GSEB Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

Question 1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) \( \frac { 1 }{ 2x } + \frac { 1 }{ 3y } = 2 \)
\( \frac { 1 }{ 3x } + \frac { 1 }{ 2y } = \frac { 13 }{ 6 } \)
(ii) \( \frac { 2 }{ \sqrt{x} } + \frac { 3 }{ \sqrt{y} } = 2 \)
\( \frac { 4 }{ \sqrt{x} } - \frac { 9 }{ \sqrt{y} } = -1 \)
(iii) \( \frac { 4 }{ x } + 3y = 14 \)
\( \frac { 3 }{ x } - 4y = 23 \)
(iv) \( \frac { 5 }{ x-1 } + \frac { 1 }{ y-2 } = 2 \)
\( \frac { 6 }{ x-1 } - \frac { 3 }{ y-2 } = 1 \)
(v) \( \frac { 7x-2y }{ xy } = 5 \)
\( \frac { 8x+7y }{ xy } = 15 \)
(vi) \( 6x + 3y = 6xy \)
\( 2x + 4y = 5xy \)
(vii) \( \frac { 10 }{ x+y } + \frac { 2 }{ x-y } = 4 \)
\( \frac { 15 }{ x+y } - \frac { 5 }{ x-y } = -2 \)
(viii) \( \frac { 1 }{ 3x+y } + \frac { 1 }{ 3x-y } = \frac { 3 }{ 4 } \)
\( \frac { 1 }{ 2(3x+y) } - \frac { 1 }{ 2(3x-y) } = -\frac { 1 }{ 8 } \)
Answer:
(i) The provided pair of equations is:
\( \frac { 1 }{ 2x } + \frac { 1 }{ 3y } = 2 \) ............... (1)
\( \frac { 1 }{ 3x } + \frac { 1 }{ 2y } = \frac { 13 }{ 6 } \) .............. (2)
Let's substitute \( \frac { 1 }{ x } = a \) and \( \frac { 1 }{ y } = b \).
Then, equations (1) and (2) can be expressed as:
\( \frac { 1 }{ 2 } a + \frac { 1 }{ 3 } b = 2 \) ............... (5)
\( \frac { 1 }{ 3 } a + \frac { 1 }{ 2 } b = \frac { 13 }{ 6 } \) .............. (6)
Multiplying equation (5) by 6, we get \( 3a + 2b = 12 \) ............... (7)
Multiplying equation (6) by 6, we get \( 2a + 3b = 13 \) ............... (8)
Now, we multiply equation (7) by 3 and equation (8) by 2:
\( 9a + 6b = 36 \) ............... (9)
\( 4a + 6b = 26 \) ............... (10)
Subtracting equation (10) from equation (9), we obtain:
\( (9a + 6b) - (4a + 6b) = 36 - 26 \)
\( 5a = 10 \)
\( a = \frac { 10 }{ 5 } = 2 \) ............... (11)
Substituting this value of \( a = 2 \) into equation (9), we get:
\( 9(2) + 6b = 36 \)
\( 18 + 6b = 36 \)
\( 6b = 36 - 18 \)
\( 6b = 18 \)
\( b = \frac { 18 }{ 6 } = 3 \) ............... (12)
From our initial substitutions \( \frac { 1 }{ x } = a \) and \( \frac { 1 }{ y } = b \):
Using equation (3) and equation (11), we have \( \frac { 1 }{ x } = 2 \), which means \( x = \frac { 1 }{ 2 } \).
Using equation (4) and equation (12), we have \( \frac { 1 }{ y } = 3 \), which means \( y = \frac { 1 }{ 3 } \).
Therefore, the solution for the given set of equations is \( x = \frac { 1 }{ 2 }, y = \frac { 1 }{ 3 } \).
Verification: By placing \( x = \frac { 1 }{ 2 } \) and \( y = \frac { 1 }{ 3 } \) into the original equations, we confirm that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 1 }{ 2(\frac{1}{2}) } + \frac { 1 }{ 3(\frac{1}{3}) } = \frac { 1 }{ 1 } + \frac { 1 }{ 1 } = 1 + 1 = 2 \)
Equation (2): \( \frac { 1 }{ 3(\frac{1}{2}) } + \frac { 1 }{ 2(\frac{1}{3}) } = \frac { 1 }{ \frac{3}{2} } + \frac { 1 }{ \frac{2}{3} } = \frac { 2 }{ 3 } + \frac { 3 }{ 2 } = \frac { 4+9 }{ 6 } = \frac { 13 }{ 6 } \)
The obtained solution is correct.
