EXERCISE 3.2
Pair Of Linear Equations In Two Variables • 3 Questions
Question 1
Hint available
Solve the following pair of linear equations by the substitution method. (i) x + y = 14 (ii) s – t = 3 x – y = 4 6 3 2 s t (iii) 3x – y = 3 (iv) 0.2x + 0.3y = 1.3 9x – 3y = 9 0.4x + 0.5y = 2.3 (v) 2 3 0 x y (vi) 3 5 2 2 3 x y 3 8 0 x y 13 3 2 6 x y
Key Idea
The substitution method involves solving one of the equations for one variable and substituting that expression into the other equation(s). This reduces the system to a single equation in one variable, which can be solved easily. After finding the value of the first variable, substitute back to obtain the second variable. For systems with more than two equations, the same idea is applied repeatedly until all variables are determined.
Step-by-Step Solution
### (i) \(x + y = 14\)
1. Solve the given equation for one variable, e.g. \(x = 14 - y\).
2. Since there is no second independent equation, the solution set is expressed as \(\{(x, y) \mid x = 14 - y\}\).
Solution: \(x = 14 - y\) (infinitely many ordered pairs).
### (ii) \(\begin{cases}s - t = 3\\2s + t = 2\end{cases}\)
1. From the first equation, \(s = t + 3\).
2. Substitute \(s = t + 3\) into the second equation:
\[2(t + 3) + t = 2 \Rightarrow 2t + 6 + t = 2 \Rightarrow 3t = -4 \Rightarrow t = -\frac{4}{3}.\]
3. Substitute \(t\) back into \(s = t + 3\):
\[s = -\frac{4}{3} + 3 = \frac{5}{3}.\]
Solution: \(s = \frac{5}{3},\; t = -\frac{4}{3}.\)
### (iii) \(\begin{cases}3x - y = 3\\9x - 3y = 9\end{cases}\)
1. Notice that the second equation is exactly three times the first; therefore both equations represent the same straight line.
2. Solve the first equation for \(y\):
\[y = 3x - 3.\]
3. The system has infinitely many solutions lying on the line \(y = 3x - 3\).
Solution: \(\{(x, y) \mid y = 3x - 3\}\).
### (iv) \(\begin{cases}0.2x + 0.3y = 1.3\\0.4x + 0.5y = 2.3\end{cases}\)
1. Multiply the first equation by 2 to make the coefficient of \(x\) the same as in the second equation:
\[0.4x + 0.6y = 2.6.\]
2. Subtract the second original equation from this new equation:
\[(0.4x + 0.6y) - (0.4x + 0.5y) = 2.6 - 2.3\]
\[0.1y = 0.3 \Rightarrow y = 3.\]
3. Substitute \(y = 3\) into the first original equation:
\[0.2x + 0.3(3) = 1.3 \Rightarrow 0.2x + 0.9 = 1.3 \Rightarrow 0.2x = 0.4 \Rightarrow x = 2.\]
Solution: \(x = 2,\; y = 3.\)
### (v) \(\begin{cases}\frac{2}{3}x + y = 0\\x - 2y = 5\end{cases}\)
1. From the first equation, express \(y\) in terms of \(x\):
\[y = -\frac{2}{3}x.\]
2. Substitute this expression for \(y\) into the second equation:
\[x - 2\left(-\frac{2}{3}x\right) = 5 \Rightarrow x + \frac{4}{3}x = 5 \Rightarrow \frac{7}{3}x = 5 \Rightarrow x = \frac{15}{7}.\]
3. Find \(y\) using \(y = -\frac{2}{3}x\):
\[y = -\frac{2}{3}\times \frac{15}{7} = -\frac{30}{21} = -\frac{10}{7}.\]
Solution: \(x = \frac{15}{7},\; y = -\frac{10}{7}.\)
### (vi) \(\begin{cases}\frac{3}{5}x - 2y = -3\\\frac{8}{3}x - y = 13\\\frac{3}{2}x + 6y = 0\end{cases}\)
1. First pair (equations 1 & 2):
- From equation 1, solve for \(x\):
\[\frac{3}{5}x = 2y - 3 \Rightarrow x = \frac{5}{3}(2y - 3) = \frac{10}{3}y - 5.\]
- Substitute this \(x\) into equation 2:
\[\frac{8}{3}\left(\frac{10}{3}y - 5\right) - y = 13\]
\[\frac{80}{9}y - \frac{40}{3} - y = 13\]
\[\left(\frac{80}{9} - \frac{9}{9}\right) y = 13 + \frac{40}{3}\]
\[\frac{71}{9}y = \frac{79}{3}\]
\[y = \frac{79}{3}\times \frac{9}{71} = \frac{237}{71}.\]
- Then \(x = \frac{10}{3}y - 5 = \frac{10}{3}\times \frac{237}{71} - 5 = \frac{2370}{213} - 5 = \frac{2370 - 1065}{213} = \frac{1305}{213} = \frac{435}{71}.\]
2. Check with the third equation:
Substitute \(x = \frac{435}{71}\) and \(y = \frac{237}{71}\) into \(\frac{3}{2}x + 6y = 0\):
\[\frac{3}{2}\times \frac{435}{71} + 6\times \frac{237}{71} = \frac{1305}{142} + \frac{1422}{71} = \frac{1305}{142} + \frac{2844}{142} = \frac{4149}{142}
eq 0.\]
Hence the three equations are inconsistent; no common solution exists for all three simultaneously.
