-4 = 1
x = -2(1) = -2
y = k = 1
Therefore, x = -2 and y = 1
(ii) 3x + 2y = 5
5x – 4y = 23
Solution:
3x + 2y = 5———(1)
5x – 4y = 23———–(2)
(3x + 2y = 5)x23⇒69x + 46y = 115
(5x – 4y = 23)x5 ⇒ 25x – 20y = 115
(-) (+) (-)
————————————————-
44x + 66y = 0
2x + 3y = 0
2x = -3y
Therefore, x/-3 = y/2 = k
x = -3k and y = 2k————(*)
Then substitute x and y in (1)
3(-3k) + 2(2k) = 5
-9k + 4k = 5
-5k = 5
k = -1
Substitute k in (*)
x = -3(-1)=3 and y = 2(-1) = -2
Therefore, x = 3 and y = -2
(iii) 2x + 3y = 13
4x + y = 11
Solution:
2x + 3y = 13………………….(1)
4x + y = 11…………………….(2)
(2x + 3y = 13)x11⇒22x + 33y = 143
(4x + y = 11)x13 ⇒52x + 13y = 143
(-) (-) (-)
———————————————–
-30x + 20y = 0
-3x + 2y = 0
3x = 2y
x/2 = y/3 = k
x = 2k and y = 3k
Substitute the value of x and y in (1)
2(2k) + 3(3k) = 13
4k +9k = 13
13k = 13
k = 13/13 = 1
Then, x = 2(1) = 2 and y = 3(1) = 3
Therefore, x = 2 and y = 3
(iv)3x + 2y = 1
2x + 3y = 4
Solution:
3x + 2y = 1…………….(1)
2x + 3y = 4……………..(2)
(3x + 2y = 1)x4⇒12x + 8y = 4
(2x + 3y = 4)x1⇒2x + 3y = 4
(-) (-) (-)
—————————————–
10x + 5y = 0
2x + y = 0
2x = -y
Thus, x/-1 = y/2 = k
x = -k and y = 2k
Substitute x and y in (1)
3x + 2y = 1
3(-k) + 2(2k) = 1
-3k + 4k = 1
k = 1
Hence, x = -k = -1
and
y = 2k = 2(1) = 2
Therefore, x = -1 and y = 2
(v) 3x + 2y = 4
2x – 3y = 7
Solution:
3x + 2y = 4……………..(1)
2x – 3y = 7……………….(2)
(3x + 2y = 4)x7⇒21x + 14y = 28
(2x – 3y = 7)x4⇒8x – 12y = 28
(-) (+) (-)
—————————————————–
13x+26y=0
x + 2y = 0
x = -2y
Thus x/-2 = y/1 = k
⇒ x = -2k and y = k
Substitute x and y in equation (1),we get
3x + 2y = 4
3(-2k) + 2(k) = 4
-6k + 2k = 4
-4k = 4
k = -1
Hence, x = -2(-1) = 2
y = k = -1
Therefore, x = 2 and y = -1
(vi) 2x + 5y = -3
5x + 2y = -18
Solution:
2x + 5y = -3………………..(1)
5x + 2y = -18………………(2)
(2x + 5y = -3)x6⇒12x + 30y = -18
(5x + 2y = -18)x1⇒5x + 2y = -18
(-) (-) (+)
———————————————-
7x + 28y = 0
x + 4y = 0
x = -4y
Thus, x/-4 = y/1 = k
⇒ x = -4k and y = k
Substitute x and y in equation(1)
2x + 5y = -3
2(-4k) + 5k = -3
-8k + 5k = -3
-3k = -3
k = 1
Hence, x = -4 and y = 1
(xii) 3x + y = 7
x – y = 5
Solution:
3x + y = 7……………..(1)
x – y = 5………………..(2)
(3x + y = 7)x5 ⇒15x +5y = 35
(x – y = 5)x7 ⇒7x – 7y = 35
(-) (+) (-)
—————————————–
8x + 12y = 0
2x + 3y = 0
2x = -3y
Thus, x/-3 = y/2 = k
⇒ x = -3k and y = 2k
Substitute x and y in eq(2)
x – y = 5
-3k – 2k = 5
-5k = 5
k = -1
Hence, x = -3(-1) = 3
y = 2(-1) = -2
Therefore, x = 3 and y = -2
(vii) 2x – y = 6
3x + y = 9
Solution:
2x – y = 6…………………(1)
3x + y = 9…………………(2)
(2x – y = 6)x3⇒6x – 3y = 18
(3x + y = 9)x2⇒ 6x + 2y = 18
(-) (-) (-)
——————————————-
0.x – 5y = 0
Thus, x/-5 = y/0 = k
⇒ x = -5k and y = 0
Substitute x and y in eq(1)
2x – y = 6
2(-5k) – 0 = 6
-10k – 0 = 6
-10k = 6
k =- 6/10 = –3/5 .
Hence, y = -5k = -5(-3/5) = 3
and x = 0.
Therefore, x = 0 and y = 3
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