Chapter 1 – Linear Equations in Two Variables
Practice set 1.1
(1) Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 …(i)
2x – 3y = 12 …(ii)
Let’s add equations (i) and (ii),
5x + 3y = 9
+ 2x – 3y = 12
________________
▢ x = ▢
x = ▢ / ▢
x = ▢
Place x = 3 in equation (i)
5 ▢ + 3y = 9
3y = 9 – ▢
3y = ▢
y = ▢ / 3
y = ▢
Ans: Solution is (x, y) = ( ▢ , ▢ )
Solution:
5x + 3y = 9 …(i)
2x – 3y = 12 …(ii)
Let’s add equations (i) and (ii),
5x + 3y = 9
+ 2x – 3y = 12
________________
7x = 21
x = \(\large \frac {21}{7}\)
x = 3
Place x = 3 in equation (i)
5 (3) + 3y = 9
3y = 9 – 15
3y = – 6
y = \(\large \frac {– \,6}{3}\)
y = – 2
Ans: Solution is (x, y) = ( 3, – 2)
2. Solve the following simultaneous equations.
(1) 3a + 5b = 26; a + 5b = 22
Solution:
Let,
3a + 5b = 26 …(i)
a + 5b = 22 …(ii)
Subtracting equation (ii) from (i),
3a + 5b = 26
a + 5b = 22
(–) (+) (–)
________________
2a + 0b = 4
∴ 2a = 4
∴ a = \(\large \frac {4}{2}\)
∴ a = 2
Substituting a = 2 in equation (ii),
a + 5b = 22
∴ 2 + 5b = 22
∴ 5b = 22 – 2
∴ 5b = 20
∴ b = \(\large \frac {20}{5}\)
∴ b = 4
Ans: (a, b) = (2, 4) is the solution to the given equations.
(2) x + 7y = 10; 3x – 2y = 7
Solution:
Let,
x + 7y = 10 …(i)
3x – 2y = 7 …(ii)
Multiplying equation (i) by 3 on both sides, we get,
3(x) + 3(7y) = 3(10)
∴ 3x + 21y = 30 …(iii)
Subtracting equation (iii) from (ii),
3x – 2y = 7
3x + 21y = 30
(–) (–) (–)
_____________________
0x – 23y = – 23
∴ – 23y = – 23
∴ y = \(\large \frac {–23}{–23}\)
∴ y = 1
Substituting y = 1 in equation (i),
x + 7y = 10
∴ x + 7(1) = 10
∴ x + 7 = 10
∴ x = 10 – 7
∴ x = 3
Ans: (x, y) = (3, 1) is the solution to the given equations.
(3) 2x – 3y = 9; 2x + y = 13
Solution:
Let,
2x – 3y = 9 …(i)
2x + y = 13 …(ii)
Subtracting equation (ii) from (i),
2x – 3y = 9
2x + y = 13
(–) (–) (–)
________________
0x – 4y = – 4
∴ – 4y = – 4
∴ y = \(\large \frac {–4}{–4}\)
∴ y = 1
Substituting y = 1 in equation (i),
2x – 3y = 9
∴ 2x – 3(1) = 9
∴ 2x – 3 = 9
∴ 2x = 9 + 3
∴ 2x = 12
∴ x = \(\large \frac {12}{2}\)
∴ x = 6
Ans: (x, y) = (6, 1) is the solution to the given equations.
(4) 5m – 3n = 19; m – 6n = – 7
Solution:
Let,
5m – 3n = 19 …(i)
m – 6n = – 7 …(ii)
∴ m = – 7 + 6n
i.e. m = 6n – 7 …(iii)
Substituting m = 6n – 7 in eqn (i), we get,
5m – 3n = 19
5 (6n – 7) – 3n = 19
∴ 30n – 35 – 3n = 19
∴ 27n – 35 = 19
∴ 27n = 19 + 35
∴ 27n = 54
∴ n = \(\large \frac {54}{27}\)
∴ n = 2
Substituting n = 2 in equation (iii),
m = 6n – 7
∴ m = 6 (2) – 7
∴ m = 12 – 7
∴ m = 5
Ans: (m, n) = (5, 2) is the solution to the given equations.
(5) 5x + 2y = –3; x + 5y = 4
Solution:
Let,
5x + 2y = –3 …(i)
x + 5y = 4 …(ii)
∴ x = 4 – 5y …(iii)
Substituting x = 4 – 5y in eqn (i), we get,
5x + 2y = – 3
5 (4 – 5y) + 2y = – 3
∴ 20 – 25y + 2y = – 3
∴ 20 – 23y = – 3
∴ – 23y = – 3 – 20
∴ – 23y = – 23
∴ y = \(\large \frac {– 23}{– 23}\)
∴ y = 1
Substituting y = 1 in equation (iii),
x = 4 – 5y
∴ x = 4 – 5(1)
∴ x = 4 – 5
∴ x = – 1
Ans: (x, y) = (– 1, 1) is the solution to the given equations.
(6) \(\large \frac {1}{3}\) x + y = \(\large \frac {10}{3}\); 2x + \(\large \frac {1}{4}\) y = \(\large \frac {11}{4}\)
Solution:
The given equations are,
\(\large \frac {1}{3}\) x + y = \(\large \frac {10}{3}\)
Multiplying the equation on both sides by 3,
3 \((\large \frac {1}{3}\) x\()\) + 3 (y) = 3 × \(\large \frac {10}{3}\)
∴ x + 3y = 10 …(i)
And,
2x + \(\large \frac {1}{4}\) y = \(\large \frac {11}{4}\)
Multiplying eqn (ii) on both sides by 4,
4 (2x) + 4 \((\large \frac {1}{4}\) y\()\) = 4 × \(\large \frac {11}{4}\)
∴ 8x + y = 11 …(ii)
∴ y = 11 – 8x …(iii)
Substituting y = 11 – 8x in eqn (i), we get,
x + 3y = 10
x + 3 (11 – 8x) = 10
x + 33 – 24x = 10
– 23x + 33 = 10
– 23x = 10 – 33
– 23x = – 23
∴ x = \(\large \frac {– 23}{– 23}\)
∴ x = 1
Substituting x = 1 in equation (iii),
y = 11 – 8x
∴ y = 11 – 8 (1)
∴ y = 11 – 8
∴ y = 3
Ans: (x, y) = (1, 3) is the solution to the given equations.
(7) 99x + 101y = 499; 101x + 99y = 501
Solution:
Let,
99x + 101y = 499 …(i)
101x + 99y = 501 …(ii)
Adding equations (i) and (ii),
99x + 101y = 499
+ 101x + 99y = 501
_______________________
200x + 200y = 1000
∴ 200 (x + y) = 1000
∴ x + y = \(\large \frac {1000}{200}\)
∴ x + y = 5 …(iii)
Subtracting equation (ii) from (i),
99x + 101y = 499
101x + 99y = 501
(–) (–) (–)
_______________________
– 2x + 2y = – 2
∴ 2 (– x + y) = – 2
∴ – x + y = \(\large \frac {– 2}{2}\)
∴ – x + y = – 1 …(iv)
Adding equations (iii) and (iv),
x + y = 5
+ (– x) + y = – 1
_______________________
0x + 2y = 4
∴ 2y = 4
∴ y = \(\large \frac {4}{2}\)
∴ y = 2
Substituting y = 2 in equation (iii),
x + y = 5
∴ x + 2 = 5
∴ x = 5 – 2
∴ x = 3
Ans: (x, y) = (3, 2) is the solution to the given equations.
