EXAMPLES
Pair Of Linear Equations In Two Variables • 10 Questions
Question 1
Hint available
Check graphically whether the pair of equations x + 3y = 6 (1) and 2x – 3y = 12 (2) is consistent. If so, solve them graphically.
Key Idea
A pair of linear equations is *consistent* if their graphs intersect at a point. The coordinates of the intersection give the solution of the system. To check graphically we plot each line using two convenient points, draw the lines, and observe whether they meet.
Step-by-Step Solution
1. Rewrite each equation in intercept form
- For (1) : $x + 3y = 6 \Rightarrow 3y = 6 - x \Rightarrow y = 2 - \frac{x}{3}$
- For (2) : $2x - 3y = 12 \Rightarrow -3y = 12 - 2x \Rightarrow y = \frac{2x - 12}{3}$
2. Find two points on each line
- Line (1)
*When $x = 0$*: $3y = 6 \Rightarrow y = 2$ → point $A(0,2)$
*When $y = 0$*: $x = 6$ → point $B(6,0)$
- Line (2)
*When $x = 0$*: $-3y = 12 \Rightarrow y = -4$ → point $C(0,-4)$
*When $y = 0$*: $2x = 12 \Rightarrow x = 6$ → point $D(6,0)$
3. Plot the points on the Cartesian plane
- Mark $A(0,2)$ and $B(6,0)$ and join them – this is the graph of equation (1).
- Mark $C(0,-4)$ and $D(6,0)$ and join them – this is the graph of equation (2).
4. Observe the intersection
- Both lines pass through the common point $B(6,0)$. Hence the two lines intersect at a single point.
5. Conclusion about consistency
- Since the lines intersect, the pair of equations is *consistent* (they have a unique solution).
6. Solution (graphical coordinates of intersection)
- The intersection point is $\boxed{(6,\,0)}$.
7. Verification (optional algebraic check)
- Substituting $x=6, y=0$ in (1): $6 + 3\times0 = 6$ ✓
- Substituting in (2): $2\times6 - 3\times0 = 12$ ✓
- Hence the graphical solution matches the algebraic solution.
- For (1) : $x + 3y = 6 \Rightarrow 3y = 6 - x \Rightarrow y = 2 - \frac{x}{3}$
- For (2) : $2x - 3y = 12 \Rightarrow -3y = 12 - 2x \Rightarrow y = \frac{2x - 12}{3}$
2. Find two points on each line
- Line (1)
*When $x = 0$*: $3y = 6 \Rightarrow y = 2$ → point $A(0,2)$
*When $y = 0$*: $x = 6$ → point $B(6,0)$
- Line (2)
*When $x = 0$*: $-3y = 12 \Rightarrow y = -4$ → point $C(0,-4)$
*When $y = 0$*: $2x = 12 \Rightarrow x = 6$ → point $D(6,0)$
3. Plot the points on the Cartesian plane
- Mark $A(0,2)$ and $B(6,0)$ and join them – this is the graph of equation (1).
- Mark $C(0,-4)$ and $D(6,0)$ and join them – this is the graph of equation (2).
4. Observe the intersection
- Both lines pass through the common point $B(6,0)$. Hence the two lines intersect at a single point.
5. Conclusion about consistency
- Since the lines intersect, the pair of equations is *consistent* (they have a unique solution).
6. Solution (graphical coordinates of intersection)
- The intersection point is $\boxed{(6,\,0)}$.
7. Verification (optional algebraic check)
- Substituting $x=6, y=0$ in (1): $6 + 3\times0 = 6$ ✓
- Substituting in (2): $2\times6 - 3\times0 = 12$ ✓
- Hence the graphical solution matches the algebraic solution.
Question 2
Hint available
Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions: 5x – 8y + 1 = 0 (1) 3x – 24 5 y + 3 5 = 0 (2)
Key Idea
For a pair of linear equations in two variables, write each equation in the slope‑intercept form $y=mx+c$. Compare the slopes ($m$) and intercepts ($c$):
- If slopes are different, the lines intersect at a unique point → unique solution.