In simple words: We changed the complex fractions into simpler letters 'a' and 'b'. Then we solved for 'a' and 'b' using basic algebra. Finally, we put 'a' and 'b' back to find 'x' and 'y', which gave us the answer.
Answer:
(ii) The provided pair of equations is:
\( \frac { 2 }{ \sqrt{x} } + \frac { 3 }{ \sqrt{y} } = 2 \) ............... (1)
\( \frac { 4 }{ \sqrt{x} } - \frac { 9 }{ \sqrt{y} } = -1 \) ............... (2)
Let's substitute \( \frac { 1 }{ \sqrt{x} } = u \) and \( \frac { 1 }{ \sqrt{y} } = v \).
Then, equations (1) and (2) can be expressed as:
\( 2u + 3v = 2 \) ............... (5)
\( 4u - 9v = -1 \) ............... (6)
Now, we multiply equation (5) by 3:
\( 3(2u + 3v) = 3(2) \)
\( 6u + 9v = 6 \) ............... (7)
Next, we add equation (6) and equation (7):
\( (4u - 9v) + (6u + 9v) = -1 + 6 \)
\( 10u = 5 \)
\( u = \frac { 5 }{ 10 } = \frac { 1 }{ 2 } \) ............... (8)
Substituting the value of \( u = \frac { 1 }{ 2 } \) into equation (5), we get:
\( 2(\frac { 1 }{ 2 }) + 3v = 2 \)
\( 1 + 3v = 2 \)
\( 3v = 2 - 1 \)
\( 3v = 1 \)
\( v = \frac { 1 }{ 3 } \) ............... (9)
From our initial substitutions \( \frac { 1 }{ \sqrt{x} } = u \) and \( \frac { 1 }{ \sqrt{y} } = v \):
Using equation (3) and equation (8), we have \( \frac { 1 }{ \sqrt{x} } = \frac { 1 }{ 2 } \), which means \( \sqrt{x} = 2 \). Squaring both sides gives \( x = 4 \).
Using equation (4) and equation (9), we have \( \frac { 1 }{ \sqrt{y} } = \frac { 1 }{ 3 } \), which means \( \sqrt{y} = 3 \). Squaring both sides gives \( y = 9 \).
Therefore, the solution for the given set of equations is \( x = 4, y = 9 \).
Verification: By placing \( x = 4 \) and \( y = 9 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 2 }{ \sqrt{4} } + \frac { 3 }{ \sqrt{9} } = \frac { 2 }{ 2 } + \frac { 3 }{ 3 } = 1 + 1 = 2 \)
Equation (2): \( \frac { 4 }{ \sqrt{4} } - \frac { 9 }{ \sqrt{9} } = \frac { 4 }{ 2 } - \frac { 9 }{ 3 } = 2 - 3 = -1 \)
The obtained solution is correct.
In simple words: We replaced the square root terms with 'u' and 'v' to make the equations simpler. We solved for 'u' and 'v', and then used those values to find 'x' and 'y' by squaring.
Answer:
(iii) The provided pair of equations is:
\( \frac { 4 }{ x } + 3y = 14 \) ............... (1)
\( \frac { 3 }{ x } - 4y = 23 \) ............... (2)
Let's substitute \( \frac { 1 }{ x } = X \).
Then, equations (1) and (2) can be expressed as:
\( 4X + 3y = 14 \) ............... (4)
\( 3X - 4y = 23 \) ............... (5)
From equation (5), we can find 'y' in terms of 'X':
\( 4y = 3X - 23 \)
\( y = \frac { 3X-23 }{ 4 } \)
Now, we substitute this expression for 'y' into equation (4):
\( 4X + 3\left( \frac { 3X-23 }{ 4 } \right) = 14 \)
Multiply the entire equation by 4 to remove the denominator:
\( 16X + 3(3X-23) = 14 \times 4 \)
\( 16X + 9X - 69 = 56 \)
\( 25X = 56 + 69 \)
\( 25X = 125 \)
\( X = \frac { 125 }{ 25 } = 5 \) ............... (7)
Substitute this value of \( X = 5 \) into the equation for 'y':
\( y = \frac { 3(5)-23 }{ 4 } \)
\( y = \frac { 15-23 }{ 4 } \)
\( y = \frac { -8 }{ 4 } = -2 \) ............... (8)
From our initial substitution \( \frac { 1 }{ x } = X \):
Using equation (3) and equation (7), we have \( \frac { 1 }{ x } = 5 \), which means \( x = \frac { 1 }{ 5 } \).