(8) 49x – 57y = 172; 57x – 49y = 252
Solution:
Let,
49x – 57y = 172 …(i)
57x – 49y = 252 …(ii)
Subtracting equations (i) and (ii), we get,
49x – 57y = 172
57x – 49y = 252
(–) (+) (–)
___________________
– 8x – 8y = – 80
∴ – 8 (x + y) = – 80
∴ x + y = \(\large \frac {– 80}{– 8}\)
∴ x + y = 10
∴ y = 10 – x …(iii)
Substituting y = 10 – x in eqn (i), we get,
49x – 57y = 172
49x – 57 (10 – x) = 172
49x – 570 + 57x = 172
106x – 570 = 172
106x = 172 + 570
106x = 742
∴ x = \(\large \frac {742}{106}\)
∴ x = 7
Substituting x = 7 in eqn (iii), we get,
y = 10 – x
y = 10 – 7
y = 3
Ans: (x, y) = (7, 3) is the solution to the given equations.
Practice set 1.2
1. Complete the following table to draw graph of the equations –
(I) x + y = 3
| x | 3 | ▢ | ▢ |
|---|---|---|---|
y | ▢ | 5 | 3 |
(x, y) | (3, 0) | ▢ | (0, 3) |
Solution:
x + y = 3
∴ y = 3 – x and x = 3 – y
when x = 3
y = 3 – 3
∴ y = 0
when y = 5
x = 3 – 5
∴ x = – 2
when y = 3
x = 3 – 3
∴ x = 0
Ans:
| x | 3 | – 2 | 0 |
|---|---|---|---|
y | 0 | 5 | 3 |
(x, y) | (3, 0) | (– 2, 5) | (0, 3) |
(II) x – y = 4
| x | ▢ | – 1 | 0 |
|---|---|---|---|
y | 0 | ▢ | – 4 |
(x, y) | ▢ | ▢ | (0, – 4) |
Solution:
x – y = 4
∴ y = x – 4 and x = 4 + y
when y = 0
x = 4 + 0
∴ x = 4
when x = – 1
y = – 1 – 4
∴ y = – 5
Ans:
| x | 4 | – 1 | 0 |
|---|---|---|---|
y | 0 | – 5 | – 4 |
(x, y) | (4, 0) | (– 1, – 5) | (0, – 4) |
2. Solve the following simultaneous equations graphically.
(1) x + y = 6 ; x – y = 4
Solution:
Equation (i)
x + y = 6
∴ y = 6 – x
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | 6 | 5 | 7 |
(x, y) | (0, 6) | (1, 5) | (– 1, 7) |
when x = 0,
y = 6 – 0
∴ y = 6
when x = 1,
y = 6 – 1
∴ y = 5
when x = – 1,
y = 6 – (–1)
∴ y = 6 + 1
∴ y = 7
Equation (ii)
x – y = 4
∴ – y = 4 – x
∴ y = x – 4
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | – 4 | – 3 | – 5 |
(x, y) | (0, – 4) | (1, – 3) | (– 1, – 5) |
when x = 0
∴ y = 0 – 4
∴ y = – 4
when x = 1
∴ y = 1 – 4
∴ y = – 3
when x = – 1
∴ y = (– 1) – 4
∴ y = – 5
The lines of the two given simultaneous equations intersect each other at (5 , 1)
Ans: The solution of given the simultaneous equations is (x , y) (5, 1).
(2) x + y = 5 ; x – y = 3
Solution:
Equation (i)
x + y = 5
∴ y = 5 – x
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | 5 | 4 | 6 |
(x, y) | (0, 5) | (1, 4) | (– 1, 6) |
when x = 0
∴ y = 5 – 0
∴ y = 5
when x = 1
∴ y = 5 – (1)
∴ y = 5 – 1
∴ y = 4
when x = – 1
∴ y = 5 – (– 1)
∴ y = 5 + 1
∴ y = 6
Equation (ii)
x – y = 3
∴ – y = 3 – x
∴ y = x – 3
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | – 3 | – 2 | – 4 |
(x, y) | (0, – 3) | (1, – 2) | (– 1, – 4) |
when x = 0
∴ y = 0 – 3
∴ y = – 3
when x = 1
∴ y = 1 – 3
∴ y = – 2
when x = – 1
∴ y = (– 1) – 3
∴ y = – 4
The lines of the two given simultaneous equations intersect each other at (4 , 1)
Ans: The solution of the given simultaneous equations is (x, y) = (4, 1).
(3) x + y = 0 ; 2x – y = 9
Solution:
Equation (i)
x + y = 0
∴ y = – x
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | 0 | – 1 | 1 |
(x, y) | (0, 0) | (1, – 1) | (– 1, 1) |
when x = 0
∴ y = (– 0)
∴ y = 0
when x = 1
∴ y = – 1
when x = – 1
∴ y = – (– 1)
∴ y = 1
Equation (ii)
2x – y = 9
∴ – y = 9 – 2x
∴ y = – (9 – 2x)
∴ y = – 9 + 2x
∴ y = 2x – 9
| x | 1 | 2 | 3 |
|---|---|---|---|
y | –7 | – 5 | – 3 |
(x, y) | (1, –7) | (2, –5) | (3, – 3) |
when x = 1
∴ y = 2(1) – 9
∴ y = 2 – 9
y = – 7
when x = 2
∴ y = 2(2) – 9
∴ y = 4 – 9
∴ y = – 5
when x = 3
∴ y = 2(3) – 9
∴ y = 6 – 9
∴ y = – 3
The lines of the two given simultaneous equations intersect each other at (3 , –3)
Ans: The solution of given simultaneous equations is (3, –3).
(4) 3x – y = 2 ; 2x – y = 3
Solution:
Equation (i)
2x – y = 3
∴ – y = 3 – 2x
∴ y = – (3 – 2x)
∴ y = – 3 + 2x
∴ y = 2x – 3
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | –3 | –1 | – 5 |
(x, y) | (0, –3) | (1, –1) | (– 1, – 5) |
when x = 0
∴ y = 2(0) – 3
∴ y = 0 – 3
∴ y = – 3
when x = 1
∴ y = 2(1) – 3
∴ y = 2 – 3
∴ y = – 1
when x = – 1
∴ y = 2(– 1) – 3
∴ y = – 2 – 3
∴ y = – 5
Equation (ii)
3x – y = 2
∴ – y = 2 – 3x
∴ y = – (2 – 3x)
∴ y = – 2 + 3x
∴ y = 3x – 2
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | –2 | 1 | – 5 |
(x, y) | (0, –2) | (1, 1) | (– 1, – 5) |
when x = 0
∴ y = 3(0) – 2
∴ y = 0 – 2
∴ y = –2
when x = 1
∴ y = 3(1) – 2
∴ y = 3 – 2
∴ y = 1
when x = – 1
∴ y = 3(– 1) – 2
∴ y = – 3 – 2
∴ y = – 5
The lines of the two given simultaneous equations intersect each other at (–1 , –5)
Ans: The solution of given simultaneous equations is (–1, –5).