3. Therefore we solve any two of them (as done above) and state that the third equation is not compatible with the first two.
Solution (using equations 1 & 2): \(x = \frac{435}{71},\; y = \frac{237}{71}.\) The system of three equations has no solution.
Summary of Answers
- (i) \(x = 14 - y\) (infinitely many solutions)
- (ii) \(s = \frac{5}{3},\; t = -\frac{4}{3}\)
- (iii) \(y = 3x - 3\) (infinitely many solutions)
- (iv) \(x = 2,\; y = 3\)
- (v) \(x = \frac{15}{7},\; y = -\frac{10}{7}\)
- (vi) The first two equations give \(x = \frac{435}{71},\; y = \frac{237}{71}\); the third equation is inconsistent, so the three‑equation system has no solution.
1. Solve the given equation for one variable, e.g. \(x = 14 - y\).
2. Since there is no second independent equation, the solution set is expressed as \(\{(x, y) \mid x = 14 - y\}\).
Solution: \(x = 14 - y\) (infinitely many ordered pairs).
### (ii) \(\begin{cases}s - t = 3\\2s + t = 2\end{cases}\)
1. From the first equation, \(s = t + 3\).
2. Substitute \(s = t + 3\) into the second equation:
\[2(t + 3) + t = 2 \Rightarrow 2t + 6 + t = 2 \Rightarrow 3t = -4 \Rightarrow t = -\frac{4}{3}.\]
3. Substitute \(t\) back into \(s = t + 3\):
\[s = -\frac{4}{3} + 3 = \frac{5}{3}.\]
Solution: \(s = \frac{5}{3},\; t = -\frac{4}{3}.\)
### (iii) \(\begin{cases}3x - y = 3\\9x - 3y = 9\end{cases}\)
1. Notice that the second equation is exactly three times the first; therefore both equations represent the same straight line.
2. Solve the first equation for \(y\):
\[y = 3x - 3.\]
3. The system has infinitely many solutions lying on the line \(y = 3x - 3\).
Solution: \(\{(x, y) \mid y = 3x - 3\}\).
### (iv) \(\begin{cases}0.2x + 0.3y = 1.3\\0.4x + 0.5y = 2.3\end{cases}\)
1. Multiply the first equation by 2 to make the coefficient of \(x\) the same as in the second equation:
\[0.4x + 0.6y = 2.6.\]
2. Subtract the second original equation from this new equation:
\[(0.4x + 0.6y) - (0.4x + 0.5y) = 2.6 - 2.3\]
\[0.1y = 0.3 \Rightarrow y = 3.\]
3. Substitute \(y = 3\) into the first original equation:
\[0.2x + 0.3(3) = 1.3 \Rightarrow 0.2x + 0.9 = 1.3 \Rightarrow 0.2x = 0.4 \Rightarrow x = 2.\]
Solution: \(x = 2,\; y = 3.\)
### (v) \(\begin{cases}\frac{2}{3}x + y = 0\\x - 2y = 5\end{cases}\)
1. From the first equation, express \(y\) in terms of \(x\):
\[y = -\frac{2}{3}x.\]
2. Substitute this expression for \(y\) into the second equation:
\[x - 2\left(-\frac{2}{3}x\right) = 5 \Rightarrow x + \frac{4}{3}x = 5 \Rightarrow \frac{7}{3}x = 5 \Rightarrow x = \frac{15}{7}.\]
3. Find \(y\) using \(y = -\frac{2}{3}x\):
\[y = -\frac{2}{3}\times \frac{15}{7} = -\frac{30}{21} = -\frac{10}{7}.\]
Solution: \(x = \frac{15}{7},\; y = -\frac{10}{7}.\)
### (vi) \(\begin{cases}\frac{3}{5}x - 2y = -3\\\frac{8}{3}x - y = 13\\\frac{3}{2}x + 6y = 0\end{cases}\)
1. First pair (equations 1 & 2):
- From equation 1, solve for \(x\):
\[\frac{3}{5}x = 2y - 3 \Rightarrow x = \frac{5}{3}(2y - 3) = \frac{10}{3}y - 5.\]
- Substitute this \(x\) into equation 2:
\[\frac{8}{3}\left(\frac{10}{3}y - 5\right) - y = 13\]
\[\frac{80}{9}y - \frac{40}{3} - y = 13\]
\[\left(\frac{80}{9} - \frac{9}{9}\right) y = 13 + \frac{40}{3}\]
\[\frac{71}{9}y = \frac{79}{3}\]
\[y = \frac{79}{3}\times \frac{9}{71} = \frac{237}{71}.\]
