Excercise 3.1
Excercise 3.2
1. Solve the following pair of linear equations by the substitution method.
(i)
\(x + y = 14\)
\(x – y = 4\)
Solution: Let,
\(x + y = 14 \quad \text{—— (i)}\)
\(x – y = 4 \quad \text{—— (ii)}\)
From equation (i)
\(x = 14 – y \quad \text{—— (iii)}\)
Now, substituting the value of \(x\) on (ii)
\((14 – y) – y = 4\)
\(\Rightarrow -2y = 4 – 14\)
\(\Rightarrow -2y = -10\)
\(\Rightarrow y = \frac{10}{2}\)
\( y = 5\)
Now, \(y = 5\) substituting on equation (iii)
We get,
\(x = 14 – 5\)
\(x = 9\)
The values of \(x = 9\) and \(y = 5\)
(ii) \(s – t = 3\)
\(\frac{s}{3} + \frac{t}{2} = 6\)
Solution: Let,
\(s – t = 3 \quad \text{—— (i)}\)
\(\frac{s}{3} + \frac{t}{2} = 6 \quad \text{—— (ii)}\)
From equation (i)
\(s – t = 3\)
\(s = 3 + t \quad \text{—— (iii)}\)
Now, substituting the value of \(s\) on equation (ii)
\(\frac{3 + t}{3} + \frac{t}{2} = 6\)
\(\frac{(3 + t)2 + 3t}{6} = 6\)
\(\Rightarrow 6 + 2t + 3t = 36\)
\(\Rightarrow 5t = 30\)
\(t = 6\)
Now, substituting the value of \(t = 6\) on (iii) we get,
\(s = 3 + 6\)
\(s = 9\)
Therefore values of \(s = 9\) and \(t = 6\)
(iii) \(3x – y = 3\)
\(9x – 3y = 9\)
Solution:
\(3x – y = 3 \quad \text{—— (i)}\)
\(9x – 3y = 9 \quad \text{—— (ii)}\)
From equation (i)
\(3x = 3 + y\)
\(x = \frac{3 + y}{3} \quad \text{—— (iii)}\)
Now, substituting the value of \(x\) on equation (ii)
\(3 \times \left( \frac{3 + y}{3} \right) – 3y = 9\)
\(\Rightarrow 9 + 3y – 3y = 9\)
\(\Rightarrow 9 = 9\)
Infinite solutions.
(iv) \(0.2x + 0.3y = 1.3\)
\(0.4x + 0.5y = 2.3\)
Solution: Given,
\(0.2x + 0.3y = 1.3\)
multiplying both sides with 10
\(2x + 3y = 13 \quad \text{—— (i)}\)
and
\(0.4x + 0.5y = 2.3\)
multiplying both sides with 10
\(4x + 5y = 23 \quad \text{—— (ii)}\)
Now, from equation (i)
\(2x = 13 – 3y\)
\(\Rightarrow x = \frac{13 – 3y}{2} \quad \text{—— (iii)}\)
Now, substituting the value of \(x\) on (ii)
\(\Rightarrow 2 \times \left( \frac{13 – 3y}{2} \right) + 5y = 23\)
\(\Rightarrow 26 – 6y + 5y = 23 \times 2\)
\(\Rightarrow -y = 23 – 26\)
\(\Rightarrow -y = -3\)
\(y = 3\)
Now, \(y = 3\) value, substituting on equation (iii)
\(x = \frac{13 – 3(3)}{2}\)
\(x = \frac{13 – 9}{2} = \frac{4}{2}\)
\(x = 2\)
Therefore, the values are \(x = 2\) and \(y = 3\).
(v) \(\sqrt{2}x + \sqrt{3}y = 0\)
\(\sqrt{3}x – \sqrt{8}y = 0\)
Solution: Let,
\(\sqrt{2}x + \sqrt{3}y = 0 \quad \text{—— (i)}\)
\(\sqrt{3}x – \sqrt{8}y = 0 \quad \text{—— (ii)}\)
From equation (i),
\(\sqrt{2}x + \sqrt{3}y = 0\)
\(\Rightarrow \sqrt{2}x = -\sqrt{3}y\)
\(\Rightarrow x = \frac{-\sqrt{3}y}{\sqrt{2}} \quad \text{—— (iii)}\)
Now, Substituting the value of \(x = \frac{-\sqrt{3}y}{\sqrt{2}}\) on (ii),
\(\sqrt{3} \times \left( \frac{-\sqrt{3}y}{\sqrt{2}} \right) – \sqrt{8}y = 0\)
\(\Rightarrow \frac{-3y}{\sqrt{2}} – \sqrt{8}y = 0\)
\(\Rightarrow \frac{-3y – \sqrt{16}y}{\sqrt{2}} = 0\)
\(\Rightarrow -3y – 4y = 0\)
\(\Rightarrow -7y = 0\)
\(\Rightarrow y = 0\)
Now, Substituting the value of \(y = 0\) on (iii),
\(x = \frac{-\sqrt{3} \times 0}{\sqrt{2}}\)
\(\Rightarrow x = 0\)
Therefore, the values are \(x = 0\) and \(y = 0\).