(5) 3x – 4y = –7 ; 5x – 2y = 0
Solution:
Equation (i)
3x – 4y = – 7
∴ – 4y = – 7 – 3x
∴ y = \(\large \frac {–\,7\,–\,3x}{–\,4}\)
∴ y = \(\large \frac {7\,+\,3x}{4}\)
| x | 1 | – 1 | 3 |
|---|---|---|---|
y | 2.5 | 1 | 4 |
(x, y) | (1, 2.5) | (– 1, 1) | (3, 4) |
when x = 1
∴ y = \(\large \frac {7\,+\,3(1)}{4}\)
∴ y = \(\large \frac {7\,+\,3}{4}\)
∴ y = \(\large \frac {10}{4}\)
∴ y = 2.5
when x = – 1
∴ y = \(\large \frac {7\,+\,3(–\,1)}{4}\)
∴ y = \(\large \frac {7\,+\,(–\,3)}{4}\)
∴ y = \(\large \frac {7\,–\,3}{4}\)
∴ y = \(\large \frac {4}{4}\)
∴ y = 1
when x = 3
∴ y = \(\large \frac {7\,+\,3(3)}{4}\)
∴ y = \(\large \frac {7\,+\,9}{4}\)
∴ y = \(\large \frac {16}{4}\)
∴ y = 4
Equation (ii)
5x – 2y = 0
∴ – 2y = – 5x
∴ y = \(\large \frac {–\,5}{–\,2}\) x
∴ y = \(\large \frac {5}{2}\) x
| x | 0 | 1 | – 2 |
|---|---|---|---|
y | 0 | 2.5 | – 5 |
(x, y) | (0, 0) | (1, 2.5) | (–2, –5) |
when x = 0
y = \(\large \frac {5}{2}\) (0)
∴ y = 0
when x = 1
y = \(\large \frac {5}{2}\) (1)
∴ y = \(\large \frac {5}{2}\)
∴ y = 2.5
when x = – 2
y = \(\large \frac {5}{2}\) (– 2)
∴ y = 5 × – 1
∴ y = – 5
The lines of the two given simultaneous equations intersect each other at (1 , 2.5)
Ans: The solution of given simultaneous equations is (1, 2.5).
(6) 2x – 3y = 4 ; 3y – x = 4
Solution:
Equation (i)
2x – 3y = 4
∴ – 3y = 4 – 2x
∴ y = \(\large \frac {4\,–\,2x}{–\,3}\)
∴ y = \(\large \frac {2x\,–\,4}{3}\)
| x | – 1 | 2 | 5 |
|---|---|---|---|
y | – 2 | 0 | 2 |
(x, y) | (– 1, – 2) | (2, 0) | (5, 2) |
when x = – 1
∴ y = \(\large \frac {2x\,–\,4}{3}\)
∴ y = \(\large \frac {2(–\,1)\,–\,4}{3}\)
∴ y = \(\large \frac {–\,2\,–\,4}{3}\)
∴ y = \(\large \frac {–\,6}{3}\)
∴ y = – 2
when x = 2
∴ y = \(\large \frac {2x\,–\,4}{3}\)
∴ y = \(\large \frac {2(2)\,–\,4}{3}\)
∴ y = \(\large \frac {4\,–\,4}{3}\)
∴ y = \(\large \frac {0}{3}\)
∴ y = 0
when x = 5
∴ y = \(\large \frac {2x\,–\,4}{3}\)
∴ y = \(\large \frac {2(5)\,–\,4}{3}\)
∴ y = \(\large \frac {10\,–\,4}{3}\)
∴ y = \(\large \frac {6}{3}\)
∴ y = 2
Equation (ii)
3y – x = 4
∴ – x = 4 – 3y
∴ x = – (4 – 3y)
∴ x = – 4 + 3y
∴ x = 3y – 4
| x | – 4 | – 1 | – 7 |
|---|---|---|---|
y | 1 | – 1 | |
(x, y) | (– 4, 0) | (– 1, 1) | (–7, –1) |
when y = 0
x = 3(0) – 4
∴ x = 0 – 4
∴ x = – 4
when y = 1
x = 3(1) – 4
∴ x = 3 – 4
∴ x = – 1
when y = – 1
x = 3(– 1) – 4
∴ x = – 3 – 4
∴ x = – 7
The lines of the two given simultaneous equations intersect each other at (8 , 4)
Ans: The solution of given simultaneous equations is (8, 4).
Practice set 1.3
1. Fill in the blanks with correct number
\(\begin{vmatrix}3&2\\4&5\end{vmatrix}\) = 3 × ▢ – ▢ × 4 = ▢ – 8 = ▢
Solution:
\(\begin{vmatrix}3&2\\4&5\end{vmatrix}\) = 3 × 5 – 2 × 4 = 15 – 8 = 5
2. Find the values of following determinants.
(1) \(\begin{vmatrix}-1&7\\2&4\end{vmatrix}\)
Solution:
\(\begin{vmatrix}-1&7\\2&4\end{vmatrix}\)
= (-1 × 4) – (7 × 2)
= – 4 – 14
= – 18
Ans: \(\begin{vmatrix}-1&7\\2&4\end{vmatrix}\) = – 18
(2) \(\begin{vmatrix}5&3\\-7&0\end{vmatrix}\)
Solution:
\(\begin{vmatrix}5&3\\-7&0\end{vmatrix}\)
= (5 × 0) – (3 × -7)
= 0 – (-21)
= 21
Ans:\(\begin{vmatrix}5&3\\-7&0\end{vmatrix}\) = 21
(2) \(\begin{vmatrix}5&3\\-7&0\end{vmatrix}\)
Solution:
\(\begin{vmatrix}5&3\\-7&0\end{vmatrix}\)
= (5 × 0) – (3 × -7)
= 0 – (-21)
= 21
Ans:\(\begin{vmatrix}5&3\\-7&0\end{vmatrix}\) = 21
(3) \(\begin{vmatrix}\large\frac73&\large\frac53\\\large\frac32&\large\frac12\end{vmatrix}\)
Solution:
\(\begin{vmatrix}\large\frac73&\large\frac53\\\large\frac32&\large\frac12\end{vmatrix}\)
\( = \left(\large\frac73\times\large\frac12\right)\;-\;\left(\large\frac53\times\large\frac32\right)\\=\;\large\frac76\;-\;\large\frac{15}6\\=\;\large\frac{-8}6\\=\;\large\frac{-4}3 \)
Ans: \(\begin{vmatrix}\large\frac73&\large\frac53\\\large\frac32&\large\frac12\end{vmatrix}\;=\;\large\frac{-4}3\)
Practice set 1.4
Solve the following simultaneous equations:
(1) \(\large \frac {2}{x}\) – \(\large \frac {3}{y}\) = 15
\(\large \frac {8}{x}\) + \(\large \frac {5}{y}\) = 77
Solution:
Let,
\(\large \frac {2}{x}\) – \(\large \frac {3}{y}\) = 15 …(i)
\(\large \frac {8}{x}\) + \(\large \frac {5}{y}\) = 77 …(ii)
Substituting \(\large \frac {1}{x}\) = m and \(\large \frac {1}{y}\) = n in (i) and (ii), we get,
2m – 3n = 15 …(iii)
8m + 5n = 77 …(iv)
Multiplying (iii) by 5 and (iv) by 3, we get,
10m – 15n = 75 …(v)
24m + 15n = 231 … (vi)
Adding (v) and (vi), we get,
10m – 15n = 75
+ 24m + 15n = 231
_______________________
34 m = 306
∴ m = \(\large \frac {306}{34}\)
∴ m = 9
Substituting m = 9 in (iii) we get,
2m – 3n = 15
∴ 2 × 9 – 3n = 15
∴ 18 – 3n = 15
∴ – 3n = 15 – 18
∴ – 3n = – 3
∴ n = \(\large \frac {-3}{-3}\)
∴ n = 1
Resubstituting the values of m and n, we get,
\(\large \frac {1}{x}\) = m
∴ \(\large \frac {1}{x}\) = 9
∴ 9x = 1
∴ x = \(\large \frac {1}{9}\)
\(\large \frac {1}{y}\) = n,
∴ \(\large \frac {1}{y}\) = 1
∴ y = 1
Ans: x = \(\large \frac {1}{9}\) and y = 1 is the solution of the given simultaneous equations.