Therefore, the solution for the given set of equations is \( x = \frac { 1 }{ 5 }, y = -2 \).
Verification: By placing \( x = \frac { 1 }{ 5 } \) and \( y = -2 \) into the original equations, we confirm that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 4 }{ \frac{1}{5} } + 3(-2) = 4 \times 5 - 6 = 20 - 6 = 14 \)
Equation (2): \( \frac { 3 }{ \frac{1}{5} } - 4(-2) = 3 \times 5 + 8 = 15 + 8 = 23 \)
The obtained solution is correct.
In simple words: We replaced \( \frac{1}{x} \) with 'X' to make the equations linear. Then we solved for 'X' and 'y' using standard methods. Finally, we used the value of 'X' to find 'x', giving us the complete solution.
Answer:
(iv) The provided pair of equations is:
\( \frac { 5 }{ x-1 } + \frac { 1 }{ y-2 } = 2 \) ............... (1)
\( \frac { 6 }{ x-1 } - \frac { 3 }{ y-2 } = 1 \) ............... (2)
Let's substitute \( \frac { 1 }{ x-1 } = u \) and \( \frac { 1 }{ y-2 } = v \).
Then, equations (1) and (2) can be expressed as:
\( 5u + v = 2 \) ............... (5)
\( 6u - 3v = 1 \) ............... (6)
Now, we multiply equation (5) by 3:
\( 3(5u + v) = 3(2) \)
\( 15u + 3v = 6 \)
Next, we add equation (6) and equation (7):
\( (6u - 3v) + (15u + 3v) = 1 + 6 \)
\( 21u = 7 \)
\( u = \frac { 7 }{ 21 } = \frac { 1 }{ 3 } \) ............... (8)
Substituting the value of \( u = \frac { 1 }{ 3 } \) into equation (5), we get:
\( 5(\frac { 1 }{ 3 }) + v = 2 \)
\( \frac { 5 }{ 3 } + v = 2 \)
\( v = 2 - \frac { 5 }{ 3 } \)
\( v = \frac { 6-5 }{ 3 } \)
\( v = \frac { 1 }{ 3 } \) ............... (9)
From our initial substitutions \( \frac { 1 }{ x-1 } = u \) and \( \frac { 1 }{ y-2 } = v \):
Using equation (3) and equation (8), we have \( \frac { 1 }{ x-1 } = \frac { 1 }{ 3 } \), which means \( x-1 = 3 \). Thus, \( x = 3 + 1 = 4 \).
Using equation (4) and equation (9), we have \( \frac { 1 }{ y-2 } = \frac { 1 }{ 3 } \), which means \( y-2 = 3 \). Thus, \( y = 3 + 2 = 5 \).
Therefore, the solution for the given set of equations is \( x = 4, y = 5 \).
Verification: By placing \( x = 4 \) and \( y = 5 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 5 }{ 4-1 } + \frac { 1 }{ 5-2 } = \frac { 5 }{ 3 } + \frac { 1 }{ 3 } = \frac { 6 }{ 3 } = 2 \)
Equation (2): \( \frac { 6 }{ 4-1 } - \frac { 3 }{ 5-2 } = \frac { 6 }{ 3 } - \frac { 3 }{ 3 } = 2 - 1 = 1 \)
The obtained solution is correct.
In simple words: We changed the complex denominators into simpler variables 'u' and 'v'. After solving for 'u' and 'v', we used those values to find 'x' and 'y'.
Answer:
(v) The provided pair of equations is:
\( \frac { 7x-2y }{ xy } = 5 \) ............... (1)
\( \frac { 8x+7y }{ xy } = 15 \) ............... (2)
We can simplify these equations by dividing each term by xy:
Equation (1): \( \frac { 7x }{ xy } - \frac { 2y }{ xy } = 5 \implies \frac { 7 }{ y } - \frac { 2 }{ x } = 5 \) ............... (3)
Equation (2): \( \frac { 8x }{ xy } + \frac { 7y }{ xy } = 15 \implies \frac { 8 }{ y } + \frac { 7 }{ x } = 15 \) ............... (4)
Let's substitute \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \).