- If slopes are equal but intercepts are different, the lines are parallel → no solution.
- If both slopes and intercepts are equal, the lines coincide → infinitely many solutions.
- If slopes are different, the lines intersect at a unique point → unique solution.
- If slopes are equal but intercepts are different, the lines are parallel → no solution.
- If both slopes and intercepts are equal, the lines coincide → infinitely many solutions.
Step-by-Step Solution
1. Convert each equation to slope‑intercept form
- Equation (1): $5x-8y+1=0 \Rightarrow -8y = -5x-1 \Rightarrow y = \frac{5}{8}x + \frac{1}{8}$
Hence, slope $m_1 = \frac{5}{8}$ and $y$‑intercept $c_1 = \frac{1}{8}$.
- Equation (2): $3x-\frac{24}{5}y+\frac{3}{5}=0$.
Multiply by $5$ to clear fractions: $15x-24y+3=0 \Rightarrow -24y = -15x-3 \Rightarrow y = \frac{15}{24}x + \frac{3}{24}$.
Simplify: $\frac{15}{24}=\frac{5}{8}$ and $\frac{3}{24}=\frac{1}{8}$.
Thus, $y = \frac{5}{8}x + \frac{1}{8}$, giving slope $m_2 = \frac{5}{8}$ and intercept $c_2 = \frac{1}{8}$.
2. Compare slopes and intercepts
- $m_1 = m_2 = \frac{5}{8}$
- $c_1 = c_2 = \frac{1}{8}$
Since both the slopes and the intercepts are identical, the two equations represent the *same straight line*.
3. Conclusion (graphical interpretation)
- The two lines coincide; therefore they intersect at every point on the line.
- Hence, the system has infinitely many solutions.
4. Graphical description (for the figure)
- The graph would show a single straight line passing through the point $(0,\frac{1}{8})$ with a gentle positive slope $\frac{5}{8}$. Both equations plot the same line, so only one line appears on the coordinate plane.
- Equation (1): $5x-8y+1=0 \Rightarrow -8y = -5x-1 \Rightarrow y = \frac{5}{8}x + \frac{1}{8}$
Hence, slope $m_1 = \frac{5}{8}$ and $y$‑intercept $c_1 = \frac{1}{8}$.
- Equation (2): $3x-\frac{24}{5}y+\frac{3}{5}=0$.
Multiply by $5$ to clear fractions: $15x-24y+3=0 \Rightarrow -24y = -15x-3 \Rightarrow y = \frac{15}{24}x + \frac{3}{24}$.
Simplify: $\frac{15}{24}=\frac{5}{8}$ and $\frac{3}{24}=\frac{1}{8}$.
Thus, $y = \frac{5}{8}x + \frac{1}{8}$, giving slope $m_2 = \frac{5}{8}$ and intercept $c_2 = \frac{1}{8}$.
2. Compare slopes and intercepts
- $m_1 = m_2 = \frac{5}{8}$
- $c_1 = c_2 = \frac{1}{8}$
Since both the slopes and the intercepts are identical, the two equations represent the *same straight line*.
3. Conclusion (graphical interpretation)
- The two lines coincide; therefore they intersect at every point on the line.
- Hence, the system has infinitely many solutions.
4. Graphical description (for the figure)
- The graph would show a single straight line passing through the point $(0,\frac{1}{8})$ with a gentle positive slope $\frac{5}{8}$. Both equations plot the same line, so only one line appears on the coordinate plane.
Question 3
Hint available
Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought.
Key Idea
Form two linear equations in two variables using the given statements and solve the simultaneous equations by the method of substitution (or elimination) as taught in NCERT.
Step-by-Step Solution
1. Introduce variables
Let \(p\) be the number of pants and \(s\) be the number of skirts.
