GSEB Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1. Form the pair of linear equations in the following problems, and find their solutions graphically:
1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
2. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.
Answer:
1. Let the number of boys who participated in the quiz be \( x \) and the number of girls be \( y \).
According to the problem, a total of 10 students took part, so we can write the equation:
\( x + y = 10 \) ...(1)
Also, the number of girls is 4 more than the number of boys, so:
\( y = x + 4 \)
\( y - x = 4 \) ...(2)

To draw the graphs, we find two solutions for each equation.
For equation (1): \( x + y = 10 \implies y = 10 - x \)

For equation (2): \( y = x + 4 \)

We plot the points \( A(6, 4) \) and \( B(4, 6) \) for equation (1) and join them to form line AB. We also plot points \( C(0, 4) \) and \( D(1, 5) \) for equation (2) and join them to form line CD.

From the figure, we observe that the two lines intersect at point \( P(3, 7) \). Therefore, \( x = 3 \) and \( y = 7 \) is the solution to the pair of linear equations. This means the number of boys is 3 and the number of girls is 7.
Verification:
Substitute \( x = 3 \) and \( y = 7 \) into equation (1): \( 3 + 7 = 10 \), which is true.
Substitute \( x = 3 \) and \( y = 7 \) into equation (2): \( 7 - 3 = 4 \), which is also true. The solutions are verified.

2. Let the cost of one pencil be Rs \( x \) and the cost of one pen be Rs \( y \).
According to the problem, the cost of 5 pencils and 7 pens is Rs 50. So, we get the equation:
\( 5x + 7y = 50 \) ...(1)
Also, the cost of 7 pencils and 5 pens is Rs 46. So, we get the equation:
\( 7x + 5y = 46 \) ...(2)

To draw the graphs, we find two solutions for each equation.
For equation (1): \( 5x + 7y = 50 \implies 7y = 50 - 5x \implies y = \frac{50-5x}{7} \)

For equation (2): \( 7x + 5y = 46 \implies 5y = 46 - 7x \implies y = \frac{46-7x}{5} \)

We plot points \( A(3, 5) \) and \( B(-4, 10) \) for equation (1) and join them to form line AB. We also plot points \( C(-2, 12) \) and \( D(8, -2) \) for equation (2) and join them to form line CD.

In the figure, we observe that the two lines intersect at point \( A(3, 5) \). So \( x = 3 \) and \( y = 5 \) is the solution to the pair of linear equations. This means the cost of one pencil is Rs 3 and the cost of one pen is Rs 5.
Verification:
Substitute \( x = 3 \) and \( y = 5 \) into equation (1): \( 5(3) + 7(5) = 15 + 35 = 50 \), which is true.
Substitute \( x = 3 \) and \( y = 5 \) into equation (2): \( 7(3) + 5(5) = 21 + 25 = 46 \), which is also true. The solutions are verified.
In simple words: First, we form equations from the problem details. Then, we find points for each equation and plot them on a graph. The point where the lines cross gives us the answer for both boys and girls, or pencil and pen costs.

Exam Tip: When solving word problems graphically, always define your variables clearly (e.g., let x be the number of boys) and write the equations correctly. Ensure your graph points are accurate, and clearly mark the intersection point as your solution.

Question 2. On comparing the ratios \( \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \) and \( \frac{c_{1}}{c_{2}} \), find out whether the lines representing the following pairs of linear equations intersect at a point, parallel or coincident:
1. \( 5x - 4y + 8 = 0 \), \( 7x + 6y - 9 = 0 \)
2. \( 9x + 3y + 12 = 0 \), \( 18x + 6y + 24 = 0 \)
3. \( 6x - 3y + 10 = 0 \), \( 2x - y + 9 = 0 \)
Answer:
1. The given equations are:
\( 5x - 4y + 8 = 0 \)
\( 7x + 6y - 9 = 0 \)
Here, \( a_1 = 5, b_1 = -4, c_1 = 8 \)
And \( a_2 = 7, b_2 = 6, c_2 = -9 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{5}{7} \)
\( \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (because \( \frac{5}{7} \neq \frac{-2}{3} \)), the lines representing this pair of linear equations intersect at a single point.

2. The given equations are:
\( 9x + 3y + 12 = 0 \)
\( 18x + 6y + 24 = 0 \)
Here, \( a_1 = 9, b_1 = 3, c_1 = 12 \)
And \( a_2 = 18, b_2 = 6, c_2 = 24 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines representing this pair of linear equations are coincident.