Then, equations (3) and (4) can be expressed as:
\( 7v - 2u = 5 \) ............... (5)
\( 8v + 7u = 15 \) ............... (6)
We can rearrange them to standard form for cross-multiplication:
\( -2u + 7v - 5 = 0 \)
\( 7u + 8v - 15 = 0 \)
Using the cross-multiplication method:
\( \frac { u }{ (7)(-15) - (8)(-5) } = \frac { v }{ (-5)(7) - (-15)(-2) } = \frac { 1 }{ (-2)(8) - (7)(7) } \)
\( \frac { u }{ -105 + 40 } = \frac { v }{ -35 - 30 } = \frac { 1 }{ -16 - 49 } \)
\( \frac { u }{ -65 } = \frac { v }{ -65 } = \frac { 1 }{ -65 } \)
This gives us:
\( \frac { u }{ -65 } = \frac { 1 }{ -65 } \implies u = 1 \) ............... (9)
\( \frac { v }{ -65 } = \frac { 1 }{ -65 } \implies v = 1 \) ............... (10)
From our initial substitutions \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \):
Using equation (5) and equation (9), we have \( \frac { 1 }{ x } = 1 \), which means \( x = 1 \).
Using equation (6) and equation (10), we have \( \frac { 1 }{ y } = 1 \), which means \( y = 1 \).
Therefore, the solution for the given set of equations is \( x = 1, y = 1 \).
Verification: By placing \( x = 1 \) and \( y = 1 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 7(1)-2(1) }{ (1)(1) } = \frac { 7-2 }{ 1 } = 5 \)
Equation (2): \( \frac { 8(1)+7(1) }{ (1)(1) } = \frac { 8+7 }{ 1 } = 15 \)
The obtained solution is correct.
In simple words: We first simplified the fractions by dividing by 'xy'. Then, we replaced \( \frac{1}{x} \) and \( \frac{1}{y} \) with 'u' and 'v' to get linear equations. We solved these using cross-multiplication to find 'u' and 'v', and then used those values to find 'x' and 'y'.
Answer:
(vi) The provided pair of equations is:
\( 6x + 3y = 6xy \) ............... (1)
\( 2x + 4y = 5xy \) ............... (2)
We can simplify these equations by dividing throughout by xy:
Equation (1): \( \frac { 6x }{ xy } + \frac { 3y }{ xy } = \frac { 6xy }{ xy } \implies \frac { 6 }{ y } + \frac { 3 }{ x } = 6 \) ............... (1)
Equation (2): \( \frac { 2x }{ xy } + \frac { 4y }{ xy } = \frac { 5xy }{ xy } \implies \frac { 2 }{ y } + \frac { 4 }{ x } = 5 \) ............... (2)
Let's substitute \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \).
Then, equations (1) and (2) can be expressed as:
\( 6v + 3u = 6 \) ............... (5)
\( 2v + 4u = 5 \) ............... (6)
Now, we multiply equation (6) by 3:
\( 3(2v + 4u) = 3(5) \)
\( 6v + 12u = 15 \) ............... (7)
Next, we subtract equation (5) from equation (7):
\( (6v + 12u) - (6v + 3u) = 15 - 6 \)
\( 9u = 9 \)
\( u = \frac { 9 }{ 9 } = 1 \)
Substitute the value of \( u = 1 \) into equation (5), we get:
\( 6v + 3(1) = 6 \)
\( 6v + 3 = 6 \)
\( 6v = 6 - 3 \)
\( 6v = 3 \)
\( v = \frac { 3 }{ 6 } = \frac { 1 }{ 2 } \)
From our initial substitutions \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \):
Using \( u = 1 \), we have \( \frac { 1 }{ x } = 1 \), which means \( x = 1 \).
Using \( v = \frac { 1 }{ 2 } \), we have \( \frac { 1 }{ y } = \frac { 1 }{ 2 } \), which means \( y = 2 \).
Therefore, the solution for the given set of equations is \( x = 1, y = 2 \).
Verification: By placing \( x = 1 \) and \( y = 2 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 6 }{ 2 } + \frac { 3 }{ 1 } = 3 + 3 = 6 \)
Equation (2): \( \frac { 2 }{ 2 } + \frac { 4 }{ 1 } = 1 + 4 = 5 \)
The obtained solution is correct.
In simple words: We first divided all terms by 'xy' to turn the equations into a simpler form. Then, we used 'u' and 'v' for \( \frac{1}{x} \) and \( \frac{1}{y} \). We solved for 'u' and 'v' and then for 'x' and 'y'.
