GSEB Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7

 

Question 1. The ages of two friends Aol and Rijo differ by 3 years, Mi's father Dharam is twice as old as Axil and Riju is twice as old as his aislar Cathy. The ages of Cathy and Pharam differ by 30 years. Find the ages of Ani and Biju.
Answer: Let the age of Ani be \( x \) years and the age of Biju be \( y \) years. Then, as per the question,
\( x - y = \pm 3 \) ......................(1)
From the given information, Ani's father Dharam's age is \( 2x \) years. Biju's sister Cathy's age is \( \frac { y }{ 2 } \) years.
According to the question, the ages of Cathy and Dharam differ by 30 years, so we can write:
\( 2x - \frac { y }{ 2 } = 30 \)
Multiplying by 2 to clear the fraction, we get:
\( 4x - y = 60 \) ......................(2)
We will solve this using two cases based on equation (1).

**Case I: When \( x - y = 3 \)**
Then, we have:
\( x - y = 3 \) (Let's call this Eq. 1a)
\( 4x - y = 60 \) (Eq. 2)
Subtracting equation (1a) from equation (2):
\( (4x - y) - (x - y) = 60 - 3 \)
\( 3x = 57 \)
\( x = \frac { 57 }{ 3 } = 19 \) years
Substituting the value of \( x \) into equation (1a):
\( 19 - y = 3 \)
\( y = 19 - 3 \)
\( y = 16 \) years
So, Ani's age is 19 years and Biju's age is 16 years.

Verification:
\( x - y = 19 - 16 = 3 \) (Correct)
\( 4x - y = 4(19) - 16 = 76 - 16 = 60 \) (Correct)
This verifies the solution for Case I.

**Case II: When \( x - y = -3 \)**
Then, we have:
\( x - y = -3 \) (Let's call this Eq. 1b)
\( 4x - y = 60 \) (Eq. 2)
Subtracting equation (1b) from equation (2):
\( (4x - y) - (x - y) = 60 - (-3) \)
\( 3x = 63 \)
\( x = \frac { 63 }{ 3 } = 21 \) years
Substituting the value of \( x \) into equation (1b):
\( 21 - y = -3 \)
\( y = 21 + 3 \)
\( y = 24 \) years
So, Ani's age is 21 years and Biju's age is 24 years.

Verification:
\( x - y = 21 - 24 = -3 \) (Correct)
\( 4x - y = 4(21) - 24 = 84 - 24 = 60 \) (Correct)
This verifies the solution for Case II.

Therefore, the possible ages for Ani and Biju are (19, 16) or (21, 24) years.
In simple words: We set up two algebraic equations based on the age differences given in the problem. Because one difference could be positive or negative, we solved the equations in two separate situations. This gave us two possible sets of ages for Ani and Biju, both of which work with the original statements.

Exam Tip: When a problem states a difference without specifying which value is larger, consider both positive and negative possibilities (e.g., \( \pm \)). Always verify your solutions by plugging the values back into the original equations.

 

Question 2. One says, "Give me a hundred, friend! I shall then becomes twice us nch as you", The othcr replies. "1f you give me ten, I shall be six times as rich as you”. Tell mu what. is the (respcctive) capital.
Answer: Let the amounts of their respective capitals be \( Rs x \) and \( Rs y \) respectively.

According to the first statement:
If the first person (with \( Rs x \)) gets \( Rs 100 \) from the second person (with \( Rs y \)), then the first person's capital becomes \( x + 100 \), and the second person's capital becomes \( y - 100 \). The first person will then have twice as much as the second.
\( x + 100 = 2(y - 100) \)
\( x + 100 = 2y - 200 \)
\( x - 2y = -200 - 100 \)
\( x - 2y = -300 \) ......................(1)

According to the second statement:
If the second person (with \( Rs y \)) gets \( Rs 10 \) from the first person (with \( Rs x \)), then the second person's capital becomes \( y + 10 \), and the first person's capital becomes \( x - 10 \). The second person will then have six times as much as the first.
\( y + 10 = 6(x - 10) \)
\( y + 10 = 6x - 60 \)
\( -6x + y = -60 - 10 \)
\( -6x + y = -70 \)
\( 6x - y = 70 \) ......................(2)

Now, we have a system of two linear equations:
1. \( x - 2y = -300 \)
2. \( 6x - y = 70 \)

From equation (1), we can express \( x \) in terms of \( y \):
\( x = 2y - 300 \) ......................(3)

Substitute this value of \( x \) into equation (2):
\( 6(2y - 300) - y = 70 \)
\( 12y - 1800 - y = 70 \)
\( 11y = 70 + 1800 \)
\( 11y = 1870 \)
\( y = \frac { 1870 }{ 11 } \)
\( y = 170 \)

Now, substitute the value of \( y \) back into equation (3) to find \( x \):
\( x = 2(170) - 300 \)
\( x = 340 - 300 \)
\( x = 40 \)

So, the solution to the equations is \( x = 40 \) and \( y = 170 \).
Hence, their respective capitals are \( Rs 40 \) and \( Rs 170 \).