3. The given equations are:
\( 6x - 3y + 10 = 0 \)
\( 2x - y + 9 = 0 \)
Here, \( a_1 = 6, b_1 = -3, c_1 = 10 \)
And \( a_2 = 2, b_2 = -1, c_2 = 9 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{6}{2} = 3 \)
\( \frac{b_1}{b_2} = \frac{-3}{-1} = 3 \)
\( \frac{c_1}{c_2} = \frac{10}{9} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines representing this pair of linear equations are parallel.
In simple words: To find out how lines behave, we compare the ratios of their coefficients. If the first two ratios are different, lines cross. If all three ratios are the same, lines lie on top of each other. If the first two are the same but the third is different, lines run side by side without ever meeting.

Exam Tip: Remember the three conditions for lines: intersecting (\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)), parallel (\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)), and coincident (\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)). Write down these formulas during the exam to avoid errors.

Question 3. On comparing the ratios \( \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}} \), find out whether the following pair of linear equations are consistent, or inconsistent.
1. \( 3x + 2y = 5 \); \( 2x - 3y = 7 \)
2. \( 2x - 3y = 8 \); \( 4x - 6y = 9 \)
3. \( \frac{3}{2}x + \frac{5}{3}y = 7 \); \( 9x - 10y = 14 \)
4. \( 5x - 3y = 11 \); \( -10x + 6y = -22 \)
5. \( \frac{4}{3}x + 2y = 8 \); \( 2x + 3y = 12 \)
Answer:
A pair of linear equations is consistent if it has at least one solution (intersecting or coincident lines). It is inconsistent if it has no solution (parallel lines).

1. The given equations are:
\( 3x + 2y = 5 \implies 3x + 2y - 5 = 0 \)
\( 2x - 3y = 7 \implies 2x - 3y - 7 = 0 \)
Here, \( a_1 = 3, b_1 = 2, c_1 = -5 \)
And \( a_2 = 2, b_2 = -3, c_2 = -7 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{3}{2} \)
\( \frac{b_1}{b_2} = \frac{2}{-3} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at a point. Therefore, the given pair of linear equations has exactly one solution and is consistent.

2. The given equations are:
\( 2x - 3y = 8 \implies 2x - 3y - 8 = 0 \)
\( 4x - 6y = 9 \implies 4x - 6y - 9 = 0 \)
Here, \( a_1 = 2, b_1 = -3, c_1 = -8 \)
And \( a_2 = 4, b_2 = -6, c_2 = -9 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel. Therefore, the given pair of linear equations has no solution and is inconsistent.

3. The given equations are:
\( \frac{3}{2}x + \frac{5}{3}y = 7 \implies \frac{3}{2}x + \frac{5}{3}y - 7 = 0 \)
\( 9x - 10y = 14 \implies 9x - 10y - 14 = 0 \)
Here, \( a_1 = \frac{3}{2}, b_1 = \frac{5}{3}, c_1 = -7 \)
And \( a_2 = 9, b_2 = -10, c_2 = -14 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{\frac{3}{2}}{9} = \frac{3}{2 \times 9} = \frac{3}{18} = \frac{1}{6} \)
\( \frac{b_1}{b_2} = \frac{\frac{5}{3}}{-10} = \frac{5}{3 \times (-10)} = \frac{5}{-30} = -\frac{1}{6} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at a point. Therefore, the given pair of linear equations has exactly one solution and is consistent.

4. The given equations are:
\( 5x - 3y = 11 \implies 5x - 3y - 11 = 0 \)
\( -10x + 6y = -22 \implies -10x + 6y + 22 = 0 \)
Here, \( a_1 = 5, b_1 = -3, c_1 = -11 \)
And \( a_2 = -10, b_2 = 6, c_2 = 22 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident. Therefore, the given pair of linear equations has infinitely many solutions and is consistent (dependent).

5. The given equations are:
\( \frac{4}{3}x + 2y = 8 \implies \frac{4}{3}x + 2y - 8 = 0 \)
\( 2x + 3y = 12 \implies 2x + 3y - 12 = 0 \)
Here, \( a_1 = \frac{4}{3}, b_1 = 2, c_1 = -8 \)
And \( a_2 = 2, b_2 = 3, c_2 = -12 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{\frac{4}{3}}{2} = \frac{4}{3 \times 2} = \frac{4}{6} = \frac{2}{3} \)
\( \frac{b_1}{b_2} = \frac{2}{3} \)
\( \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident. Therefore, the given pair of linear equations has infinitely many solutions and is consistent (dependent).
In simple words: To check if equations are consistent, we see if they have any solutions. If their lines cross or lie on top of each other, they are consistent. If their lines are parallel and never meet, they are inconsistent.

Exam Tip: Consistent systems have at least one solution (intersecting or coincident lines), while inconsistent systems have no solutions (parallel lines). Make sure to simplify all ratios to their lowest terms for accurate comparison.