Answer:
(vii) The provided pair of equations is:
\( \frac { 10 }{ x+y } + \frac { 2 }{ x-y } = 4 \) ............... (1)
\( \frac { 15 }{ x+y } - \frac { 5 }{ x-y } = -2 \) ............... (2)
Let's substitute \( \frac { 1 }{ x+y } = u \) and \( \frac { 1 }{ x-y } = v \).
Then, equations (1) and (2) can be expressed as:
\( 10u + 2v = 4 \) ............... (5)
\( 15u - 5v = -2 \) ............... (6)
We can divide equation (5) throughout by 2:
\( 5u + v = 2 \) ............... (7)
Now, we multiply equation (7) by 3:
\( 3(5u + v) = 3(2) \)
\( 15u + 3v = 6 \) ............... (8)
Next, we subtract equation (6) from equation (8):
\( (15u + 3v) - (15u - 5v) = 6 - (-2) \)
\( 15u + 3v - 15u + 5v = 6 + 2 \)
\( 8v = 8 \)
\( v = \frac { 8 }{ 8 } = 1 \)
Substitute the value of \( v = 1 \) into equation (7), we get:
\( 5u + 1 = 2 \)
\( 5u = 2 - 1 \)
\( 5u = 1 \)
\( u = \frac { 1 }{ 5 } \)
From our initial substitutions \( \frac { 1 }{ x+y } = u \) and \( \frac { 1 }{ x-y } = v \):
Using \( u = \frac { 1 }{ 5 } \), we have \( \frac { 1 }{ x+y } = \frac { 1 }{ 5 } \), which means \( x+y = 5 \) ............... (11)
Using \( v = 1 \), we have \( \frac { 1 }{ x-y } = 1 \), which means \( x-y = 1 \) ............... (12)
Now we have a new pair of linear equations for x and y. Add equation (11) and equation (12):
\( (x+y) + (x-y) = 5 + 1 \)
\( 2x = 6 \)
\( x = \frac { 6 }{ 2 } = 3 \)
Substitute this value of \( x = 3 \) into equation (11), we get:
\( 3 + y = 5 \)
\( y = 5 - 3 \)
\( y = 2 \)
Therefore, the solution for the given set of equations is \( x = 3, y = 2 \).
Verification: By placing \( x = 3 \) and \( y = 2 \) into the original equations, we confirm that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 10 }{ 3+2 } + \frac { 2 }{ 3-2 } = \frac { 10 }{ 5 } + \frac { 2 }{ 1 } = 2 + 2 = 4 \)
Equation (2): \( \frac { 15 }{ 3+2 } - \frac { 5 }{ 3-2 } = \frac { 15 }{ 5 } - \frac { 5 }{ 1 } = 3 - 5 = -2 \)
The obtained solution is correct.
In simple words: We used new variables for \( \frac{1}{x+y} \) and \( \frac{1}{x-y} \) to simplify the equations. After solving for these new variables, we got two simple equations for 'x' and 'y', which we then solved to find their values.
Answer:
(viii) The provided pair of equations is:
\( \frac { 1 }{ 3x+y } + \frac { 1 }{ 3x-y } = \frac { 3 }{ 4 } \) ............... (1)
\( \frac { 1 }{ 2(3x+y) } - \frac { 1 }{ 2(3x-y) } = -\frac { 1 }{ 8 } \) ............... (2)
Let's substitute \( \frac { 1 }{ 3x+y } = u \) and \( \frac { 1 }{ 3x-y } = v \).
