Maharashtra Board Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.6 Solutions

Choose correct alternative for each of the following questions.

Question 1. To draw graph of 4x + 5y = 19, find y when x = 1.
(a) 4
(b) 3
(c) 2
(d) -3
Answer: (b) 3
In simple words: Substitute x=1 into the equation 4x+5y=19 and solve for y to find the corresponding y-coordinate for graphing.

🎯 Exam Tip: For graphing linear equations, finding a few points by substituting x or y values is crucial. Accuracy in calculation ensures correct graph plotting and scores.

Question 2. For simultaneous equations in variables x and y, Dx = 49, Dy = - 63, D = 7 then what is x?
(a) 7
(b) -7
(c) \( \frac{1}{7} \)
(d) \( -\frac{1}{7} \)
Answer: (a) 7
In simple words: According to Cramer's rule, the value of x in a system of linear equations is found by dividing the determinant Dx by the main determinant D.

🎯 Exam Tip: Remember Cramer's rule formulas: \( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \). A common mistake is interchanging Dx and Dy or using the wrong determinant in the denominator.

Question 3. Find the value of \[ \begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix} \]
(a) -1
(b) -41
(c) 41
(d) 1
Answer: (d) 1
In simple words: To find the value of a 2x2 determinant, multiply the diagonal elements (top-left to bottom-right) and subtract the product of the off-diagonal elements (top-right to bottom-left).

🎯 Exam Tip: Pay close attention to negative signs when performing the subtraction in determinant calculations. A single sign error can lead to an incorrect answer.

Question 4. To solve x + y = 3; 3x - 2y - 4 = 0 by determinant method find D.
(a) 5
(b) 1
(c) -5
(d) -1
Answer: (c) -5
In simple words: First, rewrite the equations in the standard form \(ax + by = c\). Then, the determinant D is formed by the coefficients of x and y from both equations.

🎯 Exam Tip: Ensure equations are in the standard form \(ax + by = c\) before extracting coefficients for D, Dx, or Dy. Misplacing constants can lead to incorrect determinant values.

Question 5. ax + by = c and mx + ny = d and an \( \neq \) bm then these simultaneous equations have-
(a) Only one common solution
(b) No solution
(c) Infinite number of solutions
(d) Only two solutions.
Answer: (a) Only one common solution
In simple words: If the ratio of coefficients a/m is not equal to b/n (which is equivalent to an \( \neq \) bm), then the lines represented by the equations intersect at a single point, meaning there's a unique solution.

🎯 Exam Tip: Remember the conditions for different types of solutions: unique solution (an \( \neq \) bm), no solution (an = bm but cn \( \neq \) dm), and infinite solutions (an = bm = cn = dm).

Question 2. Complete the following table to draw the graph of 2x – 6y = 3.
Answer:

In simple words: To complete the table, substitute the given x-values into the equation \(2x - 6y = 3\) and solve for y, and vice versa, to find the missing coordinates.

🎯 Exam Tip: When completing a table for a graph, ensure precise calculations for each coordinate. These points are critical for accurately plotting the line and checking its consistency.

Question 3. Solve the following simultaneous equations graphically.
(i) 2x + 3y = 12; x - y = 1
(ii) x – 3y = 1 ; 3x – 2y + 4 = 0
(iii) 5x – 6y + 30 = 0; 5x + 4y – 20 = 0
(iv) 3x – y – 2 = 0 ; 2x + y = 8
(v) 3x + y = 10 ; x - y = 2
Answer:
(i) The given simultaneous equations are \[ 2x + 3y = 12 \]
\( \implies \) \( 3y = 12 - 2x \)
\( \implies \) \( y = \frac{12-2x}{3} \)
and \[ x - y = 1 \]
\( \implies \) \( y = x - 1 \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक ग्राफ है जो दो रैखिक समीकरणों 2x + 3y = 12 और x - y = 1 को दर्शाता है। दोनों रेखाएं बिंदु (3,2) पर प्रतिच्छेद करती हैं, जो इन समीकरणों का हल है। ग्राफ पर अक्षों को 1 सेमी = 1 इकाई के पैमाने पर चिह्नित किया गया है। The two lines intersect at point (3,2).
\( \therefore \) x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x - y = 1.
In simple words: To solve graphically, plot points for each equation by finding corresponding x and y values, then draw the lines. The point where the two lines intersect is the solution to the system.

🎯 Exam Tip: Always label both lines on your graph and clearly mark the intersection point. Using a clear scale and accurate plotting is vital for correct graphical solutions.

(ii) The given simultaneous equations are \[ x - 3y = 1 \]
\( \implies \) \( 3y = x - 1 \)
\( \implies \) \( y = \frac{x-1}{3} \)
and \[ 3x - 2y + 4 = 0 \]
\( \implies \) \( 2y = 3x + 4 \)
\( \implies \) \( y = \frac{3x+4}{2} \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ दो रैखिक समीकरणों x - 3y = 1 और 3x - 2y + 4 = 0 के समाधान को दर्शाता है। दोनों रेखाएं बिंदु (-2, -1) पर प्रतिच्छेद करती हैं, जो दिए गए समीकरणों का अद्वितीय हल है। ग्राफ पर अक्षों के लिए 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है। The two lines intersect at point (-2, -1).
\( \therefore \) x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.
In simple words: By plotting points derived from each equation, the intersection of the two lines on the graph provides the (x, y) coordinates that satisfy both equations simultaneously.