(vi) \(\frac{3x}{2} – \frac{5y}{3} = -2\)
\(\frac{x}{3} + \frac{y}{2} = \frac{13}{6}\)
Solution: Given,
\(\frac{3x}{2} – \frac{5y}{3} = -2\)
multiplying both sides with 6
\(6 \times \left( \frac{3x}{2} – \frac{5y}{3} \right) = -2 \times 6\)
\(\Rightarrow 9x – 10y = -12 \quad \text{—— (i)}\)
And,
\(\frac{x}{3} + \frac{y}{2} = \frac{13}{6}\)
multiplying both sides with 6
\(6 \times \left( \frac{x}{3} + \frac{y}{2} \right) = \frac{13}{6} \times 6\)
\(2x + 3y = 13 \quad \text{—— (ii)}\)
Now, From equation (i)
\(9x – 10y = -12\)
\(\Rightarrow 9x = -12 + 10y\)
\(\Rightarrow x = \frac{-12 + 10y}{9} \quad \text{—— (iii)}\)
Substituting the value of \(x\) on equation (ii)
\(\Rightarrow 2 \times \left( \frac{-12 + 10y}{9} \right) + 3y = 13\)
\(\Rightarrow \frac{-24 + 20y}{9} + 3y = 13\)
\(\Rightarrow \frac{-24 + 20y + 27y}{9} = 13\)
\(\Rightarrow -24 + 47y = 117\)
\(\Rightarrow 47y = 117 + 24\)
\(\Rightarrow 47y = 141\)
\(\Rightarrow y = \frac{141}{47}\)
\(\Rightarrow y = 3\)
Now, substituting the value of \(y = 3\) on equation (iii)
\(x = \frac{-12 + 10 \times 3}{9}\)
\(\Rightarrow x = \frac{-12 + 30}{9}\)
\(\Rightarrow x = \frac{18}{9}\)
\( x = 2\)
Therefore, the values are \(x = 2\) and \(y = 3\).
2. Solve \(2x + 3y = 11\) and \(2x – 4y = -24\) and hence find the value of ‘m’ for which \(y = mx + 3\)
Solution: Let,
\(2x + 3y = 11 \quad \text{—— (i)}\)
\(2x – 4y = -24 \quad \text{—— (ii)}\)
From equation (i)
\(2x + 3y = 11\)
\(\Rightarrow 2x = 11 – 3y\)
\(x = \frac{11 – 3y}{2} \quad \text{—— (iii)}\)
Now, substituting value of \(x\) on (ii)
\(2 \left( \frac{11 – 3y}{2} \right) – 4y = -24\)
\(11 – 3y – 4y = -24\)
\(-7y = -24 – 11\)
\(\Rightarrow -7y = -35\)
\(\Rightarrow y = \frac{-35}{-7}\)
\(y = 5\)
Now Substituting the value of \(y = 5\) on equation (iii)
\(x = \frac{11 – 3(5)}{2} = \frac{11 – 15}{2} = \frac{-4}{2}\)
\(x = -2\)
Now,
\(y = mx + 3\)
\(\Rightarrow 5 = m \times (-2) + 3\)
\(\Rightarrow 5 – 3 = -2m\)
\(\Rightarrow +2 = -2m\)
\(\Rightarrow m = -1\)
Therefore, the value of m is -1.
3. Form the pair linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
Solution: Let, one number be \(x\) and the other one be \(y\)
According to the question,
\(x – y = 26 \quad \text{—— (i)}\)
and, \(x = 3y \quad \text{—— (ii)}\)
From, equation (i)
\(x = 26 + y \quad \text{—— (iii)}\)
Substituting the value of \(x\) on equation (ii)
\(26 + y = 3y\)
\(\Rightarrow 26 = 2y\)
\(y = 13\)
Now, again substituting value of \(y\) on \(x\) of (ii)
We get, \(x = 26 + 13\)
\(\Rightarrow x = 39\)
Therefore, the values of the two variables are \(x = 39\) and \(y = 13\).
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution: Let, the two supplementary angles be \(x\) and \(y\).