2. Translate the statements into equations
- "The number of skirts is two less than twice the number of pants" gives \[ s = 2p - 2 \]
- "The number of skirts is four less than four times the number of pants" gives \[ s = 4p - 4 \]
3. Equate the two expressions for \(s\)
\[ 2p - 2 = 4p - 4 \]
4. Solve for \(p\)
\[ 2p - 2 = 4p - 4 \]
\[ -2p = -2 \]
\[ p = 1 \]
5. Find \(s\) using any one of the original equations
Using \(s = 2p - 2\):
\[ s = 2(1) - 2 = 0 \]
(The same value is obtained from \(s = 4p - 4\).)
6. State the answer
Champa bought 1 pant and 0 skirts.
Let \(p\) be the number of pants and \(s\) be the number of skirts.
2. Translate the statements into equations
- "The number of skirts is two less than twice the number of pants" gives \[ s = 2p - 2 \]
- "The number of skirts is four less than four times the number of pants" gives \[ s = 4p - 4 \]
3. Equate the two expressions for \(s\)
\[ 2p - 2 = 4p - 4 \]
4. Solve for \(p\)
\[ 2p - 2 = 4p - 4 \]
\[ -2p = -2 \]
\[ p = 1 \]
5. Find \(s\) using any one of the original equations
Using \(s = 2p - 2\):
\[ s = 2(1) - 2 = 0 \]
(The same value is obtained from \(s = 4p - 4\).)
6. State the answer
Champa bought 1 pant and 0 skirts.
Question 4
Hint available
Solve the following pair of equations by substitution method: 7x – 15y = 2 (1) x + 2y = 3 (2)
Key Idea
In the substitution method, one variable is expressed from one equation and substituted into the other equation, reducing the system to a single‑variable linear equation.
Step-by-Step Solution
1. Express one variable from the simpler equation (2):
$$x + 2y = 3 \;\Rightarrow\; x = 3 - 2y.$$
2. Substitute this expression for \(x\) into equation (1):
$$7x - 15y = 2 \;\Rightarrow\; 7(3 - 2y) - 15y = 2.$$
3. Simplify the resulting equation:
$$7\times3 - 7\times2y - 15y = 2$$
$$21 - 14y - 15y = 2$$
$$21 - 29y = 2.$$
4. Solve for \(y\):
$$-29y = 2 - 21$$
$$-29y = -19$$
$$y = \frac{-19}{-29} = \frac{19}{29}.$$
5. Find \(x\) using the expression from step 1:
$$x = 3 - 2y = 3 - 2\left(\frac{19}{29}\right)$$
$$x = \frac{87}{29} - \frac{38}{29} = \frac{49}{29}.$$
6. Write the ordered pair:
$$\boxed{\left( x, y \right) = \left( \frac{49}{29}, \; \frac{19}{29} \right)}.$$
7. Verification (optional):
$$7\left(\frac{49}{29}\right) - 15\left(\frac{19}{29}\right) = \frac{343 - 285}{29} = \frac{58}{29} = 2,$$
$$\left(\frac{49}{29}\right) + 2\left(\frac{19}{29}\right) = \frac{49 + 38}{29} = \frac{87}{29} = 3.$$
Both equations are satisfied, confirming the solution.
$$x + 2y = 3 \;\Rightarrow\; x = 3 - 2y.$$
2. Substitute this expression for \(x\) into equation (1):
$$7x - 15y = 2 \;\Rightarrow\; 7(3 - 2y) - 15y = 2.$$
3. Simplify the resulting equation:
$$7\times3 - 7\times2y - 15y = 2$$
$$21 - 14y - 15y = 2$$
$$21 - 29y = 2.$$
4. Solve for \(y\):
$$-29y = 2 - 21$$
$$-29y = -19$$
$$y = \frac{-19}{-29} = \frac{19}{29}.$$
5. Find \(x\) using the expression from step 1:
$$x = 3 - 2y = 3 - 2\left(\frac{19}{29}\right)$$
$$x = \frac{87}{29} - \frac{38}{29} = \frac{49}{29}.$$
6. Write the ordered pair:
$$\boxed{\left( x, y \right) = \left( \frac{49}{29}, \; \frac{19}{29} \right)}.$$
7. Verification (optional):
$$7\left(\frac{49}{29}\right) - 15\left(\frac{19}{29}\right) = \frac{343 - 285}{29} = \frac{58}{29} = 2,$$
$$\left(\frac{49}{29}\right) + 2\left(\frac{19}{29}\right) = \frac{49 + 38}{29} = \frac{87}{29} = 3.$$
Both equations are satisfied, confirming the solution.