🎯 Exam Tip: When one equation involves fractions or requires rearranging, calculate points carefully to ensure accuracy. A small error in a coordinate can shift the line and misrepresent the solution.

(iii) The given simultaneous equations are \[ 5x - 6y + 30 = 0 \]
\( \therefore \) \( 6y = 5x + 30 \)
\( \therefore \) \( y = \frac{5x+30}{6} \)
and \[ 5x + 4y - 20 = 0 \]
\( \therefore \) \( 4y = 20 - 5x \)
\( \therefore \) \( y = \frac{20-5x}{4} \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ दो समीकरणों 5x - 6y + 30 = 0 और 5x + 4y - 20 = 0 का सचित्र समाधान है। दोनों रेखाएं बिंदु (0, 5) पर प्रतिच्छेद करती हैं, जो इस प्रणाली का हल है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना दर्शाया गया है। The two lines intersect at point (0, 5).
\( \therefore \) x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.
In simple words: Graphing each linear equation by plotting points and drawing lines will visually demonstrate the solution as the unique point where they cross each other.

🎯 Exam Tip: When dealing with equations with larger coefficients, choose x-values that result in integer or easily plottable y-values to maintain accuracy and prevent plotting errors.

(iv) The given simultaneous equations are \[ 3x - y - 2 = 0 \]
\( \therefore \) \( y = 3x - 2 \)
and \[ 2x + y = 8 \]
\( \therefore \) \( y = 8 - 2x \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ 3x - y - 2 = 0 और 2x + y = 8 नामक दो रैखिक समीकरणों का समाधान दिखाता है। रेखाएं बिंदु (2, 4) पर प्रतिच्छेद करती हैं, जो इन दोनों समीकरणों को संतुष्ट करने वाला अद्वितीय हल है। ग्राफ पर अक्षों का पैमाना 1 सेमी = 1 इकाई है। The two lines intersect at point (2, 4).
\( \therefore \) x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
In simple words: Solving simultaneous equations graphically means plotting both lines on a coordinate plane; their intersection point visually represents the common solution (x, y) that satisfies both equations.

🎯 Exam Tip: Use a ruler and sharp pencil for drawing lines to ensure accuracy. Small deviations can make it hard to identify the exact intersection point, impacting the final answer.

(v) The given simultaneous equations are \[ 3x + y = 10 \]
\( \therefore \) \( y = 10 - 3x \)
and \[ x - y = 2 \]
\( \therefore \) \( y = x - 2 \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ दो रैखिक समीकरणों 3x + y = 10 और x - y = 2 का ग्राफिकल समाधान प्रदर्शित करता है। दोनों रेखाएं बिंदु (3, 1) पर प्रतिच्छेद करती हैं, जो इन समीकरणों का उभयनिष्ठ हल है। ग्राफ पर अक्षों को 1 सेमी = 1 इकाई के पैमाने पर दिखाया गया है। The two lines intersect at point (3, 1).
\( \therefore \) x = 3 and y = 1 is the solution of the simultaneous equations 3x + y = 10 and x - y = 2.
In simple words: To find the solution graphically, plot multiple points for each linear equation to accurately draw the lines; the single point where they cross each other on the graph is the common solution.

🎯 Exam Tip: Always verify your graphical solution by substituting the coordinates of the intersection point back into both original equations. Both equations must be satisfied for the solution to be correct.

Question 4. Find the values of each of the following determinants.
(i) \[ \begin{vmatrix} 4 & 3 \\ 2 & 7 \end{vmatrix} \]
(ii) \[ \begin{vmatrix} 5 & -2 \\ -3 & 1 \end{vmatrix} \]
(iii) \[ \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \]
Solution:
(i) \[ \begin{vmatrix} 4 & 3 \\ 2 & 7 \end{vmatrix} = (4 \times 7) - (3 \times 2) = 28 - 6 \]
\( \therefore \) \[ \begin{vmatrix} 4 & 3 \\ 2 & 7 \end{vmatrix} = 22 \]
(ii) \[ \begin{vmatrix} 5 & -2 \\ -3 & 1 \end{vmatrix} = (5 \times 1) - (-2 \times -3) = 5 - 6 \]
\( \therefore \) \[ \begin{vmatrix} 5 & -2 \\ -3 & 1 \end{vmatrix} = -1 \]
(iii) \[ \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} = (3 \times 4) - (-1 \times 1) \]
\( = 12 - (-1) = 12 + 1 \)
\( \therefore \) \[ \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} = 13 \]
In simple words: To evaluate a 2x2 determinant, multiply the elements on the main diagonal and subtract the product of the elements on the anti-diagonal.

🎯 Exam Tip: Be careful with signs, especially when one or more elements in the determinant are negative. A sign error is a common mistake in these calculations.