Verification:
Substitute \( x = 40 \) and \( y = 170 \) into equation (1):
\( x - 2y = 40 - 2(170) = 40 - 340 = -300 \) (Correct)
Substitute \( x = 40 \) and \( y = 170 \) into equation (2):
\( 6x - y = 6(40) - 170 = 240 - 170 = 70 \) (Correct)
The solution is correct.
In simple words: We used algebra to represent the two people's money as \( x \) and \( y \). We wrote down what they said as two math sentences. Then, we solved these two math sentences together to find out exactly how much money each person had to start with. The first person has \( Rs 40 \) and the second person has \( Rs 170 \).

Exam Tip: For word problems, carefully translate each statement into an algebraic equation. Define your variables clearly and double-check your equations before solving.

 

Question 3. A tain covered a certain diietaswe at a uniform speed. 1f the tain would have been 10 km/h faster, it would have taken 2 hours lete than the scheduled tiene. And, if the tain 10 km/h; it would have taken 3 hours more than the schedulod tiene. Find the distance covered by the train.
Answer: Let the actual speed of the train be \( x \) km/hour and the actual time taken be \( y \) hours. Then, the distance covered is given by:
Distance = Speed \( \times \) Time
Distance = \( xy \) km

According to the question, if the train were 10 km/h faster, it would take 2 hours less:
\( xy = (x + 10)(y - 2) \)
\( xy = xy - 2x + 10y - 20 \)
\( 0 = -2x + 10y - 20 \)
\( 2x - 10y + 20 = 0 \)
Dividing the entire equation by 2:
\( x - 5y + 10 = 0 \) ......................(1)

And, if the train were 10 km/h slower, it would take 3 hours more:
\( xy = (x - 10)(y + 3) \)
\( xy = xy + 3x - 10y - 30 \)
\( 0 = 3x - 10y - 30 \)
\( 3x - 10y - 30 = 0 \) ......................(2)

To solve equations (1) and (2) by the cross-multiplication method, we draw the diagram below (representing coefficients):

Now, applying the cross-multiplication formula:
\( \frac { x }{ (-5)(-30) - (-10)(10) } = \frac { y }{ (10)(3) - (-30)(1) } = \frac { 1 }{ (1)(-10) - (3)(-5) } \)

Calculating each part:
For \( x \): \( (-5)(-30) - (-10)(10) = 150 - (-100) = 150 + 100 = 250 \)
For \( y \): \( (10)(3) - (-30)(1) = 30 - (-30) = 30 + 30 = 60 \)
For the constant 1: \( (1)(-10) - (3)(-5) = -10 - (-15) = -10 + 15 = 5 \)

So, we have:
\( \frac { x }{ 250 } = \frac { y }{ 60 } = \frac { 1 }{ 5 } \)

From \( \frac { x }{ 250 } = \frac { 1 }{ 5 } \):
\( x = \frac { 250 }{ 5 } \)
\( x = 50 \)

From \( \frac { y }{ 60 } = \frac { 1 }{ 5 } \):
\( y = \frac { 60 }{ 5 } \)
\( y = 12 \)

So, the solution to the equations (1) and (2) is \( x = 50 \) and \( y = 12 \).
The speed of the train is 50 km/h and the time taken is 12 hours.

Therefore, the distance covered by the train is \( 50 \times 12 = 600 \) km.

Verification:
Substitute \( x = 50 \) and \( y = 12 \) into equation (1):
\( x - 5y + 10 = 50 - 5(12) + 10 = 50 - 60 + 10 = 0 \) (Correct)
Substitute \( x = 50 \) and \( y = 12 \) into equation (2):
\( 3x - 10y - 30 = 3(50) - 10(12) - 30 = 150 - 120 - 30 = 0 \) (Correct)
The solution is verified.
In simple words: We created two math sentences from the information about the train's speed, time, and distance. We then solved these sentences using a special method to find the train's speed and the time it traveled. Once we had those, we multiplied them to get the total distance the train covered, which is 600 km.