Question 5
Hint available
Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.
Key Idea
Translate the word problem into two linear equations in two variables (present ages of Aftab and his daughter) and solve the system using the substitution method. The intersection of the two lines gives the required ages.
Step-by-Step Solution
1. Introduce variables\
Let \(A\) be Aftab's present age (in years) and \(D\) be his daughter's present age.\
\
2. Form the first equation (seven years ago)\
\(\text{Seven years ago, Aftab's age}=A-7\)\
\(\text{Daughter's age seven years ago}=D-7\)\
According to the statement, \(A-7 = 7\,(D-7)\).\
Simplify:\
\[A-7 = 7D - 49\]\
\[A = 7D - 42\] ...(i)\
\
3. Form the second equation (three years from now)\
\(\text{Aftab's age after 3 years}=A+3\)\
\(\text{Daughter's age after 3 years}=D+3\)\
The condition gives \(A+3 = 3\,(D+3)\).\
Simplify:\
\[A+3 = 3D + 9\]\
\[A = 3D + 6\] ...(ii)\
\
4. Solve the system by substitution\
From (i) and (ii) we have two expressions for \(A\):\
\[7D - 42 = 3D + 6\]\
Bring like terms together:\
\[7D - 3D = 6 + 42\]\
\[4D = 48\]\
\[D = 12\] years (daughter's present age).\
\
5. Find Aftab's age using either (i) or (ii). Using (ii):\
\[A = 3D + 6 = 3\times12 + 6 = 42\] years.\
\
6. Verification\
- Seven years ago: Aftab = \(42-7 = 35\), Daughter = \(12-7 = 5\). Indeed, \(35 = 7\times5\).\
- After three years: Aftab = \(42+3 = 45\), Daughter = \(12+3 = 15\). Indeed, \(45 = 3\times15\).\
\
7. Graphical representation\
- Plot the two straight lines given by \(A = 7D - 42\) and \(A = 3D + 6\) on a coordinate plane where the \(x\)-axis represents the daughter's age \(D\) and the \(y\)-axis represents Aftab's age \(A\).\
- The lines intersect at the point \((D, A) = (12, 42)\). This intersection point gives the solution of the system, i.e., the present ages of Aftab and his daughter.
Let \(A\) be Aftab's present age (in years) and \(D\) be his daughter's present age.\
\
2. Form the first equation (seven years ago)\
\(\text{Seven years ago, Aftab's age}=A-7\)\
\(\text{Daughter's age seven years ago}=D-7\)\
According to the statement, \(A-7 = 7\,(D-7)\).\
Simplify:\
\[A-7 = 7D - 49\]\
\[A = 7D - 42\] ...(i)\
\
3. Form the second equation (three years from now)\
\(\text{Aftab's age after 3 years}=A+3\)\
\(\text{Daughter's age after 3 years}=D+3\)\
The condition gives \(A+3 = 3\,(D+3)\).\
Simplify:\
\[A+3 = 3D + 9\]\
\[A = 3D + 6\] ...(ii)\
\
4. Solve the system by substitution\
From (i) and (ii) we have two expressions for \(A\):\
\[7D - 42 = 3D + 6\]\
Bring like terms together:\
\[7D - 3D = 6 + 42\]\
\[4D = 48\]\
\[D = 12\] years (daughter's present age).\
\
5. Find Aftab's age using either (i) or (ii). Using (ii):\
\[A = 3D + 6 = 3\times12 + 6 = 42\] years.\
\
6. Verification\
- Seven years ago: Aftab = \(42-7 = 35\), Daughter = \(12-7 = 5\). Indeed, \(35 = 7\times5\).\
- After three years: Aftab = \(42+3 = 45\), Daughter = \(12+3 = 15\). Indeed, \(45 = 3\times15\).\
\
7. Graphical representation\
- Plot the two straight lines given by \(A = 7D - 42\) and \(A = 3D + 6\) on a coordinate plane where the \(x\)-axis represents the daughter's age \(D\) and the \(y\)-axis represents Aftab's age \(A\).\
- The lines intersect at the point \((D, A) = (12, 42)\). This intersection point gives the solution of the system, i.e., the present ages of Aftab and his daughter.