Question 5. Solve the following equations by Cramer's method.
(i) 6x-3y=-10; 3x + 5y - 8 = 0
(ii) 4m - 2n = -4; 4m + 3n = 16
(iii) \( 3x - 2y = \frac{5}{2} \); \( \frac{1}{3}x + 3y = -\frac{4}{3} \)
(iv) 7x + 3y = 15; 12y - 5x = 39
(v) \( \frac{x+y-8}{2} = \frac{x+2y-14}{3} = \frac{3x-y}{4} \)
Solution:
(i) The given simultaneous equations are
6x – 3y = -10 ...(i)
3x + 5y - 8 = 0
\( \therefore \) 3x + 5y = 8 ...(ii)
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), we get
\(a_1 = 6, b_1 = -3, c_1 = -10 \) and
\(a_2 = 3, b_2 = 5, c_2 = 8 \)
\( \therefore D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 6 & -3 \\ 3 & 5 \end{vmatrix} = (6 \times 5) - (-3 \times 3) \)
\( = 30 - (-9) \)
\( \therefore 30 + 9 = 39 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -10 & -3 \\ 8 & 5 \end{vmatrix} = (-10 \times 5) - (-3 \times 8) \)
\( = -50 - (-24) \)
\( = -50 + 24 = -26 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 6 & -10 \\ 3 & 8 \end{vmatrix} = (6 \times 8) - (-10 \times 3) \)
\( = 48 - (-30) \)
\( = 48 + 30 = 78 \)
\( \therefore \) By Cramer's rule, we get
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \therefore x = \frac{-26}{39} \) and \( y = \frac{78}{39} \)
\( \therefore x = -\frac{2}{3} \) and \( y = 2 \)
\( \therefore (x, y) = (-\frac{2}{3}, 2) \) is the solution of the given simultaneous equations.
In simple words: Cramer's method involves calculating three determinants: D (from x and y coefficients), Dx (replace x-coefficients with constants), and Dy (replace y-coefficients with constants). The solution is then found by \(x = D_x/D\) and \(y = D_y/D\).

🎯 Exam Tip: Always check that the main determinant D is not zero. If D=0, Cramer's rule cannot be applied, indicating either no solution or infinitely many solutions.

(ii) The given simultaneous equations are
4m - 2n = -4 ...(i)
4m + 3n = 16 ...(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with \(a_1m + b_1n = c_1\) and \(a_2m + b_2n = c_2\), we get
\(a_1 = 4, b_1 = -2, c_1 = -4 \) and
\(a_2 = 4, b_2 = 3, c_2 = 16 \)
\( \therefore D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 4 & -2 \\ 4 & 3 \end{vmatrix} = (4 \times 3) - (-2 \times 4) \)
\( = 12 - (-8) \)
\( = 12 + 8 = 20 \neq 0 \)
\( D_m = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -4 & -2 \\ 16 & 3 \end{vmatrix} = (-4 \times 3) - (-2 \times 16) \)
\( = -12 - (-32) \)
\( = -12 + 32 = 20 \)
\( D_n = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 4 & -4 \\ 4 & 16 \end{vmatrix} = (4 \times 16) - (-4 \times 4) \)
\( = 64 - (-16) \)
\( = 64 + 16 = 80 \)
\( \therefore \) By Cramer's rule, we get
\( m = \frac{D_m}{D} \) and \( n = \frac{D_n}{D} \)
\( \therefore m = \frac{20}{20} \) and \( n = \frac{80}{20} \)
\( \therefore m = 1 \) and \( n = 4 \)
\( \therefore (m, n) = (1, 4) \) is the solution of the given simultaneous equations.
In simple words: To use Cramer's method, convert the equations into standard form, then calculate the determinants D, Dm (for variable m), and Dn (for variable n), and finally find m and n using the ratios Dm/D and Dn/D respectively.

🎯 Exam Tip: Be meticulous with algebraic signs when calculating determinants, especially with negative numbers. A single sign error can propagate through the entire solution, leading to incorrect values for m and n.

(iii) The given simultaneous equations are
\( 3x - 2y = \frac{5}{2} \) ...(i)
\( \frac{1}{3}x + 3y = -\frac{4}{3} \) ...(ii)
[Multiplying both sides by 3]
\( \therefore x + 9y = -4 \) ...(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), we get
\(a_1 = 3, b_1 = -2, c_1 = \frac{5}{2} \) and
\(a_2 = 1, b_2 = 9, c_2 = -4 \)
\( \therefore D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 3 & -2 \\ 1 & 9 \end{vmatrix} = (3 \times 9) - (-2 \times 1) \)
\( = 27 - (-2) \)
\( = 27 + 2 = 29 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} \frac{5}{2} & -2 \\ -4 & 9 \end{vmatrix} = (\frac{5}{2} \times 9) - (-2 \times -4) \)
\( = \frac{45}{2} - 8 \)
\( = \frac{45-16}{2} = \frac{29}{2} \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 3 & \frac{5}{2} \\ 1 & -4 \end{vmatrix} = (3 \times -4) - (\frac{5}{2} \times 1) \)
\( = -12 - \frac{5}{2} \)
\( = \frac{-24-5}{2} = -\frac{29}{2} \)
\( \therefore \) By Cramer's rule, we get
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \therefore x = \frac{\frac{29}{2}}{29} \) and \( y = \frac{-\frac{29}{2}}{29} \)
\( \therefore x = \frac{1}{2} \) and \( y = -\frac{1}{2} \)
\( \therefore (x, y) = (\frac{1}{2}, -\frac{1}{2}) \) is the solution of the given simultaneous equations.
In simple words: For equations with fractions, first clear the denominators by multiplying by the LCM to get integer coefficients. Then, apply Cramer's rule by calculating D, Dx, and Dy and subsequently solving for x and y.