(2) \(\large \frac {10}{x\,+\, y}\) – \(\large \frac {2}{x\,-\,y}\) = 4
\(\large \frac {8}{x}\) + \(\large \frac {5}{y}\) = – 2
Solution:
Let,
\(\large \frac {10}{x\,+\, y}\) + \(\large \frac {2}{x\,-\,y}\) = 4 …(i)
\(\large \frac {15}{x\,+\, y}\) – \(\large \frac {5}{x\,-\, y}\) = – 2 …(ii)
Substituting \(\large \frac {1}{x\,+\, y}\) = m and \(\large \frac {5}{x\,-\, y}\) = n in (i) and (ii) we get,
10m + 2n = 4 …(iii)
15m – 5n = – 2 …(iv)
Multiplying (iii) by 5 and (iv) by 2 we get,
50m + 10n = 20 …(v)
30m – 10n = – 4 …(vi)
Adding (v) and (vi), we get
50m + 10n = 20
+ 30m – 10n = – 4
______________________
80m = 16
∴ m = \(\large \frac {16}{80}\)
∴ m = \(\large \frac {1}{5}\)
Substituting m = \(\large \frac {1}{5}\) in (iii), we get,
10m + 2n = 4
∴ 10 × \(\large \frac {1}{5}\) + 2n = 4
∴ 2 + 2n = 4
∴ 2n = 4 – 2
∴ 2n = 2
∴ n = \(\large \frac {2}{2}\)
∴ n = 1
Resubstituting the values of m and n we get,
\(\large \frac {1}{x\,+\, y}\) = m
∴ \(\large \frac {1}{x\,+\, y}\) = \(\large \frac {1}{5}\)
∴ x + y = 5 …(vii)
\(\large \frac {1}{x\,-\, y}\) = n
∴ \(\large \frac {1}{x\,-\, y}\) = 1
∴ x – y = 1 …(viii)
Adding (vii) and (viii), we get,
x + y = 5
+ x – y = 1
_____________
2x = 6
∴ x = \(\large \frac {6}{2}\)
∴ x = 3
Substituting x = 3 in equation (vii) we get,
x + y = 5
∴ 3 + y = 5
∴ y = 5 – 3
∴ y = 2
Ans: x = 3 and y = 2 is the solution of the given simultaneous equations.
(3) \(\large \frac {27}{x\,-\, 2}\) – \(\large \frac {31}{y\,+\,3}\) = 85
\(\large \frac {31}{x\,-\, 2}\) + \(\large \frac {27}{y\,+\,3}\) = 89
Solution:
Let,
\(\large \frac {27}{x\,-\, 2}\) – \(\large \frac {31}{y\,+\,3}\) = 85 …(i)
\(\large \frac {31}{x\,-\, 2}\) + \(\large \frac {27}{y\,+\,3}\) = 89 …(ii)
Substituting, \(\large \frac {1}{x\,-\, 2}\) = m and \(\large \frac {1}{y\,+\,3}\) = n in (i) and (ii), we get,
27m + 31n = 85 …(iii)
31m + 27n = 89 …(iv)
Adding (iii) and (iv), we get,
27m + 31n = 85
+ 31m + 27n = 89
__________________
58m + 58n = 174
∴ 58(m + n) = 174
∴ m + n = \(\large \frac {174}{58}\)
∴ m + n = 3 …(v)
Subtracting (iv) from (iii), we get,
27m + 31n = 85
31m + 27n = 89
(–) (–) (–)
__________________
– 4m + 4n = – 4
∴ – 4(m – n) = − 4
∴ m – n = \(\large \frac {-4}{-4}\)
∴ m – n = 1 …(vi)
Adding (v) and (vi), we get
m + n = 3
+ m – n = 1
_____________
2m = 4
∴ m = \(\large \frac {4}{2}\)
∴ m = 2
Substituting m = 2 in (v) we get,
m + n = 3
∴ 2 + n = 3
∴ n = 3 – 2
∴ n = 1
Resubstituting the values of m and n we get,
\(\large \frac {1}{x\,-\, 2}\) = m
∴ \(\large \frac {1}{x\,-\, 2}\) = 2
∴ 2(x – 2) = 1
∴ 2x – 4 = 1
∴ 2x = 1 + 4
∴ 2x = 5
∴ x = \(\large \frac {5}{2}\)
\(\large \frac {1}{y\,+\,3}\) = n
∴ \(\large \frac {1}{y\,+\,3}\) = 1
∴ y + 3 = 1
∴ y = 1 – 3
∴ y = – 2
Ans: x = \(\large \frac {5}{2}\) and y = – 2 is the solution of the given simultaneous equations.
(4) \(\large \frac {1}{3x\,+\, y}\) + \(\large \frac {1}{3x\,-\,y}\) = \(\large \frac {3}{4}\)
\(\large \frac {1}{2(3x\,+\, y)}\) – \(\large \frac {1}{2(3x\,-\,y)}\) = – \(\large \frac {1}{8}\)
Solution:
Let,
\(\large \frac {1}{3x\,+\, y}\) + \(\large \frac {1}{3x\,-\,y}\) = \(\large \frac {3}{4}\) …(i)
\(\large \frac {1}{2(3x\,+\, y)}\) – \(\large \frac {1}{2(3x\,-\,y)}\) = – \(\large \frac {1}{8}\) …(ii)
Substituting \(\large \frac {1}{3x\,+\, y}\) = m and \(\large \frac {1}{3x\,-\,y}\) = n in (i) and (ii), we get,
m + n = \(\large \frac {3}{4}\)
∴ 4(m + n) = 3
∴ 4m + 4n = 3 …(iii)
\(\large \frac {m}{2}\) – \(\large \frac {n}{2}\) = – \(\large \frac {1}{8}\)
∴ m – n = – \(\large \frac {2}{8}\)
∴ m – n = – \(\large \frac {1}{4}\)
∴ 4m – 4n = – 1 …(iv)
Adding (iii) and (iv), we get,
4m + 4n = 3
+ 4m – 4n = – 1
________________
8m = 2
∴ m = \(\large \frac {2}{8}\)
∴ m = \(\large \frac {1}{4}\)
Substituting m = \(\large \frac {1}{4}\) in (iii), we get,
4m + 4n = 3
∴ 4 × \(\large \frac {1}{4}\) + 4n = 3
∴ 1 + 4n = 3
∴ 4n = 3 – 1
∴ 4n = 2
∴ n = \(\large \frac {1}{2}\)
Resubstituting,
\(\large \frac {1}{3x\,+\, y}\) = m
∴ \(\large \frac {1}{3x\,+\, y}\) = \(\large \frac {1}{4}\)
∴ 3x + y = 4 …(v)
\(\large \frac {1}{3x\,-\,y}\) = n
∴ \(\large \frac {1}{3x\,-\, y}\) = \(\large \frac {1}{2}\)
∴ 3x – y = 2 …(vi)
Adding (v) and (vi), we get,
3x + y = 4
+ 3x – y = 2
______________
6x = 6
∴ x = 1
Substituting x = 1 in (v), we get,
3x + y = 4
∴ 3(1) + y = 4
∴ 3 + y = 4
∴ y = 4 – 3
∴ y = 1
Ans: x = 1 and y = 1 is the solution of the given simultaneous equations.
Practice set 1.5
(1) Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Solution:
Let the greater number be x and the smaller number be y.