Exam Tip: For problems involving speed, distance, and time, remember the formula Distance = Speed \( \times \) Time. Set up equations based on how speed and time changes affect the total distance. The cross-multiplication method is a useful technique for solving systems of linear equations.

 

Question 4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Answer: Let the total number of students in the class be \( x \) and the number of rows be \( y \).
Then, the number of students in each row is \( \frac { x }{ y } \).

According to the first condition:
If there are 3 extra students in a row, there would be 1 row less.
So, the number of students per row becomes \( \frac { x }{ y } + 3 \), and the number of rows becomes \( y - 1 \).
The total number of students remains the same:
\( x = \left( \frac { x }{ y } + 3 \right) (y - 1) \)
\( x = \frac { x }{ y } (y - 1) + 3(y - 1) \)
\( x = x - \frac { x }{ y } + 3y - 3 \)
\( 0 = - \frac { x }{ y } + 3y - 3 \)
Multiplying by -1 to make the leading term positive:
\( \frac { x }{ y } - 3y + 3 = 0 \) ................... (1)

According to the second condition:
If there are 3 students less in a row, there would be 2 rows more.
So, the number of students per row becomes \( \frac { x }{ y } - 3 \), and the number of rows becomes \( y + 2 \).
The total number of students remains the same:
\( x = \left( \frac { x }{ y } - 3 \right) (y + 2) \)
\( x = \frac { x }{ y } (y + 2) - 3(y + 2) \)
\( x = x + \frac { 2x }{ y } - 3y - 6 \)
\( 0 = \frac { 2x }{ y } - 3y - 6 \) ................... (2)

Let \( u = \frac { x }{ y } \) . (This is equation 3, as labeled in the source, though it's a substitution).
Substituting \( u \) into equations (1) and (2), they can be rewritten as:
From (1): \( u - 3y + 3 = 0 \) ......................(4)
From (2): \( 2u - 3y - 6 = 0 \) ......................(5)

Now, subtract equation (4) from equation (5):
\( (2u - 3y - 6) - (u - 3y + 3) = 0 - 0 \)
\( 2u - 3y - 6 - u + 3y - 3 = 0 \)
\( u - 9 = 0 \)
\( u = 9 \) ......................(6)

Substitute the value of \( u \) into equation (4):
\( 9 - 3y + 3 = 0 \)
\( 12 - 3y = 0 \)
\( 3y = 12 \)
\( y = \frac { 12 }{ 3 } \)
\( y = 4 \)

Now, using equation (3) \( u = \frac { x }{ y } \) and equation (6) \( u = 9 \), along with \( y = 4 \):
\( \frac { x }{ y } = 9 \)
\( \frac { x }{ 4 } = 9 \)
\( x = 9 \times 4 \)
\( x = 36 \)

So, the solution to the equations is \( x = 36 \) and \( y = 4 \).
Hence, the total number of students in the class is 36.

Verification:
Substitute \( x = 36 \) and \( y = 4 \) into the rewritten equations (1) and (2) (or their original forms):
For equation (1): \( \frac { x }{ y } - 3y + 3 = \frac { 36 }{ 4 } - 3(4) + 3 = 9 - 12 + 3 = 0 \) (Correct)
For equation (2): \( \frac { 2x }{ y } - 3y - 6 = \frac { 2(36) }{ 4 } - 3(4) - 6 = \frac { 72 }{ 4 } - 12 - 6 = 18 - 12 - 6 = 0 \) (Correct)
The solution is correct.
In simple words: We used two variables, \( x \) for total students and \( y \) for rows, to set up two algebraic equations based on the different ways students could stand in rows. By simplifying and solving these equations, we found that there are 36 students in the class and 4 rows.

Exam Tip: When dealing with problems involving arrangement in rows, remember that the total number of items (students) is always (number of rows) \( \times \) (items per row). Carefully set up your equations for each scenario described in the problem.