- Then \(x = \frac{10}{3}y - 5 = \frac{10}{3}\times \frac{237}{71} - 5 = \frac{2370}{213} - 5 = \frac{2370 - 1065}{213} = \frac{1305}{213} = \frac{435}{71}.\]
2. Check with the third equation:
Substitute \(x = \frac{435}{71}\) and \(y = \frac{237}{71}\) into \(\frac{3}{2}x + 6y = 0\):
\[\frac{3}{2}\times \frac{435}{71} + 6\times \frac{237}{71} = \frac{1305}{142} + \frac{1422}{71} = \frac{1305}{142} + \frac{2844}{142} = \frac{4149}{142}
eq 0.\]
Hence the three equations are inconsistent; no common solution exists for all three simultaneously.
3. Therefore we solve any two of them (as done above) and state that the third equation is not compatible with the first two.
Solution (using equations 1 & 2): \(x = \frac{435}{71},\; y = \frac{237}{71}.\) The system of three equations has no solution.
Summary of Answers
- (i) \(x = 14 - y\) (infinitely many solutions)
- (ii) \(s = \frac{5}{3},\; t = -\frac{4}{3}\)
- (iii) \(y = 3x - 3\) (infinitely many solutions)
- (iv) \(x = 2,\; y = 3\)
- (v) \(x = \frac{15}{7},\; y = -\frac{10}{7}\)
- (vi) The first two equations give \(x = \frac{435}{71},\; y = \frac{237}{71}\); the third equation is inconsistent, so the three‑equation system has no solution.
Question 2
Hint available
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Key Idea
Use the method of elimination (or substitution) to solve the pair of linear equations. Once the ordered pair (x, y) is obtained, substitute it in the given linear equation y = mx + 3 to determine the required value of m.
Step-by-Step Solution
1. Write the two equations\
\[\begin{aligned}
& (1)\; 2x + 3y = 11 \\
& (2)\; 2x - 4y = -24
\end{aligned}\]
2. Eliminate one variable – subtract equation (2) from equation (1):\
\[ (2x+3y) - (2x-4y) = 11 - (-24) \]
\[ 2x + 3y - 2x + 4y = 35 \]
\[ 7y = 35 \]
\[ y = 5 \]
3. Find x by substituting y = 5 in either equation (use (1)):\
\[ 2x + 3(5) = 11 \]
\[ 2x + 15 = 11 \]
\[ 2x = -4 \]
\[ x = -2 \]
Hence the solution of the simultaneous equations is \[ (x, y) = (-2,\;5). \]
4. Find m such that the line \( y = mx + 3 \) passes through the point \((-2,5)\). Substitute x = -2, y = 5:\
\[ 5 = m(-2) + 3 \]
\[ 5 = -2m + 3 \]
\[ -2m = 2 \]
\[ m = -1 \]
5. Result: The required value of \( m \) is \( -1 \).
\[\begin{aligned}
& (1)\; 2x + 3y = 11 \\
& (2)\; 2x - 4y = -24
\end{aligned}\]
2. Eliminate one variable – subtract equation (2) from equation (1):\
\[ (2x+3y) - (2x-4y) = 11 - (-24) \]
\[ 2x + 3y - 2x + 4y = 35 \]
\[ 7y = 35 \]
\[ y = 5 \]
3. Find x by substituting y = 5 in either equation (use (1)):\
\[ 2x + 3(5) = 11 \]
\[ 2x + 15 = 11 \]
\[ 2x = -4 \]
\[ x = -2 \]
Hence the solution of the simultaneous equations is \[ (x, y) = (-2,\;5). \]
4. Find m such that the line \( y = mx + 3 \) passes through the point \((-2,5)\). Substitute x = -2, y = 5:\
\[ 5 = m(-2) + 3 \]
\[ 5 = -2m + 3 \]
\[ -2m = 2 \]
\[ m = -1 \]
5. Result: The required value of \( m \) is \( -1 \).
Question 3
Hint available
Form the pair of 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. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? (v) A fraction becomes 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 5 6 . Find the fraction. 34 (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? 3.3.2 Elimination Method Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.