According to the first condition
x – y = 3 …(i)
Three times the greater number = 3x and twice the smaller number = 2y
According to the second condition
3x + 2y = 19 …(ii)
Multiplying (i) by 2 we get,
2x – 2y = 6 …(iii)
Adding (ii) and (iii)
3x + 2y = 19
+ 2x – 2y = 6
_______________
5x = 25
∴ x = \(\large \frac {25}{5}\)
∴ x = 5
Substituting the value of x in equation (i)
x – y = 3
∴ 5 – y = 3
∴ – y = 3 – 5
∴ – y = – 2
∴ y = 2
Ans: The greater number is 5 and smaller number is 2.
(2) Complete the following.
Solution:
In a rectangle, opposite sides are equal in length.
∴ 2x + y + 8 = 4x – y
∴ 2x – 4x + y + y = – 8
∴ – 2x + 2y = – 8
∴ 2(– x + y) = – 8
∴ – x + y = 8
∴ – x + y = – 4 …(i)
Also,
2y = x + 4
∴ – 4 = x – 2y
i.e. x – 2y = – 4 …(ii)
Adding (i) and (ii) we get,
– x + y = – 4
+ x – 2y = – 4
_______________
– y = – 8
∴ y = 8
Substituting y = 8 in (i) we get,
– x + 8 = – 4
∴ – x = – 4 – 8
∴ – x = – 12
∴ x = 12
The length of rectangle = 2x + y + 8
∴ The length of rectangle = 2(12) + 8 + 8
∴ The length of rectangle = 24 + 16
∴ The length of rectangle = 40 units
The breadth of rectangle = 2y
∴ The breadth of rectangle = 2 × 8
∴ The breadth of rectangle = 16 units
Perimeter of rectangle = 2 [length + breadth]
∴ Perimeter of rectangle = 2 [ 40 + 16 ]
∴ Perimeter of rectangle = 2 × 56
∴ Perimeter of rectangle = 112 units
Area of rectangle = length × breadth
Area of rectangle = 40 × 16
∴ Area of rectangle = 640 sq. units
Ans: Perimeter of rectangle is 112 units and area of rectangle is 640 sq. units.
(3) The sum of the father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Solution:
Let the age of son be x years and the age of father be y years.
According to the first condition,
2x + y = 70 …(i)
According to the second condition,
x + 2y = 95 …(ii)
Multiplying (i) by 2 we get,
4x + 2y = 140 …(iii)
Subtracting (iii) from (ii), we get,
x + 2y = 95
4x + 2y = 140
(–) (–) (–)
______________________
– 3x = – 45
∴ x = \(\large \frac {-\,45}{-3}\)
∴ x = 15
Substituting x = 15 in (i), we get,
2x + y = 70
∴ 2(15) + y = 70
∴ 30 + y = 70
∴ y = 70 – 30
∴ y = 40
Ans: The age of son is 15 years and the age of father is 40 years.
(4) The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Solution:
Let the numerator of the fraction be x and the denominator be y.
∴ The fraction is \(\large \frac {x}{y}\)
According to first condition,
y = 2x + 4
∴ – 2x + y = 4 …(i)
According to second condition,
(y – 6) = 12 (x – 6)
∴ y – 6 = 12x – 72
∴ – 6 + 72 = 12x – y
∴ 66 = 12x – y
∴ 12x – y = 66 …(ii)
Adding (i) and (ii) we get,
– 2x + y = 4
+ 12x – y = 66
______________
10x = 70
∴ x = \(\large \frac {70}{10}\)
∴ x = 7
Substituting x = 7 in equation (i) we get,
– 2x + y = 4
∴ – 2(7) + y = 4
∴ – 14 + y = 4
∴ y = 4 + 14
∴ y = 18
Required fraction is \(\large \frac {x}{y}\) = \(\large \frac {7}{18}\)
Ans: The required fraction is \(\large \frac {7}{18}\).
(5) Two types of boxes A, B are to be placed in a truck having a capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighs 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Solution:
1 ton = 1000 kg
∴ 10 ton = 10,000 kg
∴ Capacity of truck = 10 ton = 10,000 kg
Let the weight of type A box be x kg. and the weight of each of type B box be y kg.
According to the first condition,
150x + 100y = 10,000
∴ 15x + 10y = 1000 …(i) [Dividing both sides by 10]
According to the second condition,
260x + 40y = 10,000
∴ 65x + 10y = 2500 …(ii) [Dividing both sides by 4]
Subtracting (i) from (ii),
65x + 10y = 2500
15x + 10y = 1000
(–) (–) (–)
___________________________
50x = 1500
∴ x = \(\large \frac {1500}{50}\)
∴ x = 30
Substituting x = 30 in (i) we get,
15x + 10y = 1
∴ 15(30) + 10y = 1000
∴ 450 + 10y = 1000
∴ 10y = 1000 – 450
∴ 10y = 550
∴ y = \(\large \frac {550}{10}\)
∴ y = 55
Ans: The weight of type A box is 30 kg and the weight of type B box is 55 kg
(6) Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with an average speed of 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.
Solution:
Let the distance travelled by bus be x km and the distance travelled by aeroplane be y km.
According to the first condition.
x + y = 1900 …(i)
Time = \(\large \frac {Distance}{Speed}\)
∴ Time taken by bus to cover x km at 60 km/hr = \(\large \frac {x}{60}\) hours
∴ Time taken by aeroplane to cover y km at 700 km/hr = \(\large \frac {y}{700}\) hours
The journey was completed in 5 hours
According to second condition,
\(\large \frac {x}{60}\) + \(\large \frac {y}{700}\) = 5
LCM of 60 and 70 is 2100
∴ Multiplying both the sides of above equation by 2100, we get,
35x + 3y = 10500 …(ii)
Multiplying equation (i) by 3 we get,
∴ 3x + 3y = 5700 …(iii)
Subtracting (iii) from (ii)
35x + 3y = 10500
3x + 3y = 5700
(–) (–) (–)
_______________________
32x = 4800
∴ x = \(\large \frac {4800}{32}\)
∴ x = 150
Ans: Vishal travelled 150 km by bus.
Problem Set 1
Uploading in progress ...
1. Choose correct alternative for each of the following questions
(1) To draw graph of 4x + 5y = 19, Find y when x = 1.
(A) 4
(B) 3
(C) 2
(D) – 3
Ans: Option (D) : – 3
Solution:
4x + 5y = 19
When x = 1,
4(1) + 5y = 19
∴ 4 + 5y = 19
∴ 5y = 19 – 4
∴ 5y = 15
∴ y = \(\large \frac {15}{5}\)
∴ y = 3
(2) For simultaneous equations in variables x and y, Dx = 49, Dy = – 63, D = 7 then what is x ?
(A) 7
(B) – 7
(C) \(\large \frac {1}{7}\)
(D) \(\large \frac {–\,1}{7}\)
Ans: Option (A) : 7
Solution:
Dx = 49
D = 7
x = \(\large \frac {Dx}{D}\)
∴ x = \(\large \frac {49}{7}\)
∴ x = 7
(3) Find the value of
(A) – 1
(B) – 41
(C) 41
(D) 1
Ans: Option (D) : 1
Solution:
= [5 × (– 4) – 3 × (– 7)]
= – 20 + 21
= 1
(4) To solve x + y = 3 ; 3x – 2y – 4 = 0 by determinant method find D.
(A) 5
(B) 1
(C) – 5
(D) – 1
Ans: Option (C) : – 5
Solution:
Let,
x + y = 3 …(i)
3x – 2y – 4 = 0
∴ 3x – 2y = 4 …(ii)
D = \begin{vmatrix}1&1\\3&-2\end{vmatrix}
∴ D = [1 × (– 2) – (1 × 3)]
∴ D = – 2 – 3
∴ D = – 5
(5) ax + by = c and mx + ny = d and an ≠ bm then these simultaneous equations have
(A) Only one common solution.