Then, equation (1) can be expressed as:
\( u + v = \frac { 3 }{ 4 } \) ............... (5)
And equation (2) can be expressed as:
\( \frac { 1 }{ 2 } u - \frac { 1 }{ 2 } v = -\frac { 1 }{ 8 } \)
Multiplying both sides of this equation by 2, we get:
\( u - v = -\frac { 1 }{ 4 } \) ............... (7)
Now we have a new pair of linear equations for u and v. Let's add equation (5) and equation (7):
\( (u + v) + (u - v) = \frac { 3 }{ 4 } + (-\frac { 1 }{ 4 }) \)
\( 2u = \frac { 3 }{ 4 } - \frac { 1 }{ 4 } \)
\( 2u = \frac { 2 }{ 4 } \)
\( 2u = \frac { 1 }{ 2 } \)
\( u = \frac { 1 }{ 4 } \)
Next, subtract equation (7) from equation (5):
\( (u + v) - (u - v) = \frac { 3 }{ 4 } - (-\frac { 1 }{ 4 }) \)
\( u + v - u + v = \frac { 3 }{ 4 } + \frac { 1 }{ 4 } \)
\( 2v = \frac { 4 }{ 4 } \)
\( 2v = 1 \)
\( v = \frac { 1 }{ 2 } \)
From our initial substitutions \( \frac { 1 }{ 3x+y } = u \) and \( \frac { 1 }{ 3x-y } = v \):
Using \( u = \frac { 1 }{ 4 } \), we have \( \frac { 1 }{ 3x+y } = \frac { 1 }{ 4 } \), which means \( 3x+y = 4 \) ............... (10)
Using \( v = \frac { 1 }{ 2 } \), we have \( \frac { 1 }{ 3x-y } = \frac { 1 }{ 2 } \), which means \( 3x-y = 2 \) ............... (11)
Now we have a new pair of linear equations for x and y. Let's add equation (10) and equation (11):
\( (3x+y) + (3x-y) = 4 + 2 \)
\( 6x = 6 \)
\( x = \frac { 6 }{ 6 } = 1 \)
Substitute this value of \( x = 1 \) into equation (10), we get:
\( 3(1) + y = 4 \)
\( 3 + y = 4 \)
\( y = 4 - 3 \)
\( y = 1 \)
Therefore, the solution for the given set of equations is \( x = 1, y = 1 \).
Verification: By placing \( x = 1 \) and \( y = 1 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 1 }{ 3(1)+1 } + \frac { 1 }{ 3(1)-1 } = \frac { 1 }{ 4 } + \frac { 1 }{ 2 } = \frac { 1+2 }{ 4 } = \frac { 3 }{ 4 } \)
Equation (2): \( \frac { 1 }{ 2(3(1)+1) } - \frac { 1 }{ 2(3(1)-1) } = \frac { 1 }{ 2(4) } - \frac { 1 }{ 2(2) } = \frac { 1 }{ 8 } - \frac { 1 }{ 4 } = \frac { 1-2 }{ 8 } = -\frac { 1 }{ 8 } \)
The obtained solution is correct.
In simple words: We used new letters 'u' and 'v' to stand for the complex fractions with 'x' and 'y' in them. We solved for 'u' and 'v'. Then, using those results, we found two simpler equations for 'x' and 'y', which we solved to get the final answer.

Exam Tip: For problems involving fractions with variables, always try to simplify them by substituting new variables for the reciprocal terms or complex expressions. This converts non-linear equations into linear ones, making them much easier to solve.

Question 2. Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Answer:
(i) Let Ritu's speed of rowing in still water be \( x \) km/hour, and the speed of the current be \( y \) km/hour.
Her speed when rowing downstream will be \( (x+y) \) km/hour.
Her speed when rowing upstream will be \( (x-y) \) km/hour.
We use the formula: Time \( = \frac { \text{Distance} }{ \text{Speed} } \)
In the first situation, Ritu travels 20 km downstream in 2 hours:
\( \frac { 20 }{ x+y } = 2 \)
This simplifies to \( x+y = 10 \) ............... (1)
In the second situation, she travels 4 km upstream in 2 hours:
\( \frac { 4 }{ x-y } = 2 \)
This simplifies to \( x-y = 2 \) ............... (2)
Now, we add equation (1) and equation (2):
\( (x+y) + (x-y) = 10 + 2 \)
\( 2x = 12 \)
\( x = \frac { 12 }{ 2 } = 6 \)
Substitute this value of \( x = 6 \) into equation (1), we get:
\( 6 + y = 10 \)
\( y = 10 - 6 \)
\( y = 4 \)
Hence, Ritu's speed of rowing in still water is 6 km/hour, and the speed of the current is 4 km/hour.
Verification: By placing \( x = 6 \) and \( y = 4 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( 6 + 4 = 10 \)
Equation (2): \( 6 - 4 = 2 \)
The obtained solution is correct.
In simple words: We used 'x' for Ritu's speed and 'y' for the current's speed. Then we wrote equations for her speed going with and against the current. Solving these equations gave us both her speed and the current's speed.

Exam Tip: Remember that when rowing downstream, speeds add up (boat + current), and when rowing upstream, speeds subtract (boat - current). Always clearly define your variables and use the distance = speed × time formula.