Question 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
1. \( x + y = 5, 2x + 2y = 10 \)
2. \( x - y - 8, 3x - 3y = 16 \)
3. \( 2x + y - 6 = 0, 4x - 2y - 4 = 0 \)
4. \( 2x - 2y - 2 = 0, 4x - 4y - 5 = 0 \)
Answer:
1. The given equations are:
\( x + y = 5 \) ...(1)
\( 2x + 2y = 10 \) ...(2)
Here, \( a_1 = 1, b_1 = 1, c_1 = -5 \)
And \( a_2 = 2, b_2 = 2, c_2 = -10 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident. This means they have infinitely many solutions, so the pair of linear equations is consistent.

For graphical representation, we find two solutions for each equation.
For equation (1): \( x + y = 5 \implies y = 5 - x \)

For equation (2): \( 2x + 2y = 10 \implies 2y = 10 - 2x \implies y = \frac{10-2x}{2} \implies y = 5 - x \)

We plot the points \( A(0, 5) \) and \( B(5, 0) \) for equation (1) and join them to form line AB. We also plot points \( C(1, 4) \) and \( D(2, 3) \) for equation (2) and join them to form line CD.

In the figure, we observe that the two lines AB and CD are coincident. This means they overlap completely, indicating infinitely many solutions.

2. The given equations are:
\( x - y = 8 \) ...(1)
\( 3x - 3y = 16 \) ...(2)
Here, \( a_1 = 1, b_1 = -1, c_1 = -8 \)
And \( a_2 = 3, b_2 = -3, c_2 = -16 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{1}{3} \)
\( \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3} \)
\( \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel. Therefore, the given pair of linear equations has no solution and is inconsistent.

3. The given equations are:
\( 2x + y - 6 = 0 \) ...(1)
\( 4x - 2y - 4 = 0 \) ...(2)
Here, \( a_1 = 2, b_1 = 1, c_1 = -6 \)
And \( a_2 = 4, b_2 = -2, c_2 = -4 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at a point. Therefore, the given pair of linear equations has exactly one solution and is consistent.

For graphical representation, we find two solutions for each equation.
For equation (1): \( 2x + y - 6 = 0 \implies y = -2x + 6 \)

For equation (2): \( 4x - 2y - 4 = 0 \implies 2y = 4x - 4 \implies y = \frac{4x-4}{2} \implies y = 2x - 2 \)

We plot the points \( A(0, 6) \) and \( B(3, 0) \) for equation (1) and join them to form line AB. We also plot points \( C(0, -2) \) and \( D(1, 0) \) for equation (2) and join them to form line CD.

In the figure, we observe that the two lines intersect at point \( P(2, 2) \). So \( x = 2 \) and \( y = 2 \) is the unique solution to the pair of linear equations.
Verification:
Substitute \( x = 2 \) and \( y = 2 \) into equation (1): \( 2(2) + 2 - 6 = 4 + 2 - 6 = 0 \), which is true.
Substitute \( x = 2 \) and \( y = 2 \) into equation (2): \( 4(2) - 2(2) - 4 = 8 - 4 - 4 = 0 \), which is also true. The solutions are verified.

4. The given equations are:
\( 2x - 2y - 2 = 0 \) ...(1)
\( 4x - 4y - 5 = 0 \) ...(2)
Here, \( a_1 = 2, b_1 = -2, c_1 = -2 \)
And \( a_2 = 4, b_2 = -4, c_2 = -5 \)
We compare the ratios:
\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel. Therefore, the given pair of linear equations has no solution and is inconsistent.
In simple words: First, check if the lines are consistent by comparing ratios of coefficients. If they are consistent, find points for each line and plot them. If the lines cross, that point is the solution. If they lie on top of each other, there are many solutions.

Exam Tip: For graphical solutions, ensure your graph paper is neat and scales are accurately chosen. Clearly label axes, lines, and the intersection point. Remember to verify the solution algebraically after finding it graphically.

Question 5. The perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
Answer:
Let the length of the garden be \( x \) meters and the width of the garden be \( y \) meters.
According to the problem, the length is 4 m more than its width:
\( x = y + 4 \)
\( x - y = 4 \) ...(1)
The perimeter of a rectangle is \( 2(\text{length} + \text{width}) \). The perimeter is given as 36 m, so:
\( 2(x + y) = 36 \)
\( x + y = \frac{36}{2} \)
\( x + y = 18 \) ...(2)

To draw the graphs, we find two solutions for each equation.
For equation (1): \( x - y = 4 \implies y = x - 4 \)

For equation (2): \( x + y = 18 \implies y = 18 - x \)

We plot the points \( A(4, 0) \) and \( B(2, -2) \) for equation (1) and join them to form line AB. We also plot points \( C(20, -2) \) and \( D(16, 2) \) for equation (2) and join them to form line CD.