 

Question 5. In a \( \Delta ABC \), \( \angle C = 3 \angle B = 2 \left(\angle A+\angle B \right) \) Find the three angles.
Answer: We are given the relationship between the angles:
\( \angle C = 3 \angle B = 2 \left(\angle A+\angle B \right) \) ......................(1)

We also know that the sum of the measures of the three angles in any triangle is \( 180^\circ \):
\( \angle A + \angle B + \angle C = 180^\circ \) ......................(2)

From (1), we have two main relationships:
(a) \( \angle C = 3 \angle B \)
(b) \( 3 \angle B = 2 \left(\angle A+\angle B \right) \)

Let's use (b) to find a relationship between \( \angle A \) and \( \angle B \):
\( 3 \angle B = 2 \angle A + 2 \angle B \)
\( 3 \angle B - 2 \angle B = 2 \angle A \)
\( \angle B = 2 \angle A \) ......................(3)

Now we have relationships for \( \angle B \) and \( \angle C \) in terms of \( \angle A \):
From (3), \( \angle B = 2 \angle A \)
From (a), \( \angle C = 3 \angle B = 3(2 \angle A) = 6 \angle A \)

Substitute these expressions into the sum of angles equation (2):
\( \angle A + \angle B + \angle C = 180^\circ \)
\( \angle A + (2 \angle A) + (6 \angle A) = 180^\circ \)
\( 9 \angle A = 180^\circ \)
\( \angle A = \frac { 180^\circ }{ 9 } \)
\( \angle A = 20^\circ \)

Now, find \( \angle B \) and \( \angle C \) using \( \angle A = 20^\circ \):
\( \angle B = 2 \angle A = 2(20^\circ) = 40^\circ \)
\( \angle C = 6 \angle A = 6(20^\circ) = 120^\circ \)

So, the three angles of the \( \Delta ABC \) are \( \angle A = 20^\circ \), \( \angle B = 40^\circ \), and \( \angle C = 120^\circ \).

Verification:
1. Check the sum of angles: \( 20^\circ + 40^\circ + 120^\circ = 180^\circ \) (Correct)
2. Check the given relationship \( \angle C = 3 \angle B \): \( 120^\circ = 3(40^\circ) = 120^\circ \) (Correct)
3. Check the given relationship \( 3 \angle B = 2 \left(\angle A+\angle B \right) \):
\( 3(40^\circ) = 120^\circ \)
\( 2(20^\circ + 40^\circ) = 2(60^\circ) = 120^\circ \)
So, \( 120^\circ = 120^\circ \) (Correct)
All conditions are satisfied, confirming the solution.
In simple words: We started with the special rules given for how the angles in the triangle relate to each other. We also remembered that all angles in a triangle must add up to 180 degrees. By using these two pieces of information, we solved for each angle one by one. We found the angles are 20, 40, and 120 degrees.

Exam Tip: For problems involving angles of a triangle, always start with the fundamental rule that the sum of all interior angles is \( 180^\circ \). Break down complex relationships into simpler equations and solve systematically.

 

Question 6. Draw the graphs of the equations \( 5x – y = 5 \) and \( 3x – y = 3 \), Determine lo co-ordinate of the vertices of the triangle formed by these lines end they-axis.
Answer: The given equations are:
1. \( 5x - y = 5 \)
2. \( 3x - y = 3 \)

Let's find two solutions for each equation to draw their graphs.

For equation (1): \( 5x - y = 5 \implies y = 5x - 5 \)

The points are A(1, 0) and B(2, 5).

For equation (2): \( 3x - y = 3 \implies y = 3x - 3 \)
The points are C(2, 3) and D(3, 6).

Now, we plot these points on a graph and draw lines through them.
* Plot points A(1,0) and B(2,5) and join them to form the line representing \( 5x - y = 5 \).
* Plot points C(2,3) and D(3,6) and join them to form the line representing \( 3x - y = 3 \).

The lines intersect at point A(1,0). Both lines intersect the y-axis at different points.
For \( y = 5x - 5 \), if \( x = 0 \), then \( y = -5 \). So, the y-intercept is E(0,-5).
For \( y = 3x - 3 \), if \( x = 0 \), then \( y = -3 \). So, the y-intercept is F(0,-3).

The triangle is formed by the two lines and the y-axis. Its vertices are the intersection point of the two lines and their respective y-intercepts.
Upon observing the figure, we notice that the coordinates of the vertices of the triangle AEF are:
A: Intersection of the two lines = (1, 0)
E: Y-intercept of the line \( 5x - y = 5 \) = (0, -5)
F: Y-intercept of the line \( 3x - y = 3 \) = (0, -3)
In simple words: We found points for each of the two given equations and drew their lines on a graph. These two lines meet at one point, and each line also crosses the vertical y-axis at its own point. These three special points (where the lines cross each other and where they each cross the y-axis) form a triangle. The corners of this triangle are (1,0), (0,-5), and (0,-3).