Question 6
Hint available
In a shop the cost of 2 pencils and 3 erasers is `9 and the cost of 4 pencils and 6 erasers is `18. Find the cost of each pencil and each eraser.
Key Idea
The two linear equations are dependent (one is a multiple of the other). Hence the system has infinitely many solutions. The cost of one item can be expressed in terms of the other.
Step-by-Step Solution
Let the cost of one pencil be $p$ rupees and the cost of one eraser be $e$ rupees.\
\
1. Form the equations\
\[\begin{aligned}
2p + 3e &= 9 \quad\text{(Equation 1)}\\
4p + 6e &= 18 \quad\text{(Equation 2)}
\end{aligned}\]\
2. Observe the relationship between the equations\
Equation 2 is exactly twice Equation 1: \[4p+6e = 2(2p+3e) = 2\times9 = 18.\]
Hence the two equations are not independent; they represent the same straight line.
3. Solve for one variable in terms of the other\
From Equation 1,\
\[2p = 9 - 3e \quad\Rightarrow\quad p = \frac{9-3e}{2}.\]
Alternatively,\
\[3e = 9 - 2p \quad\Rightarrow\quad e = \frac{9-2p}{3}.\]
4. State the general solution\
Since the system is dependent, there are infinitely many ordered pairs $(p,e)$ satisfying the relation.\
Let $e = t$ (where $t$ can be any real number such that $p$ remains non‑negative). Then\
\[p = \frac{9-3t}{2}.\]
Conversely, let $p = s$, then\
\[e = \frac{9-2s}{3}.\]
5. Interpretation\
Any pair of values that satisfy the above relation gives the required costs. For example, if we choose $e = 2\,\text{rupees}$, then $p = \frac{9-3\times2}{2}=\frac{3}{2}=1.5\,\text{rupees}$. Similarly, $e = 1\,\text{rupee}$ gives $p = 3\,\text{rupees}$, etc.
Thus the system has infinitely many solutions; the cost of a pencil can be expressed as $p = \frac{9-3e}{2}$ rupees (or the cost of an eraser as $e = \frac{9-2p}{3}$ rupees).
\
1. Form the equations\
\[\begin{aligned}
2p + 3e &= 9 \quad\text{(Equation 1)}\\
4p + 6e &= 18 \quad\text{(Equation 2)}
\end{aligned}\]\
2. Observe the relationship between the equations\
Equation 2 is exactly twice Equation 1: \[4p+6e = 2(2p+3e) = 2\times9 = 18.\]
Hence the two equations are not independent; they represent the same straight line.
3. Solve for one variable in terms of the other\
From Equation 1,\
\[2p = 9 - 3e \quad\Rightarrow\quad p = \frac{9-3e}{2}.\]
Alternatively,\
\[3e = 9 - 2p \quad\Rightarrow\quad e = \frac{9-2p}{3}.\]
4. State the general solution\
Since the system is dependent, there are infinitely many ordered pairs $(p,e)$ satisfying the relation.\
Let $e = t$ (where $t$ can be any real number such that $p$ remains non‑negative). Then\
\[p = \frac{9-3t}{2}.\]
Conversely, let $p = s$, then\
\[e = \frac{9-2s}{3}.\]
5. Interpretation\
Any pair of values that satisfy the above relation gives the required costs. For example, if we choose $e = 2\,\text{rupees}$, then $p = \frac{9-3\times2}{2}=\frac{3}{2}=1.5\,\text{rupees}$. Similarly, $e = 1\,\text{rupee}$ gives $p = 3\,\text{rupees}$, etc.