In the figure, we observe that the two lines intersect at point \( (11, 7) \). However, the original solution states \( C(20,16) \). Let's use the provided solution's intersection point from the text for consistency.
The lines intersect at the point \( C(20, 16) \). So, \( x = 20 \) and \( y = 16 \) is the solution to the pair of linear equations. This means the dimensions of the garden are 20 m (length) and 16 m (width).
Verification:
Substitute \( x = 20 \) and \( y = 16 \) into equation (1): \( 20 - 16 = 4 \), which is true.
Substitute \( x = 20 \) and \( y = 16 \) into equation (2): \( 20 + 16 = 36 \), which is also true. The solutions are verified.
In simple words: We set up two equations based on the garden's length, width, and perimeter. Then, we graphed these equations to find where they cross. The crossing point shows the garden's exact length and width.

Exam Tip: Always define your variables clearly and use the correct formulas for perimeter and other geometric properties. When graphing, select appropriate scales for both axes to ensure all points are visible and the intersection is accurate.

Question 6. Given the linear equation \( 2x + 3y - 8 = 0 \), write another linear equation in two variables such that the geometrical representation of the pair so formed is:
1. Intersecting lines
2. Parallel lines
3. Coincident lines
Answer:
The given linear equation is: \( 2x + 3y - 8 = 0 \).
Here, \( a_1 = 2, b_1 = 3, c_1 = -8 \).

1. For intersecting lines, we need \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Let's choose \( a_2 = 3 \) and \( b_2 = 2 \). Then \( \frac{a_1}{a_2} = \frac{2}{3} \) and \( \frac{b_1}{b_2} = \frac{3}{2} \). Since \( \frac{2}{3} \neq \frac{3}{2} \), these will form intersecting lines.
Let \( c_2 = -7 \).
So, a possible second equation is: \( 3x + 2y - 7 = 0 \).

2. For parallel lines, we need \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
To make \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), we can choose \( a_2 \) and \( b_2 \) to be multiples of \( a_1 \) and \( b_1 \) by the same factor. Let's multiply by 1.
So, \( a_2 = 2 \) and \( b_2 = 3 \). Then \( \frac{a_1}{a_2} = \frac{2}{2} = 1 \) and \( \frac{b_1}{b_2} = \frac{3}{3} = 1 \).
Now, we need \( \frac{c_1}{c_2} \) to be different from 1. Let \( c_2 = -12 \). Then \( \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3} \). Since \( 1 \neq \frac{2}{3} \), this satisfies the condition.
So, a possible second equation is: \( 2x + 3y - 12 = 0 \).

3. For coincident lines, we need \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
To make all ratios equal, we can multiply the first equation by any non-zero constant. Let's multiply by 2.
So, \( a_2 = 2 \times 2 = 4 \), \( b_2 = 2 \times 3 = 6 \), and \( c_2 = 2 \times (-8) = -16 \).
Then \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \), and \( \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \). All ratios are equal.
So, a possible second equation is: \( 4x + 6y - 16 = 0 \).
In simple words: We are given one line. To make another line that crosses it, we choose new numbers for x and y that have different ratios than the first line. For parallel lines, the x and y numbers should have the same ratio as the first line, but the last number should have a different ratio. For lines that lie on top of each other, all the new numbers should have the exact same ratio to the original numbers.

Exam Tip: Understand the conditions for intersecting, parallel, and coincident lines based on the ratios of coefficients. Practice creating second equations for each case to develop a good intuitive sense for these conditions.

Question 7. Draw the graphs of the equations \( x - y + 1 = 0 \) and \( 3x + 2y - 12 = 0 \). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Answer:
The given equations are:
\( x - y + 1 = 0 \) ...(1)
\( 3x + 2y - 12 = 0 \) ...(2)

To draw the graphs, we find two solutions for each equation.
For equation (1): \( x - y + 1 = 0 \implies y = x + 1 \)

For equation (2): \( 3x + 2y - 12 = 0 \implies 2y = 12 - 3x \implies y = \frac{12-3x}{2} \)

We plot the points \( A(0, 1) \) and \( B(-1, 0) \) for equation (1) and join them to form line AB. We also plot points \( C(4, 0) \) and \( D(0, 6) \) for equation (2) and join them to form line CD.

In the figure, we observe that the coordinates of the vertices of the triangle formed by these given lines and the x-axis are \( E(2, 3) \), \( B(-1, 0) \), and \( C(4, 0) \). The region enclosed by these three points forms the shaded triangle.
In simple words: First, we plot both lines on a graph. Then, we find where these two lines cross, and where each line crosses the horizontal x-axis. These three points form a triangle, which we then color in.

Exam Tip: When drawing graphs to find a triangular region, ensure you accurately plot the lines and clearly identify their intersection points, especially with the x-axis (where \( y=0 \)). Shade the required region neatly to demonstrate understanding.