Exam Tip: To draw a line, you only need two points. For two linear equations, their intersection point is the common solution. When forming a triangle with the y-axis, the vertices will be the intersection of the lines and the y-intercepts of each line.

 

Question 7. Solve the following pair of hnear equationa:
(i) \( px + qy = p-q \)
\( qx - py = p + q \)
(ii) \( ax + by = c \)
\( bx + ay = 1 + c \)
(iii) \( \frac { x }{ a } + \frac { y }{ b } = 0 \)
\( ax + by = a^2 + b^2 \)
(iv) \( (a - b)x + (a + b)y = a^2-2ab - b^2 \)
\( (a + b) (x + y) = a^2 + b^2 \)
(v) \( 152x – 378y = -74 \)
\( -378x + 152y = -604 \)
Answer:
(i) The given pair of linear equations is:
1. \( px + qy = p-q \) ......................(1)
2. \( qx - py = p+q \) ......................(2)

To eliminate \( y \), multiply equation (1) by \( p \) and equation (2) by \( q \):
From (1) \( \times p \): \( p^2x + pqy = p^2 - pq \) ......................(3)
From (2) \( \times q \): \( q^2x - pqy = pq + q^2 \) ......................(4)

Adding equation (3) and equation (4):
\( (p^2x + pqy) + (q^2x - pqy) = (p^2 - pq) + (pq + q^2) \)
\( p^2x + q^2x = p^2 + q^2 \)
\( (p^2 + q^2)x = p^2 + q^2 \)
\( x = \frac { p^2 + q^2 }{ p^2 + q^2 } \)
\( x = 1 \)

Substitute this value of \( x \) into equation (1):
\( p(1) + qy = p-q \)
\( p + qy = p-q \)
\( qy = p-q-p \)
\( qy = -q \)
\( y = \frac { -q }{ q } \)
\( y = -1 \)

So, the solution to the given pair of linear equations is \( x = 1, y = -1 \).

Verification:
Substitute \( x = 1, y = -1 \) into equation (1):
\( p(1) + q(-1) = p - q \) (Correct)
Substitute \( x = 1, y = -1 \) into equation (2):
\( q(1) - p(-1) = q + p \) (Correct)
The solution is correct.

(ii) The given pair of linear equations is:
1. \( ax + by = c \implies ax + by - c = 0 \) ......................(3)
2. \( bx + ay = 1 + c \implies bx + ay - (1 + c) = 0 \) ......................(4)

To solve these equations by the cross-multiplication method, we draw the diagram below (representing coefficients):

Applying the cross-multiplication formula:
\( \frac { x }{ b(-(1+c)) - a(-c) } = \frac { y }{ (-c)(b) - (-(1+c))(a) } = \frac { 1 }{ (a)(a) - (b)(b) } \)

Calculating each part:
For \( x \): \( -b(1+c) + ac = -b - bc + ac \)
For \( y \): \( -cb + a(1+c) = -bc + a + ac \)
For the constant 1: \( a^2 - b^2 \)

So, we have:
\( \frac { x }{ -b - bc + ac } = \frac { y }{ -bc + a + ac } = \frac { 1 }{ a^2 - b^2 } \)

Therefore, the solution is:
\( x = \frac { -b - bc + ac }{ a^2 - b^2 } \)
\( y = \frac { -bc + a + ac }{ a^2 - b^2 } \)

Verification:
Substitute \( x = \frac { -b - bc + ac }{ a^2 - b^2 } \) and \( y = \frac { -bc + a + ac }{ a^2 - b^2 } \) into the original equations.
For \( ax + by \):
\( a\left( \frac { -b - bc + ac }{ a^2 - b^2 } \right) + b\left( \frac { -bc + a + ac }{ a^2 - b^2 } \right) \)
\( = \frac { -ab - abc + a^2c - b^2c + ab + a^2c }{ a^2 - b^2 } \)
\( = \frac { a^2c - b^2c }{ a^2 - b^2 } = \frac { c(a^2 - b^2) }{ a^2 - b^2 } = c \) (Correct)

For \( bx + ay \):
\( b\left( \frac { -b - bc + ac }{ a^2 - b^2 } \right) + a\left( \frac { -bc + a + ac }{ a^2 - b^2 } \right) \)
\( = \frac { -b^2 - b^2c + abc - abc + a^2 + a^2c }{ a^2 - b^2 } \)
\( = \frac { a^2 - b^2 + a^2c - b^2c }{ a^2 - b^2 } = \frac { (a^2 - b^2) + c(a^2 - b^2) }{ a^2 - b^2 } = \frac { (a^2 - b^2)(1+c) }{ a^2 - b^2 } = 1 + c \) (Correct)
The solution is verified.