According to the questions
\(x + y = 180^{\circ} \quad \text{—— (i)}\)
and \(x = y + 18^{\circ} \quad \text{—— (ii)}\)
The value of \(x\) of equation (ii) substituting on equation (i), we get
\(y + 18^{\circ} + y = 180^{\circ}\)
\(\Rightarrow 2y = 180^{\circ} – 18^{\circ}\)
\(\Rightarrow 2y = 162^{\circ}\)
\(\Rightarrow y = \frac{162^{\circ}}{2}\)
\(\Rightarrow y = 81^{\circ}\)
Now, Substituting the value of \(y\) on equation (ii)
We get \(x = 81^{\circ} + 18^{\circ}\)
\(x = 99^{\circ}\)
Therefore, the two supplementary angles are \(x = 99^{\circ}\) and \(y = 81^{\circ}\) respectively.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for 1750. Find the cost of each bat and each ball.
Solution: Let, one cricket bat costs \(x\) and one cricket ball costs \(y\)
According to the questions
\(7x + 6y = 3800 \quad \text{—— (i)}\)
and \(3x + 5y = 1750 \quad \text{—— (ii)}\)
From equation (i), we get
\(7x + 6y = 3800\)
\(\Rightarrow 7x = 3800 – 6y\)
\(x = \frac{3800 – 6y}{7} \quad \text{—— (iii)}\)
Now, Substituting the value of \(x\) on equation (ii), we get
\(3 \times \left( \frac{3800 – 6y}{7} \right) + 5y = 1750\)
\(\Rightarrow 11400 – 18y + 35y = 12250\)
\(\Rightarrow 17y = 12250 – 11400\)
\(\Rightarrow 17y = 850\)
\(y = \frac{850}{17}\)
\(y = 50\)
Now, again substituting the value of \(y\) on equation (iii) we get,
\(x = \frac{3800 – 6 \times 50}{7}\)
\(x = \frac{3800 – 300}{7} = \frac{3500}{7}\)
\(x = 500\)
Therefore, the cost of each bat is 500 rupees and the cost of each ball is 50 rupees.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10km, the charge paid is Rs. 105 and for a journey of 15km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25km?
Solution: Let, the fixed charge be \(x\) and the charge per a distance be \(y\)
According to the question,
\(x + 10y = 105 \quad \text{—— (i)}\)
and, \(x + 15y = 155 \quad \text{—— (ii)}\)
From equation (i)
we get, \(x = 105 – 10y \quad \text{—— (iii)}\)
Now, substituting the value of \(x\) on equation (ii) we get,
\((105 – 10y) + 15y = 155\)
\(\Rightarrow 105 – 10y + 15y = 155\)
\(\Rightarrow 5y = 155 – 105\)
\(y = \frac{50}{5}\)
\(y = 10\)
Now, again substituting the value of \(y\) on (iii) we get
\(x = 105 – 10 \times 10\)
\(x = 5\)
Therefore, the fare of the taxi the fixed charge \(x = 5\) rupees and the charge for distance covered \(y = 10\) rupees per km.
(v) A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
Solution: Let, the numerator be \(x\) and the denominator be \(y\)
According to the question,
\(\frac{x + 2}{y + 2} = \frac{9}{11}\)
\(\ (x + 2) \times 11 = 9 \times (y + 2)\)
\(\Rightarrow 11x + 22 = 9y + 18\)
\(\Rightarrow 11x – 9y + 4 = 0 \quad \text{—— (i)}\)
And,
\(\frac{x + 3}{y + 3} = \frac{5}{6}\)
\((x + 3) \times 6 = 5(y + 3)\)
\(\Rightarrow 6x + 18 = 5y + 15\)
\(\Rightarrow 6x – 5y + 3 = 0 \quad \text{—— (ii)}\)
From equation (i) we get
\(11x = 9y – 4\)
\(x = \frac{9y – 4}{11} \quad \text{—— (iii)}\)
Substituting the value of \(x\) on equation (ii) we get,
\(6 \times \left( \frac{9y – 4}{11} \right) – 5y + 3 = 0\)
\(\Rightarrow 54y – 24 – 55y + 33 = 0\)
\(\Rightarrow -y = -9\)
\(\Rightarrow y = 9\)
Now, the value of \(y\) substituting on equation (iii) we get,
\(x = \frac{9 \times 9 – 4}{11}\)
\(\Rightarrow x = \frac{81 – 4}{11}\)
\(\Rightarrow x = \frac{77}{11}\)
\(\Rightarrow x = 7\)
The value of numerator is \(x = 7\)
denominator is \(y = 9\)
And the fraction is \(\frac{7}{9}\).
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution: Let, the present age of Jacob be \(x\)
and the present age of his son be \(y\).