🎯 Exam Tip: When equations involve fractions, convert them to integer coefficient form first. This simplifies determinant calculations and reduces the chances of arithmetic errors.

(iv) The given simultaneous equations are
7x + 3y = 15 ...(i)
12y - 5x = 39
i.e. -5x + 12y = 39 ...(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), we get
\(a_1 = 7, b_1 = 3, c_1 = 15 \) and
\(a_2 = -5, b_2 = 12, c_2 = 39 \)
\( \therefore D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 7 & 3 \\ -5 & 12 \end{vmatrix} = (7 \times 12) - (3 \times -5) \)
\( = 84 - (-15) \)
\( = 84 + 15 = 99 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} 15 & 3 \\ 39 & 12 \end{vmatrix} = (15 \times 12) - (3 \times 39) \)
\( = 180 - 117 = 63 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 7 & 15 \\ -5 & 39 \end{vmatrix} = (7 \times 39) - (15 \times -5) \)
\( = 273 - (-75) \)
\( = 273 + 75 = 348 \)
\( \therefore \) By Cramer's rule, we get
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \therefore x = \frac{63}{99} \) and \( y = \frac{348}{99} \)
\( \therefore x = \frac{7}{11} \) and \( y = \frac{116}{33} \)
\( \therefore (x, y) = (\frac{7}{11}, \frac{116}{33}) \) is the solution of the given simultaneous equations.
In simple words: To solve by Cramer's rule, first ensure both equations are in the standard \(ax+by=c\) form. Then, compute the determinant of coefficients D, and the determinants Dx and Dy by replacing the respective variable columns with the constant terms. Finally, calculate x and y.

🎯 Exam Tip: Re-arranging equations to \(ax+by=c\) form is a critical first step for Cramer's rule. Incorrect placement of terms, especially the constant, will lead to wrong determinant values and an incorrect solution.

(v) The given simultaneous equations are
\( \frac{x+y-8}{2} = \frac{x+2y-14}{3} = \frac{3x-y}{4} \)
\( \therefore \frac{x+y-8}{2} = \frac{x+2y-14}{3} \)
\( \implies 3(x+y-8) = 2(x+2y-14) \)
\( \implies 3x + 3y - 24 = 2x + 4y - 28 \)
\( \implies 3x - 2x + 3y - 4y = -28 + 24 \)
\( \implies x - y = -4 \) ...(i)
And
\( \frac{x+y-8}{2} = \frac{3x-y}{4} \)
\( \therefore 4(x + y - 8) = 2(3x - y) \)
\( \therefore 4x + 4y - 32 = 6x - 2y \)
\( \therefore 6x - 4x - 2y - 4y = -32 \)
\( \therefore 2x - 6y = -32 \)
\( \therefore x - 3y = -16 \) ...(ii)[Dividing both sides by 2]
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), we get
\(a_1 = 1, b_1 = -1, c_1 = -4 \) and
\(a_2 = 1, b_2 = -3, c_2 = -16 \)
\( \therefore D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 1 & -1 \\ 1 & -3 \end{vmatrix} = (1 \times -3) - (-1 \times 1) \)
\( = -3 - (-1) \)
\( = -3 + 1 = -2 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -4 & -1 \\ -16 & -3 \end{vmatrix} = (-4 \times -3) - (-1 \times -16) \)
\( = 12 - 16 = -4 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 1 & -4 \\ 1 & -16 \end{vmatrix} = (1 \times -16) - (-4 \times 1) \)
\( = -16 - (-4) \)
\( = -16 + 4 = -12 \)
\( \therefore \) By Cramer's rule, we get
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \therefore x = \frac{-4}{-2} \) and \( y = \frac{-12}{-2} \)
\( \therefore x = 2 \) and \( y = 6 \)
\( \therefore (x, y) = (2, 6) \) is the solution of the given simultaneous equations.
In simple words: When given a triple equality of expressions, create two separate pairs of equalities to form two linear equations. Simplify these equations into the standard \(ax+by=c\) form and then apply Cramer's rule to find the values of x and y.

🎯 Exam Tip: For problems with multiple equalities, strategically pick two pairs to form two distinct linear equations. Always simplify these equations to the standard form before proceeding with Cramer's rule to avoid errors.