(iii) The given pair of linear equations is:
1. \( \frac { x }{ a } + \frac { y }{ b } = 0 \) ......................(1)
2. \( ax + by = a^2 + b^2 \) ......................(2)

From equation (1), we can express \( y \) in terms of \( x \):
\( \frac { y }{ b } = - \frac { x }{ a } \)
\( y = - \frac { b }{ a } x \) ......................(3)

Substitute this value of \( y \) into equation (2):
\( ax + b \left( - \frac { b }{ a } x \right) = a^2 + b^2 \)
\( ax - \frac { b^2 }{ a } x = a^2 + b^2 \)
To remove the fraction, multiply the entire equation by \( a \):
\( a^2x - b^2x = a(a^2 + b^2) \)
\( (a^2 - b^2)x = a(a^2 + b^2) \)
\( x = \frac { a(a^2 + b^2) }{ a^2 - b^2 } \)
However, the source solution shows \( x = a \). Let's recheck the first equation. If it was \( \frac{x}{a} - \frac{y}{b} = 0 \) or if \( \frac{x}{a} = \frac{y}{b} \). The source has \( \frac { y }{ b } = \frac { x }{ a } \implies y = \frac { b }{ a } x \). Let's follow this.

Let's restart (iii) using the solution's first step: From equation (1), \( \frac { y }{ b } = \frac { x }{ a } \). This means \( y = \frac { b }{ a } x \) ......................(3)

Substitute this value of \( y \) into equation (2):
\( ax + b \left(\frac { b }{ a } x \right) = a^2 + b^2 \)
\( ax + \frac { b^2 }{ a } x = a^2 + b^2 \)
\( \frac { a^2x + b^2x }{ a } = a^2 + b^2 \)
\( \frac { (a^2 + b^2)x }{ a } = a^2 + b^2 \)
\( x = \frac { a(a^2 + b^2) }{ a^2 + b^2 } \)
\( x = a \)

Substitute this value of \( x \) into equation (3):
\( y = \frac { b }{ a } (a) \)
\( y = b \)

Hence, the solution to the given pair of linear equations is \( x = a, y = b \).

Verification:
Substitute \( x = a, y = b \) into equation (1):
\( \frac { a }{ a } + \frac { b }{ b } = 1 + 1 = 2 \). This does NOT match the equation \( \frac { x }{ a } + \frac { y }{ b } = 0 \).
The OCR might have incorrectly transcribed \( \frac { x }{ a } + \frac { y }{ b } = 0 \) when it should have been \( \frac { x }{ a } - \frac { y }{ b } = 0 \). Let's assume the source's derivation of \( y = \frac{b}{a}x \) is correct, which implies \( \frac{x}{a} - \frac{y}{b} = 0 \).
Let's re-verify with the assumed equation \( \frac { x }{ a } - \frac { y }{ b } = 0 \):
Substitute \( x = a, y = b \) into \( \frac { x }{ a } - \frac { y }{ b } = 0 \):
\( \frac { a }{ a } - \frac { b }{ b } = 1 - 1 = 0 \) (Correct)
Substitute \( x = a, y = b \) into equation (2):
\( ax + by = a(a) + b(b) = a^2 + b^2 \) (Correct)
So, the solution \( x=a, y=b \) is correct if equation (1) was \( \frac { x }{ a } - \frac { y }{ b } = 0 \). I will present the solution based on the source's derivation which matches this modified first equation.

(iii) The given pair of linear equations is:
1. \( \frac { x }{ a } - \frac { y }{ b } = 0 \) (Derived from source steps, not initial OCR)
2. \( ax + by = a^2 + b^2 \) ......................(2)

From equation (1), we get \( \frac { x }{ a } = \frac { y }{ b } \), which means \( y = \frac { b }{ a } x \) ......................(3)

Substitute this value of \( y \) into equation (2):
\( ax + b \left(\frac { b }{ a } x \right) = a^2 + b^2 \)
\( ax + \frac { b^2 }{ a } x = a^2 + b^2 \)
\( \frac { a^2x + b^2x }{ a } = a^2 + b^2 \)
\( \frac { (a^2 + b^2)x }{ a } = a^2 + b^2 \)
\( x = \frac { a(a^2 + b^2) }{ a^2 + b^2 } \)
\( x = a \)

Substitute this value of \( x \) into equation (3):
\( y = \frac { b }{ a } (a) \)
\( y = b \)

Hence, the solution to the given pair of linear equations is \( x = a, y = b \).
This verifies the solution.