Key Idea
Introduce variables for the unknown quantities, translate the word statements into linear equations, and then solve the resulting pair of equations using the substitution method as described in the NCERT textbook.
Step-by-Step Solution
### (i) Two numbers
Let the two numbers be $x$ and $y$.
- Difference: $x - y = 26$ … (1)
- One number is three times the other: $x = 3y$ … (2)
Substitution: From (2), $x = 3y$. Substitute in (1):
$$3y - y = 26 \ \ 2y = 26 \ y = 13$$
Then $x = 3y = 39$.\
Answer: The numbers are $39$ and $13$.
### (ii) Supplementary angles
Let the smaller angle be $x^{\circ}$ and the larger be $y^{\circ}$.
- Supplementary: $x + y = 180$ … (1)
- Larger exceeds smaller by $18^{\circ}$: $y = x + 18$ … (2)
Substitution: Put (2) in (1):
$$x + (x + 18) = 180 \ 2x + 18 = 180 \ 2x = 162 \ x = 81$$
Hence $y = 81 + 18 = 99$.\
Answer: Angles are $81^{\circ}$ and $99^{\circ}$.
### (iii) Cost of bats and balls
Let cost of one bat = $b$ rupees, cost of one ball = $c$ rupees.
- First purchase: $7b + 6c = 3800$ … (1)
- Second purchase: $3b + 5c = 1750$ … (2)
Substitution: From (2), express $b$:
$$3b = 1750 - 5c \ b = \frac{1750 - 5c}{3}$$
Substitute in (1):
$$7\left(\frac{1750 - 5c}{3}\right) + 6c = 3800$$
Multiply by 3:
$$7(1750 - 5c) + 18c = 11400$$
$$12250 - 35c + 18c = 11400$$
$$-17c = -850 \ c = 50$$
Then $b = \frac{1750 - 5\times 50}{3} = \frac{1750 - 250}{3} = \frac{1500}{3} = 500$.\
Answer: One bat costs Rs. 500 and one ball costs Rs. 50.
### (iv) Taxi charges
Let fixed charge = $F$ rupees, charge per km = $p$ rupees.
- For 10 km: $F + 10p = 105$ … (1)
- For 15 km: $F + 15p = 155$ … (2)
Substitution: From (1), $F = 105 - 10p$. Substitute in (2):
$$105 - 10p + 15p = 155 \ 105 + 5p = 155 \ 5p = 50 \ p = 10$$
Then $F = 105 - 10\times 10 = 5$.\
Charges: Fixed charge = Rs. 5, charge per km = Rs. 10.
Cost for 25 km: $F + 25p = 5 + 25\times 10 = 5 + 250 = 255$ rupees.
### (v) Fraction problem
Let the required fraction be $\frac{x}{y}$ (with $x, y$ positive integers).
- Adding 2 to numerator and denominator gives $\frac{x+2}{y+2}=\frac{9}{11}$ … (1)
- Adding 3 gives $\frac{x+3}{y+3}=\frac{5}{6}$ … (2)
From (1): Cross‑multiply:
$$11(x+2) = 9(y+2) \ 11x + 22 = 9y + 18 \ 11x - 9y = -4 \ \ (A)$$
From (2):
$$6(x+3) = 5(y+3) \ 6x + 18 = 5y + 15 \ 6x - 5y = -3 \ \ (B)$$
Substitution: Solve (B) for $x$: $6x = 5y - 3 \ x = \frac{5y - 3}{6}$.
Substitute in (A):
$$11\left(\frac{5y - 3}{6}\right) - 9y = -4$$
Multiply by 6:
$$11(5y - 3) - 54y = -24 \ 55y - 33 - 54y = -24 \ y - 33 = -24 \ y = 9$$
Then $x = \frac{5\times 9 - 3}{6} = \frac{45 - 3}{6} = \frac{42}{6} = 7$.
Answer: The fraction is $\frac{7}{9}$.
### (vi) Ages of Jacob and his son
Let Jacob’s present age = $J$ years, son’s present age = $S$ years.
- Five years hence: $J+5 = 3(S+5)$ … (1)
- Five years ago: $J-5 = 7(S-5)$ … (2)
From (1): $J+5 = 3S + 15 \ J = 3S + 10$ … (3)
Substitution: Put (3) in (2):
$$(3S + 10) - 5 = 7(S - 5) \ 3S + 5 = 7S - 35 \ 5 + 35 = 7S - 3S \ 40 = 4S \ S = 10$$
Then $J = 3\times 10 + 10 = 40$.\
Answer: Jacob is 40 years old and his son is 10 years old.