Five years hence,
Jacob’s age will be \(x + 5\)
His son’s age will be \(3(y + 5) = (x + 5)\)
\(\Rightarrow 3y + 15 = x + 5 \quad \text{—— (i)}\)
Five years ago,
Jacob’s age was \(x – 5\)
His son’s age was \(7(y – 5) = (x – 5)\)
\(\Rightarrow 7y – 35 = x – 5 \quad \text{—— (ii)}\)
Now, From equation (i):
\(3y = x + 5 – 15\)
\(\Rightarrow y = \frac{x – 10}{3} \quad \text{—— (iii)}\)
Now, Substituting the value of \(y\) on equation (ii):
\(7 \times \left( \frac{x – 10}{3} \right) – 35 = x – 5\)
\(\Rightarrow \frac{7x – 70 – 105}{3} = x – 5\)
\(\Rightarrow 7x – 175 = 3(x – 5)\)
\(\Rightarrow 7x – 175 = 3x – 15\)
\(\Rightarrow 4x = 160\)
\(\Rightarrow x = \frac{160}{4}\)
\(\Rightarrow x = 40\)
Now, the value of \(x\) substituting on equation (iii) we get,
\(y = \frac{40 – 10}{3}\)
\(\Rightarrow y = \frac{30}{3}\)
\(\Rightarrow y = 10\)
Therefore, the present age of Jacob is \(x = 40\) years and his son’s age is \(y = 10\) years.
4. If \(x + 3\) is a factor of \(x^3 + ax^2 – bx + 6\) and \(a + b = 7\), find the values of \(a\) and \(b\).
Solution: Given, \(x + 3\) is a factor of \(x^3 + ax^2 – bx + 6\)
Since, \(x + 3 = 0 \Rightarrow x = -3\)
Substituting the value of \(x\) on the equation, we get:
latex^3 + a(-3)^2 – b(-3) + 6 = 0[/latex]
\(\Rightarrow -27 + 9a + 3b + 6 = 0\)
\(\Rightarrow -21 + 3(3a + b) = 0\)
\(\Rightarrow 3(3a + b) = +21\)
\(\Rightarrow 3a + b = 7 \quad \text{—— (i)}\)
Given that,
\(a + b = 7 \quad \text{—— (ii)}\)
Subtracting equation (ii) from equation (i):
\((3a + b) – (a + b) = 7 – 7\)
\(\Rightarrow 2a = 0\)
\(\Rightarrow a = 0\)
Substituting the value of \(a\) on equation (ii):
\(3(0) + b = 7\)
\(\Rightarrow b = 7\)
Therefore, the values are \(a = 0\) and \(b = 7\).
5. Consider the following pair of linear equations of two variables, then choose the correct option from the given alternatives.
\(2x = y \quad \text{and} \quad -5x + 2y – 3 = 0\)
(i) The graphs of the equations intersect at a point.
(ii) The graphs of the equations are parallel to each other.
(iii) The graphs of the equations coincide.
(iv) The equations have a unique solution.
(a) Both (i) and (ii) are true
(b) Both (i) and (iii) are true
(c) Both (ii) and (iii) are true
(d) Both (i) and (iv) are true
Ans: (d) Both (i) and (iv) are true
6. Match the items of column I with column II and then choose the correct option.
| Column I | Column II |
|---|---|
A. \(x – y = 0\) B. \(2x – 3y = 5\) and \(x – y = 1\) C. \(x + 2y = 6\) and \(4x + 8y = 24\) D. \(2x + 3y = 6\) and \(4x + 6y = 10\) | (i) unique solution (ii) linear equation in two variables (iii) No solution (iv) Infinitely many solutions and lines coincide |
(a) A \(\rightarrow\) (ii), B \(\rightarrow\) (i), C \(\rightarrow\) (iv), D \(\rightarrow\) (iii)
(b) A \(\rightarrow\) (ii), B \(\rightarrow\) (iv), C \(\rightarrow\) (i), D \(\rightarrow\) (iii)
(c) A \(\rightarrow\) (iv), B \(\rightarrow\) (i), C \(\rightarrow\) (ii), D \(\rightarrow\) (iii)
(d) A \(\rightarrow\) (iv), B \(\rightarrow\) (ii), C \(\rightarrow\) (i), D \(\rightarrow\) (iii)
Ans: (a) A \(\rightarrow\) (ii), B \(\rightarrow\) (i), C \(\rightarrow\) (iv), D \(\rightarrow\) (iii)
7. The value of \(x\) for which the pair \((x, 4)\) satisfies the equation \(3x + y = 19\) is:
(a) 6
(b) 5
(c) 3
(d) 4
Ans: (b) 5
8. Two equations are given as \(2x + 4y = 10\) and \(kx + 8y = 20\), for which value of \(k\) will the system have infinitely many solutions?
(a) 4
(b) 3
(c) 2
(d) 1
Ans: (a) 4