(B) No solution.
(C) Infinite number of solutions.
(D) Only two solutions.
Ans: Option (A) : Only one common solution
Solution:
ax + by = c
mx + ny = d
D = \begin{vmatrix}a&b\\m&n\end{vmatrix}
∴ D = [a × n – b × m]
∴ D = an – bm
∴ D ≠ 0 …[an ≠ bm, Given]
Hence, the given equations have a unique solution or only one common solution.
2. Complete the following table to draw the graph of 2x – 6y = 3
| x | – 5 | ▢ |
|---|---|---|
y | ▢ | 0 |
(x, y) | ▢ | ▢ |
Solution:
2x – 6y = 3
When x = – 5,
2(– 5) – 6y = 3
∴ – 10 – 6y = 3
∴ – 6y = 3 + 10
∴ – 6y = 13
∴ y = – \(\large \frac {13}{6}\)
When y = 0,
2x – 6(0) = 3
∴ 2x – 0 = 3
∴ 2x = 3
∴ x = \(\large \frac {3}{2}\)
| x | – 5 | \(\large \frac {3}{2}\) |
|---|---|---|
y | – \(\large \frac {13}{6}\) | 0 |
(x, y) | (– 5 , – \(\large \frac {13}{6}\)) | (\(\large \frac {3}{2}\), 0) |
3. Solve the following simultaneous equations graphically.
(1) 2x + 3y = 12 ; x – y = 1
Solution:
Equation (i)
2x + 3y = 12
∴ 3y = 12 – 2x
Dividing by 3 on both sides, we get,
∴ y = 4 – \(\large \frac {2}{3}\)x
| x | 0 | 3 | – 3 |
|---|---|---|---|
y | 4 | 2 | 6 |
(x, y) | (0, 4) | (3, 2) | (– 3, 6) |
when x = 0,
y = 4 – \(\large \frac {2}{3}\) (0)
∴ y = 4
when x = 3,
y = 4 – \(\large \frac {2}{3}\) (3)
∴ y = 4 – 2
∴ y = 2
when x = – 3,
y = 4 – \(\large \frac {2}{3}\) (– 3)
∴ y = 4 – 2(– 1)
∴ y = 4 + 2
∴ y = 6
Equation (ii)
x – y = 1
∴ – y = 1 – x
∴ y = x – 1
| x | 0 | 1 | – 1 |
|---|---|---|---|
y | – 1 | 0 | – 2 |
(x, y) | (0, – 1) | (1, 0) | (– 1, – 2) |
when x = 0,
∴ y = 0 – 1
∴ y = – 1
when x = 1,
∴ y = 1 – 1
∴ y = 0
when x = – 1,
y = – 1 – 1
∴ y = – 2
The lines of the two given simultaneous equations intersect each other at (3 , 2)
Ans: The solution of given the simultaneous equations is (x , y) = (3, 2).
(2) x – 3y = 1 ; 3x – 2y + 4 = 0
Solution:
Equation (i)
x – 3y = 1
∴ x = 1 + 3y
| x | 1 | 4 | 7 |
|---|---|---|---|
y | 1 | 2 | |
(x, y) | (1, 0) | (4, 1) | (7, 2) |
when y = 0,
x = 1 + 3(0)
∴ x = 1 + 0
∴ x = 1
when y = 1,
x = 1 + 3(1)
∴ x = 1 + 3
∴ x = 4
when y = 2,
x = 1 + 3(2)
∴ x = 1 + 6
∴ x = 7
Equation (ii)
3x – 2y + 4 = 0
∴ – 2y = 4 – 3x
Dividing by – 2y on both sides, we get,
∴ y = – 2 + \(\large \frac {3}{2}\)x
∴ y = \(\large \frac {3}{2}\)x – 2
| x | 0 | 2 | 4 |
|---|---|---|---|
y | – 2 | 1 | 4 |
(x, y) | (0, – 2) | (2, 1) | (4, 4) |
when x = 0,
∴ y = \(\large \frac {3}{2}\) (0) – 2
∴ y = 0 – 2
∴ y = – 2
when x = 2,
∴ y = \(\large \frac {3}{2}\) (2) – 2
∴ y = 3 – 2
∴ y = 1
when x = 4,
∴ y = \(\large \frac {3}{2}\) (4) – 2
∴ y = 3(2) – 2
∴ y = 6 – 2
∴ y = 4
The lines of the two given simultaneous equations intersect each other at (– 2, – 1)
Ans: The solution of given the simultaneous equations is (x , y) = (– 2, – 1).
6. Solve the following simultaneous equations.
(1) \(\large \frac {2}{x}\) + \(\large \frac {2}{3y}\) = \(\large \frac {1}{6}\) ; \(\large \frac {3}{x}\) + \(\large \frac {2}{y}\) = 0
Solution:
Let,
\(\large \frac {2}{x}\) + \(\large \frac {2}{3y}\) = \(\large \frac {1}{6}\) …(i)
\(\large \frac {3}{x}\) + \(\large \frac {2}{y}\) = 0 …(ii)
Substituting \(\large \frac {1}{x}\) = m and \(\large \frac {1}{y}\) = n in equations (i) and (ii), we get,
2m + \(\large \frac {2}{3}\)n = \(\large \frac {1}{6}\)
Multiplying both sides by 6, we get,
12m + 4n = 1 …(iii)
3m + 2n = 0 …(iv)
Multiplying both sides of (iv) by 2 we get,
6m + 4n = 0 …(v)
Subtracting equation (v) from (iii), we get,
12m + 4n = 1
6m + 4n = 0
(–) (–) (–)
_________________
6m = 1
∴ m = \(\large \frac {1}{6}\)
Substituting m = \(\large \frac {1}{6}\) in equation (iii) we get,
12 \(\large (\frac {1}{6})\) + 4n = 1
∴ 2 + 4n = 1
∴ 4n = 1 – 2
∴ 4n = –1
∴ n = \(\large \frac {-\,1}{4}\)
Resubstituting the values of m and n we get,
\(\large \frac {1}{x}\) = \(\large \frac {1}{6}\)
∴ x = 6
\(\large \frac {1}{y}\) = \(\large \frac {-\,1}{4}\)
∴ y = – 4
Ans: x = 6, y = – 4 is the solution of the given simultaneous equations.
(2) \(\large \frac {7}{2x \,+\, 1}\) + \(\large \frac {13}{y\,+\, 2}\) = 27 ; \(\large \frac {13}{2x \,+\, 1}\) + \(\large \frac {7}{y\,+\, 2}\) = 33
Solution:
Let,
\(\large \frac {7}{2x \,+\, 1}\) + \(\large \frac {13}{y\,+\, 2}\) = 27 …(i)
\(\large \frac {13}{2x \,+\, 1}\) + \(\large \frac {7}{y\,+\, 2}\) = 33 …(ii)
Substituting \(\large \frac {1}{2x \,+\, 1}\) = m and \(\large \frac {1}{y\,+\, 2}\) = n in equations (i) and (ii), we get,
7m + 13n = 27 …(iii)
13m + 7n = 33 …(iv)
Adding (iii) and (iv) we get,
7m + 13n = 27
+ 13m + 7n = 33
___________________
20m + 20n = 60
Dividing throughout by 20, we get,
m + n = 3 …(v)
Subtracting (iv) from (iii), we get
7m + 13n = 27
13m + 7n = 33
(–) (–) (–)
________________
– 6m + 6n = – 6
Dividing throughout by – 6
m – n = 1 …(vi)
Adding (v) and (vi) we get,
m + n = 3
+ m – n = 1
___________
2m = 4
∴ m = \(\large \frac {4}{2}\)
∴ m = 2
Substituting m = 2 in (v) we get,
m + n = 3
∴ 2 + n = 3
∴ n = 3 – 2
∴ n = 1
Resubstituting the values of m and n we get,
\(\large \frac {1}{2x \,+\, 1}\) = m
∴ \(\large \frac {1}{2x \,+\, 1}\) = 2
∴ 2(2x + 1) = 1
∴ 4x + 2 = 1
∴ 4x = 1 – 2
∴ 4x = – 1
∴ x = \(\large \frac {-\,1}{4}\)
\(\large \frac {1}{y\,+\, 2}\) = n
∴ \(\large \frac {1}{y\,+\, 2}\) = 1
∴ y + 2 = 1
∴ y = 1 – 2
∴ y = – 1
Ans: x = \(\large \frac {-\,1}{4}\) and y = – 1 is the solution of the given simultaneous equations.