Thus the system has infinitely many solutions; the cost of a pencil can be expressed as $p = \frac{9-3e}{2}$ rupees (or the cost of an eraser as $e = \frac{9-2p}{3}$ rupees).
Question 7
Hint available
Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other?
Key Idea
For two linear equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, the lines intersect if $\frac{a_1}{a_2}
eq\frac{b_1}{b_2}$. If $\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}$ the lines are parallel (distinct) and do not meet. If all three ratios are equal, the lines coincide.
eq\frac{b_1}{b_2}$. If $\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}$ the lines are parallel (distinct) and do not meet. If all three ratios are equal, the lines coincide.
Step-by-Step Solution
1. Write the given equations in the standard form $a x + b y + c = 0$:
\[\begin{aligned}
&\text{(i)}\; x + 2y - 4 = 0 \quad\Rightarrow\; a_1=1,\; b_1=2,\; c_1=-4,\\
&\text{(ii)}\; 2x + 4y - 12 = 0 \quad\Rightarrow\; a_2=2,\; b_2=4,\; c_2=-12.\end{aligned}\]
2. Compute the ratios of the coefficients:
\[\frac{a_1}{a_2}=\frac{1}{2},\qquad \frac{b_1}{b_2}=\frac{2}{4}=\frac{1}{2},\qquad \frac{c_1}{c_2}=\frac{-4}{-12}=\frac{1}{3}.\]
3. Compare the ratios:
- Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) the two lines have the same slope; they are either coincident or parallel.
- However \(\frac{c_1}{c_2}=\frac{1}{3}
eq\frac{1}{2}=\frac{a_1}{a_2}\). Hence the third ratio is different.
4. Conclusion based on the condition:
- When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), the lines are parallel distinct.
- Therefore the two rails do not intersect each other.
Answer: The rails do not cross; they are parallel distinct lines.
\[\begin{aligned}
&\text{(i)}\; x + 2y - 4 = 0 \quad\Rightarrow\; a_1=1,\; b_1=2,\; c_1=-4,\\
&\text{(ii)}\; 2x + 4y - 12 = 0 \quad\Rightarrow\; a_2=2,\; b_2=4,\; c_2=-12.\end{aligned}\]
2. Compute the ratios of the coefficients:
\[\frac{a_1}{a_2}=\frac{1}{2},\qquad \frac{b_1}{b_2}=\frac{2}{4}=\frac{1}{2},\qquad \frac{c_1}{c_2}=\frac{-4}{-12}=\frac{1}{3}.\]
3. Compare the ratios:
- Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) the two lines have the same slope; they are either coincident or parallel.
- However \(\frac{c_1}{c_2}=\frac{1}{3}
eq\frac{1}{2}=\frac{a_1}{a_2}\). Hence the third ratio is different.
4. Conclusion based on the condition:
- When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), the lines are parallel distinct.
- Therefore the two rails do not intersect each other.
Answer: The rails do not cross; they are parallel distinct lines.
Question 8
Hint available
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ` 2000 per month, find their monthly incomes.
Key Idea
Use the given ratios to express incomes and expenditures in terms of common multiples, then form two linear equations using the fact that (Income – Expenditure) = Savings = 2000 for each person. Solve the simultaneous equations by elimination (or substitution) to obtain the actual incomes.