(iv) The given pair of linear equations is:
1. \( (a - b)x + (a + b)y = a^2 - 2ab - b^2 \) ......................(1)
2. \( (a + b)(x + y) = a^2 + b^2 \)
Expanding equation (2):
\( (a + b)x + (a + b)y = a^2 + b^2 \) ......................(2)

Subtract equation (2) from equation (1):
\( [(a - b)x + (a + b)y] - [(a + b)x + (a + b)y] = (a^2 - 2ab - b^2) - (a^2 + b^2) \)
\( (a - b)x - (a + b)x = a^2 - 2ab - b^2 - a^2 - b^2 \)
\( (a - b - (a + b))x = -2ab - 2b^2 \)
\( (a - b - a - b)x = -2ab - 2b^2 \)
\( (-2b)x = -2ab - 2b^2 \)
\( x = \frac { -2ab - 2b^2 }{ -2b } \)
\( x = \frac { -2b(a + b) }{ -2b } \)
\( x = a + b \)

Substitute this value of \( x \) into equation (1):
\( (a - b)(a + b) + (a + b)y = a^2 - 2ab - b^2 \)
Using the difference of squares formula, \( (a - b)(a + b) = a^2 - b^2 \):
\( a^2 - b^2 + (a + b)y = a^2 - 2ab - b^2 \)
\( (a + b)y = a^2 - 2ab - b^2 - (a^2 - b^2) \)
\( (a + b)y = a^2 - 2ab - b^2 - a^2 + b^2 \)
\( (a + b)y = -2ab \)
\( y = \frac { -2ab }{ a + b } \)

Hence, the solution to the given pair of linear equations is \( x = a + b \) and \( y = \frac { -2ab }{ a + b } \).

Verification:
Substitute \( x = a + b \) and \( y = \frac { -2ab }{ a + b } \) into equation (1):
\( (a - b)x + (a + b)y = (a - b)(a + b) + (a + b)\left(\frac { -2ab }{ a + b } \right) \)
\( = (a^2 - b^2) - 2ab \)
\( = a^2 - 2ab - b^2 \) (Correct)

Substitute \( x = a + b \) and \( y = \frac { -2ab }{ a + b } \) into equation (2):
\( (a + b)(x + y) = (a + b)\left( (a + b) + \frac { -2ab }{ a + b } \right) \)
\( = (a + b)^2 + (a + b)\left(\frac { -2ab }{ a + b } \right) \)
\( = (a + b)^2 - 2ab \)
\( = a^2 + 2ab + b^2 - 2ab \)
\( = a^2 + b^2 \) (Correct)
The solution is verified.

(v) The given pair of linear equations is:
1. \( 152x - 378y = -74 \) ......................(1)
2. \( -378x + 152y = -604 \) ......................(2)

Notice that the coefficients of \( x \) and \( y \) are swapped in the two equations. This indicates a special method for solving. Add and subtract the equations.

Adding equation (1) and equation (2):
\( (152x - 378y) + (-378x + 152y) = -74 + (-604) \)
\( (152 - 378)x + (-378 + 152)y = -678 \)
\( -226x - 226y = -678 \)
Divide the entire equation by -226:
\( x + y = \frac { -678 }{ -226 } \)
\( x + y = 3 \) ......................(3)

Subtracting equation (2) from equation (1):
\( (152x - 378y) - (-378x + 152y) = -74 - (-604) \)
\( (152 + 378)x + (-378 - 152)y = -74 + 604 \)
\( 530x - 530y = 530 \)
Divide the entire equation by 530:
\( x - y = \frac { 530 }{ 530 } \)
\( x - y = 1 \) ......................(4)

Now we have a simpler system of equations:
3. \( x + y = 3 \)
4. \( x - y = 1 \)

Adding equation (3) and equation (4):
\( (x + y) + (x - y) = 3 + 1 \)
\( 2x = 4 \)
\( x = \frac { 4 }{ 2 } \)
\( x = 2 \)

Substitute the value of \( x \) into equation (3):
\( 2 + y = 3 \)
\( y = 3 - 2 \)
\( y = 1 \)

Hence, the solution to the given pair of linear equations is \( x = 2, y = 1 \).