(3) \(\large \frac {148}{x}\) + \(\large \frac {231}{y}\) = \(\large \frac {527}{xy}\) ; \(\large \frac {231}{x}\) + \(\large \frac {148}{y}\) = \(\large \frac {610}{xy}\)
Solution:
Let,
\(\large \frac {148}{x}\) + \(\large \frac {231}{y}\) = \(\large \frac {527}{xy}\) …(i)
\(\large \frac {231}{x}\) + \(\large \frac {148}{y}\) = \(\large \frac {610}{xy}\) …(ii)
Multiplying both the sides of equation (i) and equation (ii) by xy we get,
148y + 231x = 527
i.e. 231x + 148y = 527 …(iii)
231y + 148x = 610
i.e. 148x + 231y = 610 …(iv)
Adding (iii) and (iv) we get
231x + 148y = 527
+ 148x + 231y = 610
_______________________
379x + 379y = 1137
Dividing throughout by 379,
x + y = \(\large \frac {1137}{379}\)
∴ x + y = 3 …(v)
Subtracting (iv) from (iii) we get,
231x + 148y = 527
148x + 231y = 610
(–) (–) (–)
_______________________
83x – 83y = – 83
Dividing throughout by 83
x – y = –1 …(vi)
Adding (v) and (vi) we get,
x + y = 3
+ x – y = – 1
______________
2x = 2
∴ x = 1
Substituting x = 1 in equation (v) we get
x + y = 3
∴ 1 + y = 3
∴ y = 3 – 1
∴ y = 2
Ans: x = 1 and y = 2 is the solution of the given simultaneous equations.
(4) \(\large \frac {7x \,-\, 2y}{xy}\) = 5 ; \(\large \frac {8x\,+\, 7y}{xy}\) = 15
Solution:
\(\large \frac {7x \,-\, 2y}{xy}\) = 5
∴ \(\large \frac {7x}{xy}\) – \(\large \frac {2y}{xy}\)= 5
∴ \(\large \frac {7}{y}\) – \(\large \frac {2}{x}\) = 5
i.e. – \(\large \frac {2}{x}\) + \(\large \frac {7}{y}\) = 5 …(i)
\(\large \frac {8x\,+\, 7y}{xy}\) = 15
∴ \(\large \frac {8x}{xy}\) + \(\large \frac {7y}{xy}\)= 15
∴ \(\large \frac {8}{y}\) + \(\large \frac {7}{x}\) = 15
i.e. \(\large \frac {7}{x}\) + \(\large \frac {8}{y}\) = 15 …(ii)
Substituting \(\large \frac {1}{x}\) = m and \(\large \frac {1}{y}\) = n in equations (i) and (ii), we get,
– 2m + 7n = 5 …(iii)
7m + 8n = 15 …(iv)
Multiplying (iii) by 7 and (iv) by 2 we get,
– 14m + 49n = 35 …(v)
14m + 16n = 30 …(vi)
Adding (v) and (vi) we get,
– 14m + 49n = 35
+ 14m + 16n = 30
________________
65n = 65
∴ n = \(\large \frac {65}{65}\)
∴ n = 1
Substituting n = 1 in (iv) we get,
7m + 8n = 15
∴ 7m + 8 (1) = 15
∴ 7m + 8 = 15
∴ 7m = 15 – 8
∴ 7m = 7
∴ m = \(\large \frac {7}{7}\)
∴ m = 1
Resubstituting the values of m and n we get,
\(\large \frac {1}{x}\) = m
∴ \(\large \frac {1}{x}\) = 1
\(\large \frac {1}{y}\) = n
∴ \(\large \frac {1}{y}\) = 1
Ans: x = 1 and y = 1 is the solution of the given simultaneous equation.
(5) \(\large \frac {1}{2(3x \,+\, 4y)}\) + \(\large \frac {1}{5(2x\,-\, 3y)}\) = \(\large \frac {1}{4}\) ; \(\large \frac {5}{3x \,+\, 4y}\) – \(\large \frac {2}{2x\,-\, 3y}\) = – \(\large \frac {3}{2}\)
Solution:
\(\large \frac {1}{2(3x \,+\, 4y)}\) + \(\large \frac {1}{5(2x\,-\, 3y)}\) = \(\large \frac {1}{4}\)
Multiplying both sides by 20, we get,
\(\large \frac {20}{2(3x \,+\, 4y)}\) + \(\large \frac {20}{5(2x\,-\, 3y)}\) = \(\large \frac {20}{4}\)
∴ \(\large \frac {10}{3x \,+\, 4y}\) + \(\large \frac {4}{2x\,-\, 3y}\) = 5 …(i)
\(\large \frac {5}{3x \,+\, 4y}\) – \(\large \frac {2}{2x\,-\, 3y}\) = – \(\large \frac {3}{2}\)
Multiplying both sides by 2, we get,
∴ \(\large \frac {10}{3x \,+\, 4y}\) – \(\large \frac {4}{2x\,-\, 3y}\) = – 3 …(ii)
Substituting \(\large \frac {1}{3x \,+\, 4y}\) = m and \(\large \frac {1}{2x\,-\, 3y}\) = n in equations (i) and (ii), we get,
10m + 4n = 5 …(iii)
10m – 4n = – 3 …(iv)
Adding (iii) and (iv) we get
10m + 4n = 5
+ 10m – 4n = – 3
____________________
20m = 2
∴ m = \(\large \frac {2}{20}\)
∴ m = \(\large \frac {1}{10}\)
Substituting m = \(\large \frac {1}{10}\) in equation (iii), we get,
10m + 4n = 5
∴ 10 \(\large (\frac {1}{10})\) + 4n = 5
∴ 1 + 4n = 5
∴ 4n = 5 – 1
∴ 4n = 4
∴ n = \(\large \frac {4}{4}\)
∴ n = 1
Resubstituting the values of m and n, we get,
\(\large \frac {1}{3x \,+\, 4y}\) = m
∴ \(\large \frac {1}{3x \,+\, 4y}\) = \(\large \frac {1}{10}\)
∴ 3x + 4y = 10 …(v)
\(\large \frac {1}{2x\,-\, 3y}\) = n
∴ \(\large \frac {1}{2x\,-\, 3y}\) = 1
∴ 2x – 3y = 1 …(vi)
Multiplying equation (v) by 3 and equation (vi) by 4, we get,
9x + 12y = 30 (vii)
8x – 12y = 4 (viii)
Adding (vii) and (viii) we get
9x + 12y = 30
+ 8x – 12y = 4
_______________
17x = 34
∴ x = \(\large \frac {34}{17}\)
∴ x = 2
Substituting x = 2 in equation (vi) we get,
3x + 4y = 10
∴ 3(2) + 4y = 10
∴ 6 + 4y = 10
∴ 4y = 10 – 6
∴ 4y = 4
∴ y = \(\large \frac {4}{4}\)
∴ y = 1
Ans: x = 2 and y = 1 is the solution of the given simultaneous equations.