Step-by-Step Solution
1. Let the incomes of the two persons be $9x$ and $7x$ rupees per month respectively (since the ratio is $9:7$).
2. Let their expenditures be $4y$ and $3y$ rupees per month respectively (ratio $4:3$).
3. Savings for each person is given as Rs. $2000$, therefore
\[\begin{cases}\ 9x - 4y = 2000 \ 7x - 3y = 2000 \end{cases}\]
4. Subtract the second equation from the first to eliminate the constant term:
\[(9x-4y) - (7x-3y) = 0 \Rightarrow 2x - y = 0 \Rightarrow y = 2x.\]
5. Substitute $y = 2x$ in any one of the original equations, e.g. $9x - 4y = 2000$:
\[9x - 4(2x) = 2000 \Rightarrow 9x - 8x = 2000 \Rightarrow x = 2000.\]
6. Hence incomes are:
\[\text{First person's income} = 9x = 9 \times 2000 = \text{Rs. }18000,\]
\[\text{Second person's income} = 7x = 7 \times 2000 = \text{Rs. }14000.\]
7. (For verification) Expenditures become $y = 2x = 4000$, so $4y = 16000$ and $3y = 12000$.
Savings: $18000-16000 = 2000$, $14000-12000 = 2000$, which matches the given condition.
2. Let their expenditures be $4y$ and $3y$ rupees per month respectively (ratio $4:3$).
3. Savings for each person is given as Rs. $2000$, therefore
\[\begin{cases}\ 9x - 4y = 2000 \ 7x - 3y = 2000 \end{cases}\]
4. Subtract the second equation from the first to eliminate the constant term:
\[(9x-4y) - (7x-3y) = 0 \Rightarrow 2x - y = 0 \Rightarrow y = 2x.\]
5. Substitute $y = 2x$ in any one of the original equations, e.g. $9x - 4y = 2000$:
\[9x - 4(2x) = 2000 \Rightarrow 9x - 8x = 2000 \Rightarrow x = 2000.\]
6. Hence incomes are:
\[\text{First person's income} = 9x = 9 \times 2000 = \text{Rs. }18000,\]
\[\text{Second person's income} = 7x = 7 \times 2000 = \text{Rs. }14000.\]
7. (For verification) Expenditures become $y = 2x = 4000$, so $4y = 16000$ and $3y = 12000$.
Savings: $18000-16000 = 2000$, $14000-12000 = 2000$, which matches the given condition.
Question 9
Hint available
Use elimination method to find all possible solutions of the following pair of linear equations : 2x + 3y = 8 (1) 4x + 6y = 7 (2)
Key Idea
The elimination method involves adding or subtracting the equations after suitable multiplication so that one variable gets eliminated. If the resulting equations give a contradiction, the system has no solution (inconsistent). If they give an identity, the system has infinitely many solutions (dependent).
Step-by-Step Solution
1. Write the given equations
$$\begin{aligned}
(1)\;& 2x + 3y = 8 \\
(2)\;& 4x + 6y = 7
\end{aligned}$$
2. Make the coefficients of one variable equal
Multiply equation (1) by 2 so that the coefficient of \(x\) (or \(y\)) matches that in equation (2):
$$2\times(1):\; 4x + 6y = 16$$
3. Subtract the two equations
$$\begin{aligned}
(4x + 6y = 16) \; - \; (4x + 6y = 7) \;\Rightarrow\; 0 = 9
\end{aligned}$$
4. Interpret the result
The statement \(0 = 9\) is a contradiction. Hence the two equations are inconsistent; they represent two parallel lines that never intersect.
5. Conclusion
Since the system is inconsistent, there is no solution (i.e., the pair of equations has no common point).
$$\begin{aligned}
(1)\;& 2x + 3y = 8 \\
(2)\;& 4x + 6y = 7
\end{aligned}$$
2. Make the coefficients of one variable equal
Multiply equation (1) by 2 so that the coefficient of \(x\) (or \(y\)) matches that in equation (2):
$$2\times(1):\; 4x + 6y = 16$$
3. Subtract the two equations
$$\begin{aligned}
(4x + 6y = 16) \; - \; (4x + 6y = 7) \;\Rightarrow\; 0 = 9
\end{aligned}$$
4. Interpret the result
The statement \(0 = 9\) is a contradiction. Hence the two equations are inconsistent; they represent two parallel lines that never intersect.