Question 6. Solve the following simultaneous equations:
(i) \( \frac{2}{x} + \frac{2}{3y} = \frac{1}{6}; \frac{3}{x} + \frac{2}{y} = 0 \)
Answer:
Let \( \frac{1}{x} = p \) and \( \frac{1}{y} = q \).
The given equations become:
\( \frac{2}{x} + \frac{2}{3y} = \frac{1}{6} \implies 2p + \frac{2}{3}q = \frac{1}{6} \)
Multiplying by 3:
\( 6p + 2q = \frac{1}{2} \) ...(iii)
\( \frac{3}{x} + \frac{2}{y} = 0 \implies 3p + 2q = 0 \) ...(iv)

Subtracting equation (iv) from (iii), we get:
\( (6p + 2q) - (3p + 2q) = \frac{1}{2} - 0 \)
\( 3p = \frac{1}{2} \)
\( p = \frac{1}{6} \)

Substituting \( p = \frac{1}{6} \) in equation (iv), we get:
\( 3(\frac{1}{6}) + 2q = 0 \)
\( \frac{1}{2} + 2q = 0 \)
\( 2q = -\frac{1}{2} \)
\( q = -\frac{1}{4} \)

Resubstituting the values of p and q, we get:
\( \frac{1}{x} = p = \frac{1}{6} \implies x = 6 \)
\( \frac{1}{y} = q = -\frac{1}{4} \implies y = -4 \)
Therefore, \( (x, y) = (6, -4) \) is the solution of the given simultaneous equations.
In simple words: We used substitution to convert the fractional equations into linear form, solved for the substituted variables, and then back-substituted to find x and y.

🎯 Exam Tip: When dealing with fractional equations where variables are in the denominator, substitution is a key strategy. Pay close attention to arithmetic during addition/subtraction of equations.

Question 6.
(ii) \( \frac{7}{2x + 1} + \frac{13}{y + 2} = 27; \frac{13}{2x + 1} + \frac{7}{y + 2} = 33 \)
Answer:
Let \( \frac{1}{2x + 1} = p \) and \( \frac{1}{y + 2} = q \).
The given simultaneous equations become:
\( 7p + 13q = 27 \) ...(iii)
\( 13p + 7q = 33 \) ...(iv)

Adding equations (iii) and (iv), we get:
\( (7p + 13q) + (13p + 7q) = 27 + 33 \)
\( 20p + 20q = 60 \)
Dividing both sides by 20:
\( p + q = 3 \) ...(v)

Subtracting equation (iv) from (iii), we get:
\( (7p + 13q) - (13p + 7q) = 27 - 33 \)
\( -6p + 6q = -6 \)
Dividing both sides by -6:
\( p - q = 1 \) ...(vi)

Adding equations (v) and (vi), we get:
\( (p + q) + (p - q) = 3 + 1 \)
\( 2p = 4 \)
\( p = 2 \)

Substituting \( p = 2 \) in equation (v), we get:
\( 2 + q = 3 \)
\( q = 3 - 2 \)
\( q = 1 \)

Resubstituting the values of p and q, we get:
\( \frac{1}{2x + 1} = p = 2 \implies 2(2x + 1) = 1 \)
\( 4x + 2 = 1 \)
\( 4x = 1 - 2 \)
\( 4x = -1 \)
\( x = -\frac{1}{4} \)

\( \frac{1}{y + 2} = q = 1 \implies y + 2 = 1 \)
\( y = 1 - 2 \)
\( y = -1 \)
Therefore, \( (x, y) = (-\frac{1}{4}, -1) \) is the solution of the given simultaneous equations.
In simple words: This problem involves a special type of simultaneous equations where the coefficients are interchanged. We solve them by adding and subtracting the equations to get simpler linear equations, then finding the values of the substituted variables and finally the original variables.

🎯 Exam Tip: For equations where coefficients are swapped (e.g., `ap + bq = c` and `bp + aq = d`), the most efficient method is to add and subtract the equations to form two simpler linear equations.

Question 6.
(iii) \( \frac{148}{x} + \frac{231}{y} = \frac{527}{xy}; \frac{231}{x} + \frac{148}{y} = \frac{610}{xy} \)
Answer:
The given simultaneous equations are:
\( \frac{148}{x} + \frac{231}{y} = \frac{527}{xy} \)
Multiplying both sides by xy:
\( 148y + 231x = 527 \)
i.e., \( 231x + 148y = 527 \) ...(i)

\( \frac{231}{x} + \frac{148}{y} = \frac{610}{xy} \)
Multiplying both sides by xy:
\( 231y + 148x = 610 \)
i.e., \( 148x + 231y = 610 \) ...(ii)

Adding equations (i) and (ii), we get:
\( (231x + 148y) + (148x + 231y) = 527 + 610 \)
\( 379x + 379y = 1137 \)
Dividing both sides by 379:
\( x + y = \frac{1137}{379} \)
\( x + y = 3 \) ...(iii)

Subtracting equation (ii) from (i), we get:
\( (231x + 148y) - (148x + 231y) = 527 - 610 \)
\( 83x - 83y = -83 \)
Dividing both sides by 83:
\( x - y = -1 \) ...(iv)

Adding equations (iii) and (iv), we get:
\( (x + y) + (x - y) = 3 + (-1) \)
\( 2x = 2 \)
\( x = 1 \)

Substituting \( x = 1 \) in equation (iii), we get:
\( 1 + y = 3 \)
\( y = 3 - 1 \)
\( y = 2 \)
Therefore, \( (x, y) = (1, 2) \) is the solution of the given simultaneous equations.
In simple words: First, we converted the equations by multiplying by xy to clear denominators, resulting in a system with interchanged coefficients, which we solved by adding and subtracting.

🎯 Exam Tip: Always look for common multipliers or divisors to simplify equations before solving. This reduces computational complexity and minimizes error potential.