Question 2. (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Answer:
(ii) Let the time taken by 1 woman alone to complete the embroidery work be \( x \) days.
Let the time taken by 1 man alone to complete the embroidery work be \( y \) days.
The work done by 1 woman in 1 day is \( \frac { 1 }{ x } \).
The work done by 1 man in 1 day is \( \frac { 1 }{ y } \).
So, the work done by 2 women in 1 day is \( \frac { 2 }{ x } \).
And the work done by 5 men in 1 day is \( \frac { 5 }{ y } \).
Given that 2 women and 5 men together finish the work in 4 days, their combined 1-day work is \( \frac { 1 }{ 4 } \).
So, we have the equation: \( \frac { 2 }{ x } + \frac { 5 }{ y } = \frac { 1 }{ 4 } \) ............... (1)
Similarly, the work done by 3 women in 1 day is \( \frac { 3 }{ x } \).
And the work done by 6 men in 1 day is \( \frac { 6 }{ y } \).
Given that 3 women and 6 men together finish the work in 3 days, their combined 1-day work is \( \frac { 1 }{ 3 } \).
So, we have the equation: \( \frac { 3 }{ x } + \frac { 6 }{ y } = \frac { 1 }{ 3 } \) ............... (2)
Let's substitute \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \).
Then, equations (1) and (2) can be expressed as:
\( 2u + 5v = \frac { 1 }{ 4 } \) ............... (5)
\( 3u + 6v = \frac { 1 }{ 3 } \) ............... (6)
Now, we multiply equation (5) by 3 and equation (6) by 2:
Multiplying (5) by 3: \( 3(2u + 5v) = 3(\frac { 1 }{ 4 }) \implies 6u + 15v = \frac { 3 }{ 4 } \) ............... (7)
Multiplying (6) by 2: \( 2(3u + 6v) = 2(\frac { 1 }{ 3 }) \implies 6u + 12v = \frac { 2 }{ 3 } \) ............... (8)
Next, we subtract equation (8) from equation (7):
\( (6u + 15v) - (6u + 12v) = \frac { 3 }{ 4 } - \frac { 2 }{ 3 } \)
\( 3v = \frac { 9 - 8 }{ 12 } \)
\( 3v = \frac { 1 }{ 12 } \)
\( v = \frac { 1 }{ 36 } \)
Substitute this value of \( v = \frac { 1 }{ 36 } \) into equation (5), we get:
\( 2u + 5(\frac { 1 }{ 36 }) = \frac { 1 }{ 4 } \)
\( 2u + \frac { 5 }{ 36 } = \frac { 1 }{ 4 } \)
\( 2u = \frac { 1 }{ 4 } - \frac { 5 }{ 36 } \)
\( 2u = \frac { 9 - 5 }{ 36 } \)
\( 2u = \frac { 4 }{ 36 } \)
\( 2u = \frac { 1 }{ 9 } \)
\( u = \frac { 1 }{ 18 } \)
From our initial substitutions \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \):
Using \( u = \frac { 1 }{ 18 } \), we have \( \frac { 1 }{ x } = \frac { 1 }{ 18 } \), which means \( x = 18 \).
Using \( v = \frac { 1 }{ 36 } \), we have \( \frac { 1 }{ y } = \frac { 1 }{ 36 } \), which means \( y = 36 \).
Therefore, the time taken by 1 woman alone to complete the work is 18 days, and the time taken by 1 man alone to complete the work is 36 days.
Verification: By placing \( x = 18 \) and \( y = 36 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 2 }{ 18 } + \frac { 5 }{ 36 } = \frac { 1 }{ 9 } + \frac { 5 }{ 36 } = \frac { 4+5 }{ 36 } = \frac { 9 }{ 36 } = \frac { 1 }{ 4 } \)
Equation (2): \( \frac { 3 }{ 18 } + \frac { 6 }{ 36 } = \frac { 1 }{ 6 } + \frac { 1 }{ 6 } = \frac { 2 }{ 6 } = \frac { 1 }{ 3 } \)
The obtained solution is correct.
In simple words: We figured out how much work one woman or one man does in a single day. Then, we set up equations based on their combined work and solved for how many days each would take to do the job alone.

Exam Tip: For work-related problems, always think about the "rate of work" (work done per day) as the reciprocal of the total time taken. This converts the problem into linear equations involving fractions, which can then be solved by substitution.