7. Solve the following word problems.
(1) A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
Let the digit in unit’s place is x and that in the ten’s place is y
∴ the number = □ y + x
The number obtained by interchanging the digits is □ x + y
According to first condition two digit number + the number obtained by interchanging the digits = 143
10y + x + □ = 143
□ x + □ y = 143
x + y = □ . . . . . (I)
From the second condition,
digit in unit’s place = digit in the ten’s place + 3
x = □ + 3
∴ x – y = 3 . . . . . (II)
Adding equations (I) and (II)
2x = □
x = 8
Putting this value of x in equation (I)
x + y = 13
8 + □ = 13
∴ y = □
The original number is 10 y + x
= □ + 8
= 58
Solution:
Let the digit in unit’s place is x and that in the ten’s place is y
∴ the number = 10 y + x
The number obtained by interchanging the digits is 10 x + y
According to first condition two digit number + the number obtained by interchanging the digits = 143
10y + x + 10x + y = 143
11 x + 11 y = 143
x + y = 13 . . . . . (I)
From the second condition,
digit in unit’s place = digit in the ten’s place + 3
x = y + 3
∴ x – y = 3 . . . . . (II)
Adding equations (I) and (II)
2x = 16
x = 8
Putting this value of x in equation (I)
x + y = 13
8 + y = 13
∴ y = 5
The original number is 10 y + x
= 50 + 8
= 58
(2) Kantabai bought 1 \(\large \frac {1}{2}\) kg tea and 5 kg sugar from a shop. She paid ₹ 50 as return fare for rickshaw. Total expense was ₹ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ₹ 880 for that. Find the rate of sugar and tea per kg.
Solution:
Let the cost of 1 kg of tea be ₹ x and 1 kg of sugar be ₹ y
Amount paid as Rickshaw fare = ₹ 50
∴ Total Amount spent by Kantabai = ₹ 700
∴ Total cost of tea and sugar = ₹ 700 – ₹ 50 = ₹ 650
According to first condition,
\(\large \frac {3}{2}\)x + 5y = 650
Multiplying both sides by 2
∴ 3x + 10y = 1300 …(i)
According to second condition,
2x + 7y = 880 …(ii)
Multiplying (i) by 2 and (ii) by 3 we get,
6x + 20y = 2600 …(iii)
6x + 21y = 2640 …(iv)
Subtracting equation (iv) from equation (iii), we get,
6x + 20y = 2600
6x + 21y = 2640
(–) (–) (–)
_______________________
– y = – 40
∴ y = 40
Substituting y = 40 in equation (ii), we get,
2x + 7y = 880
∴ 2x + 7(40) = 880
∴ 2x + 280 = 880
∴ 2x = 880 – 280
∴ 2x = 600
∴ x = \(\large \frac {600}{2}\)
∴ x = 300
Ans: The cost of 1 kg tea is ₹ 300 and 1 kg sugar is ₹ 40.
(3) To find the number of notes that Anushka had, complete the following activity.
Solution:
Let,
100x + 50y = 2500 …(i)
50x + 100y = 2000 …(ii)
Multiplying (ii) by 2 we get,
100x + 200y = 4000 …(iii)
Subtracting equation (ii) from equation (i)
100x + 50y = 2500
100x + 200y = 4000
(–) (–) (–)
_________________________
– 150 y = – 1500
150 y = 1500
∴ y = \(\large \frac {1500}{150}\)
∴ y = 10
Substituting y = 10 in equation (i) we get,
100x + 50y = 2500
∴ 100x + 50(10) = 2500
∴ 100x + 500 = 2500
∴ 100x = 2500 – 500
∴ 100x = 2000
∴ x = \(\large \frac {2000}{100}\)
∴ x = 20
Ans: Number of 100 rupees notes is 20 and number of 50 rupees notes is 10.
(4) Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Solution:
Let the present age of Manish be x years and the present age of Savita y years.
According to the first condition
x + y = 31 …(i)
Three years ago,
Manish’s Age = (x – 3) years and
Savita’s age = (y – 3) years
According to second condition,
x – 3 = 4 (y – 3)
∴ x – 3 = 4y – 12
∴ x – 4y = – 12 + 3
∴ x – 4y = – 9 …(ii)
Subtracting equation (ii) from equation (i) we get,
x + y = 31
x – 4y = – 9
(–) (+) (+)
________________
5y = 40
∴ y = \(\large \frac {40}{8}\)
∴ y = 8
Substituting y = 8 in equation (i) we get,
x + y = 31
∴ x + 8 = 31
∴ x = 31 – 8
∴ x = 23
Ans: The present age of Manisha is 23 years and the present age of Savita is 8 years.
(5) In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary for one day for both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Solution:
Let the per day salary of skilled workers be ₹ x and per day salary of unskilled workers be ₹ y.
According to the first condition
\(\large \frac {x}{y}\)x = \(\large \frac {5}{3}\)
3x = 5y
3x – 5y = 0 …(i)
According to the second condition
x + y = 720 …(ii)
Multiplying (ii) by 5 we get
5x + 5y = 3600 …(iii)
Adding (i) and (iii) we get,
5x + 5y = 3600
+ 3x – 5y = 0
_________________
8x = 3600
∴ x = \(\large \frac {3600}{7}\)
∴ x = 450
Substituting x = 450 in equation (ii), we get,
x + y = 720
∴ 450 + y = 720
∴ y = 720 – 450
∴ y = 270
Ans: Daily wages of skilled workers is ₹ 450 and that of unskilled workers is ₹ 270.
(6) Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Solution:
Let the speed at which Hamid is driving his motorcycle be x km/hr. and the speed at which Joseph is driving his motorcycle be y km/hr. [x > y]
Case I :
When they move towards each other.
In this case, both motorcycles are moving towards each other and they meet at C after 20 minutes i.e. \(\large \frac {20}{60}\) = \(\large \frac {1}{3}\) hours
Then AC + BC = AB i.e. distance covered by Hamid in \(\large \frac {1}{3}\) hours + distance covered by Joseph in \(\large \frac {1}{3}\) hours = 30 km.
According to the first condition,
\(\large \frac {1}{3}\)x + \(\large \frac {1}{3}\)y = 30
∴ \(\large \frac {1}{3}\) (x + y) = 30
∴ x + y = 30 × 3
∴ x + y = 90 …(i)
Case II :
When Joseph moves in the opposite direction. (Instead of towards A)
As shown in figure, both are moving in the same direction and they meet at C after 3 hours.
Then AC – BC = AB i.e. Distance covered by Hamid in 3 hours – distance covered by Joseph in 3 hours = 30km.
We know that,
Distance = speed × time
∴ 3x – 3y = 30
∴ x – y = 10 …(ii)
Adding (i) and (ii) we get,
x + y = 90
+ x – y = 10
_____________
2x = 100
∴ x = \(\large \frac {100}{2}\)
∴ x = 50
Substituting x = 50 in equation (i) we get
∴ x + y = 90
∴ 50 + y = 90
∴ y = 90 – 50
∴ y = 40
Ans: Speed of Hamid’s motorcycle = 50 km/hr and that of Joseph’s Motorcycle is 40 km/hr.