Question 6.
(iv) \( \frac{7x - 2y}{xy} = 5; \frac{8x + 7y}{xy} = 15 \)
Answer:
The given simultaneous equations are:
\( \frac{7x - 2y}{xy} = 5 \implies \frac{7x}{xy} - \frac{2y}{xy} = 5 \implies \frac{7}{y} - \frac{2}{x} = 5 \) ...(i)
\( \frac{8x + 7y}{xy} = 15 \implies \frac{8x}{xy} + \frac{7y}{xy} = 15 \implies \frac{8}{y} + \frac{7}{x} = 15 \) ...(ii)

Let \( \frac{1}{x} = p \) and \( \frac{1}{y} = q \).
Equations (i) and (ii) become:
\( 7q - 2p = 5 \) ...(iii)
\( 8q + 7p = 15 \) ...(iv)

Multiplying equation (iii) by 7, we get:
\( 7(7q - 2p) = 7(5) \implies 49q - 14p = 35 \) ...(v)
Multiplying equation (iv) by 2, we get:
\( 2(8q + 7p) = 2(15) \implies 16q + 14p = 30 \) ...(vi)

Adding equations (v) and (vi), we get:
\( (49q - 14p) + (16q + 14p) = 35 + 30 \)
\( 65q = 65 \)
\( q = 1 \)

Substituting \( q = 1 \) in equation (iv), we get:
\( 8(1) + 7p = 15 \)
\( 8 + 7p = 15 \)
\( 7p = 15 - 8 \)
\( 7p = 7 \)
\( p = 1 \)

Resubstituting the values of p and q, we get:
\( \frac{1}{x} = p = 1 \implies x = 1 \)
\( \frac{1}{y} = q = 1 \implies y = 1 \)
Therefore, \( (x, y) = (1, 1) \) is the solution of the given simultaneous equations.
In simple words: We first separated the fractions to convert the original equations into a linear form by substituting variables, then solved the new system and found the values of x and y.

🎯 Exam Tip: Remember to correctly separate terms when dividing by `xy` to form expressions like `7/y - 2/x`. This is crucial for simplifying complex fractional equations.

Question 6.
(v) \( \frac{1}{2(3x + 4y)} + \frac{1}{5(2x - 3y)} = \frac{1}{4}; \frac{5}{3x + 4y} - \frac{2}{2x - 3y} = \frac{3}{2} \)
Answer:
Let \( \frac{1}{3x + 4y} = p \) and \( \frac{1}{2x - 3y} = q \).
Equations (i) and (ii) become:
\( \frac{1}{2}p + \frac{1}{5}q = \frac{1}{4} \)
Multiplying both sides by 20:
\( 10p + 4q = 5 \) ...(iii)

\( 5p - 2q = \frac{3}{2} \)
Multiplying both sides by 2:
\( 10p - 4q = 3 \) ...(iv)

Adding equations (iii) and (iv), we get:
\( (10p + 4q) + (10p - 4q) = 5 + 3 \)
\( 20p = 8 \)
\( p = \frac{8}{20} = \frac{2}{5} \)

Substituting \( p = \frac{2}{5} \) in equation (iii), we get:
\( 10(\frac{2}{5}) + 4q = 5 \)
\( 4 + 4q = 5 \)
\( 4q = 5 - 4 \)
\( 4q = 1 \)
\( q = \frac{1}{4} \)

Resubstituting the values of p and q, we get:
\( \frac{1}{3x + 4y} = p = \frac{2}{5} \implies 3x + 4y = \frac{5}{2} \) ...(v)
\( \frac{1}{2x - 3y} = q = \frac{1}{4} \implies 2x - 3y = 4 \) ...(vi)

Multiplying equation (v) by 3, we get:
\( 3(3x + 4y) = 3(\frac{5}{2}) \implies 9x + 12y = \frac{15}{2} \) ...(vii)
Multiplying equation (vi) by 4, we get:
\( 4(2x - 3y) = 4(4) \implies 8x - 12y = 16 \) ...(viii)

Adding equations (vii) and (viii), we get:
\( (9x + 12y) + (8x - 12y) = \frac{15}{2} + 16 \)
\( 17x = \frac{15}{2} + \frac{32}{2} \)
\( 17x = \frac{47}{2} \)
\( x = \frac{47}{34} \)

Substituting \( x = \frac{47}{34} \) in equation (vi), we get:
\( 2(\frac{47}{34}) - 3y = 4 \)
\( \frac{47}{17} - 3y = 4 \)
\( -3y = 4 - \frac{47}{17} \)
\( -3y = \frac{68 - 47}{17} \)
\( -3y = \frac{21}{17} \)
\( y = -\frac{21}{17 \times 3} \)
\( y = -\frac{7}{17} \)
Therefore, \( (x, y) = (\frac{47}{34}, -\frac{7}{17}) \) is the solution of the given simultaneous equations.
In simple words: This complex problem was simplified using two layers of substitution: first for expressions involving x and y to get linear equations in p and q, and then solving for x and y using the results.

🎯 Exam Tip: Problems with nested expressions require careful step-by-step substitution. It's vital to re-substitute correctly and solve the resulting simpler equations accurately.