Question 2. (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Answer:
(iii) Let the speed of the train be \( x \) km/hour and the speed of the bus be \( y \) km/hour.
We use the formula: Time \( = \frac { \text{Distance} }{ \text{Speed} } \)
Scenario 1: Roohi travels 60 km by train and the remaining distance by bus.
Remaining distance by bus \( = 300 - 60 = 240 \) km.
Total time taken \( = 4 \) hours.
So, \( \frac { 60 }{ x } + \frac { 240 }{ y } = 4 \)
Dividing the entire equation by 60, we get:
\( \frac { 1 }{ x } + \frac { 4 }{ y } = \frac { 4 }{ 60 } \implies \frac { 1 }{ x } + \frac { 4 }{ y } = \frac { 1 }{ 15 } \) ............... (1)
Scenario 2: She travels 100 km by train and the remaining distance by bus.
Remaining distance by bus \( = 300 - 100 = 200 \) km.
Total time taken \( = 4 \) hours 10 minutes. Convert this to hours: \( 4 + \frac { 10 }{ 60 } = 4 + \frac { 1 }{ 6 } = \frac { 24+1 }{ 6 } = \frac { 25 }{ 6 } \) hours.
So, \( \frac { 100 }{ x } + \frac { 200 }{ y } = \frac { 25 }{ 6 } \)
Dividing the entire equation by 25, we get:
\( \frac { 4 }{ x } + \frac { 8 }{ y } = \frac { 1 }{ 6 } \) ............... (2)
Let's substitute \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \).
Then, equations (1) and (2) can be expressed as:
\( u + 4v = \frac { 1 }{ 15 } \) ............... (3)
\( 4u + 8v = \frac { 1 }{ 6 } \) ............... (4)
Now, we multiply equation (3) by 2:
\( 2(u + 4v) = 2(\frac { 1 }{ 15 }) \implies 2u + 8v = \frac { 2 }{ 15 } \) ............... (5)
Next, we subtract equation (5) from equation (4):
\( (4u + 8v) - (2u + 8v) = \frac { 1 }{ 6 } - \frac { 2 }{ 15 } \)
\( 2u = \frac { 5 - 4 }{ 30 } \)
\( 2u = \frac { 1 }{ 30 } \)
\( u = \frac { 1 }{ 60 } \)
Substitute this value of \( u = \frac { 1 }{ 60 } \) into equation (3), we get:
\( \frac { 1 }{ 60 } + 4v = \frac { 1 }{ 15 } \)
\( 4v = \frac { 1 }{ 15 } - \frac { 1 }{ 60 } \)
\( 4v = \frac { 4 - 1 }{ 60 } \)
\( 4v = \frac { 3 }{ 60 } \)
\( 4v = \frac { 1 }{ 20 } \)
\( v = \frac { 1 }{ 80 } \)
From our initial substitutions \( \frac { 1 }{ x } = u \) and \( \frac { 1 }{ y } = v \):
Using \( u = \frac { 1 }{ 60 } \), we have \( \frac { 1 }{ x } = \frac { 1 }{ 60 } \), which means \( x = 60 \).
Using \( v = \frac { 1 }{ 80 } \), we have \( \frac { 1 }{ y } = \frac { 1 }{ 80 } \), which means \( y = 80 \).
Therefore, the speed of the train is 60 km/hour, and the speed of the bus is 80 km/hour.
Verification: By placing \( x = 60 \) and \( y = 80 \) into the original equations, we find that both equations (1) and (2) are satisfied:
Equation (1): \( \frac { 1 }{ 60 } + \frac { 4 }{ 80 } = \frac { 1 }{ 60 } + \frac { 1 }{ 20 } = \frac { 1+3 }{ 60 } = \frac { 4 }{ 60 } = \frac { 1 }{ 15 } \)
Equation (2): \( \frac { 4 }{ 60 } + \frac { 8 }{ 80 } = \frac { 1 }{ 15 } + \frac { 1 }{ 10 } = \frac { 2+3 }{ 30 } = \frac { 5 }{ 30 } = \frac { 1 }{ 6 } \)
The obtained solution is correct.
In simple words: We used 'x' and 'y' for the speeds of the train and bus. We wrote equations for the time taken in two different travel plans. After solving these equations, we found the individual speeds of the train and bus.

Exam Tip: Always be careful with units, especially converting minutes to hours for time calculations. Clearly define variables for speeds and distances, and use the formula time = distance/speed consistently to set up your equations.