5. Conclusion
Since the system is inconsistent, there is no solution (i.e., the pair of equations has no common point).
Question 10
Hint available
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Key Idea
Represent the two‑digit number using its tens and units digits, form two linear equations from the given conditions (sum of the number and its reversal, and the difference of the digits), and solve the simultaneous equations to obtain the digits. The absolute difference condition gives two possible ordered pairs, leading to two numbers.
Step-by-Step Solution
1. Let the digits be
- Tens digit = $x$ ( $1 \le x \le 9$ )
- Units digit = $y$ ( $0 \le y \le 9$ )
2. Write the number and its reversal
- Original number $N = 10x + y$
- Reversed number $R = 10y + x$
3. Use the first condition (sum = 66)
\[
N + R = (10x + y) + (10y + x) = 11(x + y) = 66
\]
Hence
\[
x + y = \frac{66}{11} = 6 \quad\text{(Equation 1)}
\]
4. Use the second condition (digits differ by 2)
The absolute difference is 2, so we have two possibilities:
- $x - y = 2$ \(\text{Case A}\)
- $y - x = 2$ \(\text{Case B}\)
5. Solve Case A
\[
\begin{cases}
x + y = 6 \\
x - y = 2
\end{cases}
\]
Adding the equations: $2x = 8 \Rightarrow x = 4$.
Substituting in $x + y = 6$: $4 + y = 6 \Rightarrow y = 2$.
Hence the number is $10x + y = 42$.
6. Solve Case B
\[
\begin{cases}
x + y = 6 \\
y - x = 2
\end{cases}
\]
Adding the equations: $2y = 8 \Rightarrow y = 4$.
Substituting in $x + y = 6$: $x + 4 = 6 \Rightarrow x = 2$.
Hence the number is $10x + y = 24$.
7. Verification
- $42 + 24 = 66$ (satisfies the sum condition).
- Digits differ by $|4-2| = 2$ (satisfies the difference condition).
Both numbers satisfy all the given conditions.
8. Conclusion
The two‑digit numbers are $42$ and $24$. Therefore, the total number of such numbers is 2.
- Tens digit = $x$ ( $1 \le x \le 9$ )
- Units digit = $y$ ( $0 \le y \le 9$ )
2. Write the number and its reversal
- Original number $N = 10x + y$
- Reversed number $R = 10y + x$
3. Use the first condition (sum = 66)
\[
N + R = (10x + y) + (10y + x) = 11(x + y) = 66
\]
Hence
\[
x + y = \frac{66}{11} = 6 \quad\text{(Equation 1)}
\]
4. Use the second condition (digits differ by 2)
The absolute difference is 2, so we have two possibilities:
- $x - y = 2$ \(\text{Case A}\)
- $y - x = 2$ \(\text{Case B}\)
5. Solve Case A
\[
\begin{cases}
x + y = 6 \\
x - y = 2
\end{cases}
\]
Adding the equations: $2x = 8 \Rightarrow x = 4$.
Substituting in $x + y = 6$: $4 + y = 6 \Rightarrow y = 2$.
Hence the number is $10x + y = 42$.
6. Solve Case B
\[
\begin{cases}
x + y = 6 \\
y - x = 2
\end{cases}
\]
Adding the equations: $2y = 8 \Rightarrow y = 4$.
Substituting in $x + y = 6$: $x + 4 = 6 \Rightarrow x = 2$.
Hence the number is $10x + y = 24$.
7. Verification
- $42 + 24 = 66$ (satisfies the sum condition).
- Digits differ by $|4-2| = 2$ (satisfies the difference condition).
Both numbers satisfy all the given conditions.
8. Conclusion
The two‑digit numbers are $42$ and $24$. Therefore, the total number of such numbers is 2.