Question 7. Solve the following word problems, i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit's place is 3 more than the digit in the ten's place. Find the original number.
Answer:
Solution:
Let the digit in unit's place be x and that in the ten's place be y.
The original number = \( 10y + x \).
The number obtained by interchanging the digits = \( 10x + y \).

According to the first condition,
Two digit number + the number obtained by interchanging the digits = 143
\( (10y + x) + (10x + y) = 143 \)
\( 11x + 11y = 143 \)
Dividing both sides by 11:
\( x + y = 13 \) ...(i)

According to the second condition,
Digit in unit's place = digit in the ten's place + 3
\( x = y + 3 \)
\( x - y = 3 \) ...(ii)

Adding equations (i) and (ii), we get:
\( (x + y) + (x - y) = 13 + 3 \)
\( 2x = 16 \)
\( x = 8 \)

Putting this value of x in equation (i), we get:
\( 8 + y = 13 \)
\( y = 13 - 8 \)
\( y = 5 \)

The original number is \( 10y + x = 10(5) + 8 = 50 + 8 = 58 \).
In simple words: We formed two linear equations based on the two conditions given for the two-digit number and its interchanged version, then solved them to find the digits and reconstruct the original number.

🎯 Exam Tip: For two-digit number problems, remember the standard representation: `10 * (tens digit) + (units digit)`. This formulation is key to setting up correct equations.

Question 7.
(ii) Kantabai bought \( 1 \frac{1}{2} \) kg tea and 5 kg sugar from a shop. She paid Rs. 50 as return fare for rickshaw. Total expense was Rs. 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So, next month she placed the order online for 2 kg tea and 7 kg sugar. She paid Rs. 880 for that. Find the rate of sugar and tea per kg. Solution:
Answer:
Let the rate of tea be Rs. x per kg and that of sugar be Rs. y per kg.

According to the first condition,
Cost of \( 1 \frac{1}{2} \) kg tea + cost of 5 kg sugar + fare for rickshaw = total expense
\( \frac{3}{2}x + 5y + 50 = 700 \)
\( \frac{3}{2}x + 5y = 700 - 50 \)
\( \frac{3}{2}x + 5y = 650 \)
Multiplying both sides by 2:
\( 3x + 10y = 1300 \) ...(i)

According to the second condition (online order),
Cost of 2 kg tea + cost of 7 kg sugar = total expense
\( 2x + 7y = 880 \) ...(ii)

Multiplying equation (i) by 2, we get:
\( 2(3x + 10y) = 2(1300) \implies 6x + 20y = 2600 \) ...(iii)
Multiplying equation (ii) by 3, we get:
\( 3(2x + 7y) = 3(880) \implies 6x + 21y = 2640 \) ...(iv)

Subtracting equation (iii) from (iv), we get:
\( (6x + 21y) - (6x + 20y) = 2640 - 2600 \)
\( y = 40 \)

Substituting \( y = 40 \) in equation (i), we get:
\( 3x + 10(40) = 1300 \)
\( 3x + 400 = 1300 \)
\( 3x = 1300 - 400 \)
\( 3x = 900 \)
\( x = \frac{900}{3} \)
\( x = 300 \)
The rate of tea is Rs. 300 per kg and that of sugar is Rs. 40 per kg.
In simple words: We set up two linear equations representing Kantabai's two shopping scenarios, considering the cost of tea, sugar, and additional expenses like rickshaw fare, then solved the system to find the per-kg rates for both items.

🎯 Exam Tip: Clearly define variables for unknown quantities. Ensure all costs and conditions are correctly translated into algebraic expressions to avoid errors in equation formation.

Question 7.
(iii) To find number of notes that Anushka had, complete the following activity.
Suppose that Anushka had x notes of Rs. 100 and y notes of Rs. 50 each

The no. of notes (\( \square, \square \))
Answer:
Solution:
Anushka had x notes of Rs. 100 and y notes of Rs. 50.
According to the first condition,
Total amount = Rs. 2500
\( 100x + 50y = 2500 \)
Dividing both sides by 50:
\( 2x + y = 50 \) ...(i)

According to the second condition,
If the number of notes were interchanged (y notes of Rs. 100 and x notes of Rs. 50), the amount would be Rs. 500 less than the previous amount (2500).
So, new amount = \( 2500 - 500 = 2000 \).
\( 50x + 100y = 2000 \)
Dividing both sides by 50:
\( x + 2y = 40 \) ...(ii)

To solve equations (i) and (ii):
Multiply equation (ii) by 2:
\( 2(x + 2y) = 2(40) \implies 2x + 4y = 80 \) ...(iii)

Subtracting equation (i) from (iii), we get:
\( (2x + 4y) - (2x + y) = 80 - 50 \)
\( 3y = 30 \)
\( y = 10 \)

Substituting \( y = 10 \) in equation (i), we get:
\( 2x + 10 = 50 \)
\( 2x = 50 - 10 \)
\( 2x = 40 \)
\( x = \frac{40}{2} \)
\( x = 20 \)

Anushka had 20 notes of Rs. 100 and 10 notes of Rs. 50.
In simple words: We formulated two linear equations from the given conditions about the total value of notes and a hypothetical scenario with interchanged notes, then solved these equations to find the count of 100-Rupee and 50-Rupee notes.