Verification:
Substitute \( x = 2, y = 1 \) into equation (1):
\( 152(2) - 378(1) = 304 - 378 = -74 \) (Correct)
Substitute \( x = 2, y = 1 \) into equation (2):
\( -378(2) + 152(1) = -756 + 152 = -604 \) (Correct)
The solution is verified.
In simple words: For each set of equations, we used methods like substitution, elimination, or cross-multiplication. The goal was always to find values for \( x \) and \( y \) that make both equations true. We showed step-by-step how to simplify and solve each pair, and then checked our answers by putting them back into the original equations.

Exam Tip: For solving systems of linear equations, choose the most efficient method (substitution, elimination, or cross-multiplication) based on the coefficients. For equations with swapped coefficients like in part (v), adding and subtracting the equations can simplify them quickly. Always verify your solutions.

 

Question 8. ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.
Answer: We know that the opposite angles of a cyclic quadrilateral are supplementary (their sum is \( 180^\circ \)).
From the given diagram, the angles are:
\( \angle A = 4y + 20^\circ \)
\( \angle B = 3y - 5^\circ \)
\( \angle C = -4x^\circ \)
\( \angle D = -7x + 5^\circ \)

Using the property of cyclic quadrilaterals:
\( \angle A + \angle C = 180^\circ \)
\( (4y + 20^\circ) + (-4x^\circ) = 180^\circ \)
\( 4y - 4x + 20^\circ = 180^\circ \)
\( 4y - 4x = 160^\circ \)
Divide the entire equation by 4:
\( y - x = 40^\circ \) ......................(1)

Also,
\( \angle B + \angle D = 180^\circ \)
\( (3y - 5^\circ) + (-7x + 5^\circ) = 180^\circ \)
\( 3y - 7x - 5^\circ + 5^\circ = 180^\circ \)
\( 3y - 7x = 180^\circ \) ......................(2)

Now we have a system of two linear equations:
1. \( y - x = 40 \)
2. \( 3y - 7x = 180 \)

From equation (1), express \( y \) in terms of \( x \):
\( y = 40 + x \) ......................(3)

Substitute this value of \( y \) into equation (2):
\( 3(40 + x) - 7x = 180 \)
\( 120 + 3x - 7x = 180 \)
\( 120 - 4x = 180 \)
\( -4x = 180 - 120 \)
\( -4x = 60 \)
\( x = \frac { 60 }{ -4 } \)
\( x = -15^\circ \)

Substitute the value of \( x \) into equation (3):
\( y = 40 + (-15) \)
\( y = 40 - 15 \)
\( y = 25^\circ \)

Now, find the measure of each angle using \( x = -15^\circ \) and \( y = 25^\circ \):
\( \angle A = 4y + 20^\circ = 4(25^\circ) + 20^\circ = 100^\circ + 20^\circ = 120^\circ \)
\( \angle B = 3y - 5^\circ = 3(25^\circ) - 5^\circ = 75^\circ - 5^\circ = 70^\circ \)
\( \angle C = -4x^\circ = -4(-15^\circ) = 60^\circ \)
\( \angle D = -7x + 5^\circ = -7(-15^\circ) + 5^\circ = 105^\circ + 5^\circ = 110^\circ \)

Hence, the angles of the cyclic quadrilateral are \( \angle A = 120^\circ \), \( \angle B = 70^\circ \), \( \angle C = 60^\circ \), and \( \angle D = 110^\circ \).

Verification:
Check opposite angles:
\( \angle A + \angle C = 120^\circ + 60^\circ = 180^\circ \) (Correct)
\( \angle B + \angle D = 70^\circ + 110^\circ = 180^\circ \) (Correct)
The solution is correct.
In simple words: We know that opposite angles in a cyclic quadrilateral (a four-sided shape inside a circle) always add up to 180 degrees. We used the given expressions for the angles to create two math equations. By solving these equations, we found the values for \( x \) and \( y \), and then calculated each angle. The angles are 120, 70, 60, and 110 degrees.

Exam Tip: Remember the key property of cyclic quadrilaterals: opposite angles are supplementary. This property allows you to set up two linear equations, which can then be solved using standard algebraic methods to find unknown angle measures.