Let the two numbers be $x$ and $y$.
- Difference: $x - y = 26$ … (1)
- One number is three times the other: $x = 3y$ … (2)
Substitution: From (2), $x = 3y$. Substitute in (1):
$$3y - y = 26 \ \ 2y = 26 \ y = 13$$
Then $x = 3y = 39$.\
Answer: The numbers are $39$ and $13$.
### (ii) Supplementary angles
Let the smaller angle be $x^{\circ}$ and the larger be $y^{\circ}$.
- Supplementary: $x + y = 180$ … (1)
- Larger exceeds smaller by $18^{\circ}$: $y = x + 18$ … (2)
Substitution: Put (2) in (1):
$$x + (x + 18) = 180 \ 2x + 18 = 180 \ 2x = 162 \ x = 81$$
Hence $y = 81 + 18 = 99$.\
Answer: Angles are $81^{\circ}$ and $99^{\circ}$.
### (iii) Cost of bats and balls
Let cost of one bat = $b$ rupees, cost of one ball = $c$ rupees.
- First purchase: $7b + 6c = 3800$ … (1)
- Second purchase: $3b + 5c = 1750$ … (2)
Substitution: From (2), express $b$:
$$3b = 1750 - 5c \ b = \frac{1750 - 5c}{3}$$
Substitute in (1):
$$7\left(\frac{1750 - 5c}{3}\right) + 6c = 3800$$
Multiply by 3:
$$7(1750 - 5c) + 18c = 11400$$
$$12250 - 35c + 18c = 11400$$
$$-17c = -850 \ c = 50$$
Then $b = \frac{1750 - 5\times 50}{3} = \frac{1750 - 250}{3} = \frac{1500}{3} = 500$.\
Answer: One bat costs Rs. 500 and one ball costs Rs. 50.
### (iv) Taxi charges
Let fixed charge = $F$ rupees, charge per km = $p$ rupees.
- For 10 km: $F + 10p = 105$ … (1)
- For 15 km: $F + 15p = 155$ … (2)
Substitution: From (1), $F = 105 - 10p$. Substitute in (2):
$$105 - 10p + 15p = 155 \ 105 + 5p = 155 \ 5p = 50 \ p = 10$$
Then $F = 105 - 10\times 10 = 5$.\
Charges: Fixed charge = Rs. 5, charge per km = Rs. 10.
Cost for 25 km: $F + 25p = 5 + 25\times 10 = 5 + 250 = 255$ rupees.
### (v) Fraction problem
Let the required fraction be $\frac{x}{y}$ (with $x, y$ positive integers).
- Adding 2 to numerator and denominator gives $\frac{x+2}{y+2}=\frac{9}{11}$ … (1)
- Adding 3 gives $\frac{x+3}{y+3}=\frac{5}{6}$ … (2)
From (1): Cross‑multiply:
$$11(x+2) = 9(y+2) \ 11x + 22 = 9y + 18 \ 11x - 9y = -4 \ \ (A)$$
From (2):
$$6(x+3) = 5(y+3) \ 6x + 18 = 5y + 15 \ 6x - 5y = -3 \ \ (B)$$
Substitution: Solve (B) for $x$: $6x = 5y - 3 \ x = \frac{5y - 3}{6}$.
Substitute in (A):
$$11\left(\frac{5y - 3}{6}\right) - 9y = -4$$
Multiply by 6:
$$11(5y - 3) - 54y = -24 \ 55y - 33 - 54y = -24 \ y - 33 = -24 \ y = 9$$
Then $x = \frac{5\times 9 - 3}{6} = \frac{45 - 3}{6} = \frac{42}{6} = 7$.
Answer: The fraction is $\frac{7}{9}$.
### (vi) Ages of Jacob and his son
Let Jacob’s present age = $J$ years, son’s present age = $S$ years.
- Five years hence: $J+5 = 3(S+5)$ … (1)
- Five years ago: $J-5 = 7(S-5)$ … (2)
From (1): $J+5 = 3S + 15 \ J = 3S + 10$ … (3)
Substitution: Put (3) in (2):
$$(3S + 10) - 5 = 7(S - 5) \ 3S + 5 = 7S - 35 \ 5 + 35 = 7S - 3S \ 40 = 4S \ S = 10$$
Then $J = 3\times 10 + 10 = 40$.\
Answer: Jacob is 40 years old and his son is 10 years old.