🎯 Exam Tip: For problems involving currency, clearly distinguish between the number of notes and their value. Carefully set up the total value equation for each scenario.

Question 7.
(iv) Sum of the present ages of Manish and Savita is 31, Manish's age 3 years ago was 4 times the age of Savita. Find their present ages.
Answer:
Solution:
Let the present ages of Manish and Savita be x years and y years respectively.

According to the first condition,
\( x + y = 31 \) ...(i)

3 years ago:
Manish's age = \( (x - 3) \) years
Savita's age = \( (y - 3) \) years

According to the second condition,
\( (x - 3) = 4(y - 3) \)
\( x - 3 = 4y - 12 \)
\( x - 4y = -12 + 3 \)
\( x - 4y = -9 \) ...(ii)

Subtracting equation (ii) from (i), we get:
\( (x + y) - (x - 4y) = 31 - (-9) \)
\( x + y - x + 4y = 31 + 9 \)
\( 5y = 40 \)
\( y = \frac{40}{5} \)
\( y = 8 \)

Substituting \( y = 8 \) in equation (i), we get:
\( x + 8 = 31 \)
\( x = 31 - 8 \)
\( x = 23 \)
The present ages of Manish and Savita are 23 years and 8 years respectively.
In simple words: We translated the conditions about the sum of ages and a past age relationship into two linear equations, which were then solved to determine Manish's and Savita's current ages.

🎯 Exam Tip: When dealing with age-related problems, clearly define variables for present ages. Always remember to adjust ages correctly for past or future scenarios (e.g., `x-3` for 3 years ago).

Question 7.
(v) In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is Rs. 720. Find daily wages of skilled and unskilled workers. Solution:
Answer:
Let the daily wages of skilled workers be Rs. x and that of unskilled workers be Rs. y.

According to the first condition,
Ratio of salary of skilled and unskilled workers is 5 : 3
\( \frac{x}{y} = \frac{5}{3} \)
\( 3x = 5y \)
\( 3x - 5y = 0 \) ...(i)

According to the second condition,
Total salary of one day of both of them is Rs. 720
\( x + y = 720 \) ...(ii)

Multiplying equation (ii) by 5:
\( 5(x + y) = 5(720) \implies 5x + 5y = 3600 \) ...(iii)

Adding equations (i) and (iii), we get:
\( (3x - 5y) + (5x + 5y) = 0 + 3600 \)
\( 8x = 3600 \)
\( x = \frac{3600}{8} \)
\( x = 450 \)

Substituting \( x = 450 \) in equation (ii), we get:
\( 450 + y = 720 \)
\( y = 720 - 450 \)
\( y = 270 \)
The daily wages of skilled workers is Rs. 450 and that of unskilled workers is Rs. 270.
In simple words: We formulated two equations from the given ratio of salaries and the total daily salary, then solved this system to find the individual daily wages for skilled and unskilled workers.

🎯 Exam Tip: Ratio problems can be converted into linear equations by cross-multiplication. Always ensure units are consistent (e.g., daily wages for both conditions).

Question 7.
(vi) Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A), Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph. Solution:
Answer:
Let the speeds of Hamid and Joseph be x km/hr and y km/hr respectively.

According to the first condition (traveling towards each other):
Distance between A and B = 30 km.
Time taken to meet = 20 minutes = \( \frac{20}{60} \) hours = \( \frac{1}{3} \) hours.
Distance travelled by Hamid = \( x \times \frac{1}{3} = \frac{x}{3} \) km.
Distance travelled by Joseph = \( y \times \frac{1}{3} = \frac{y}{3} \) km.
Since they travel towards each other and meet, the sum of their distances equals the total distance AB.
\( \frac{x}{3} + \frac{y}{3} = 30 \)
Multiplying both sides by 3:
\( x + y = 90 \) ...(i)

According to the second condition (traveling in the same direction, Hamid catches Joseph):
Time taken for Hamid to catch Joseph = 3 hours.
Distance travelled by Hamid = \( 3x \) km.
Distance travelled by Joseph = \( 3y \) km.
When Hamid (starting from A) catches Joseph (starting from B and going in the same direction as Hamid), Hamid has covered 30 km more than Joseph.
\( 3x = 3y + 30 \)
Dividing both sides by 3:
\( x = y + 10 \)
\( x - y = 10 \) ...(ii)

Adding equations (i) and (ii), we get:
\( (x + y) + (x - y) = 90 + 10 \)
\( 2x = 100 \)
\( x = \frac{100}{2} \)
\( x = 50 \)

Substituting \( x = 50 \) in equation (i), we get:
\( 50 + y = 90 \)
\( y = 90 - 50 \)
\( y = 40 \)
The speeds of Hamid and Joseph are 50 km/hr and 40 km/hr respectively.
In simple words: We set up equations for two relative speed scenarios: when Hamid and Joseph move towards each other (sum of speeds equals distance/time) and when Hamid overtakes Joseph (difference of speeds equals distance/time), then solved to find their individual speeds.

🎯 Exam Tip: For distance-speed-time problems, convert all time units to hours for consistency. Remember that relative speed is sum when moving towards each other, and difference when one overtakes the other.

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