Maharashtra Board Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.3 Solutions

Question 1. Fill in the blanks with correct number.
\[ \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} = 3 \times \text{___} - \text{___} \times 4 = \text{___} - 8 = \text{___} \]
Answer:
\[ \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} = 3 \times \mathbf{5} - \mathbf{2} \times 4 = \mathbf{15} - 8 = \mathbf{7} \]
The filled values are 5, 2, 15, and 7 respectively to complete the determinant evaluation.
In simple words: To solve a 2x2 determinant, we cross-multiply the numbers. We multiply the top-left number by the bottom-right number, and then subtract the product of the other two numbers.

🎯 Exam Tip: Always write down the cross-multiplication steps clearly before calculating the final subtraction to avoid simple calculation mistakes with negative signs.

Practice Set 1.3

Question 1. Fill in the blanks with correct numbers to solve the determinant:
\[ \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} = 3 \times \boxed{\text{ }} - \boxed{\text{ }} \times 4 = \boxed{\text{ }} - 8 = \boxed{\text{ }} \]
Answer:
\[ \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} = 3 \times \boxed{5} - \boxed{2} \times 4 \]
\[ = \boxed{15} - 8 \]
\[ = \boxed{7} \]
This systematic cross-multiplication method is fundamental to solving linear equations using Cramer's Rule.
In simple words: To solve a determinant, we multiply the top-left number by the bottom-right number, and then subtract the product of the top-right and bottom-left numbers.

🎯 Exam Tip: Always write down the intermediate multiplication steps clearly to avoid simple calculation errors under exam pressure.

Question 2. Find the values of following determinants.
(i) \( \begin{vmatrix} -1 & 7 \\ 2 & 4 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 5 & 3 \\ -7 & 0 \end{vmatrix} \)
(iii) \( \begin{vmatrix} \frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2} \end{vmatrix} \)
Answer:
(i) \( \begin{vmatrix} -1 & 7 \\ 2 & 4 \end{vmatrix} \)
\( = (-1 \times 4) - (7 \times 2) \)
\( = -4 - 14 \)
\( = -18 \)

(ii) \( \begin{vmatrix} 5 & 3 \\ -7 & 0 \end{vmatrix} \)
\( = (5 \times 0) - (3 \times -7) \)
\( = 0 - (-21) \)
\( = 0 + 21 \)
\( = 21 \)

(iii) \( \begin{vmatrix} \frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2} \end{vmatrix} \)
\( = \left(\frac{7}{3} \times \frac{1}{2}\right) - \left(\frac{5}{3} \times \frac{3}{2}\right) \)
\( = \frac{7}{6} - \frac{15}{6} \)
\( = \frac{7 - 15}{6} \)
\( = \frac{-8}{6} \)
\( = -\frac{4}{3} \)

Simplifying fractions to their lowest terms ensures the final answer is mathematically elegant and complete.
In simple words: We find the value of each determinant by cross-multiplying the diagonal terms and subtracting the second product from the first.

🎯 Exam Tip: Pay close attention to negative signs during subtraction, as subtracting a negative number turns it into addition (e.g., \(0 - (-21) = 21\)).

Question 2. Find the values of the following determinants:
(i) \( \begin{vmatrix} -1 & 7 \\ 2 & 4 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 5 & 3 \\ -7 & 0 \end{vmatrix} \)
(iii) \( \begin{vmatrix} \frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2} \end{vmatrix} \)
Answer:
(i) \( \begin{vmatrix} -1 & 7 \\ 2 & 4 \end{vmatrix} = (-1 \times 4) - (7 \times 2) \)
\( = -4 - 14 \)
\( \therefore \begin{vmatrix} -1 & 7 \\ 2 & 4 \end{vmatrix} = -18 \)

(ii) \( \begin{vmatrix} 5 & 3 \\ -7 & 0 \end{vmatrix} = (5 \times 0) - (3 \times -7) \)
\( = 0 - (-21) \)
\( \therefore \begin{vmatrix} 5 & 3 \\ -7 & 0 \end{vmatrix} = 21 \)

(iii) \( \begin{vmatrix} \frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2} \end{vmatrix} = \left(\frac{7}{3} \times \frac{1}{2}\right) - \left(\frac{5}{3} \times \frac{3}{2}\right) \)
\( = \frac{7}{6} - \frac{15}{6} \)
\( = \frac{7-15}{6} \)
\( = \frac{-8}{6} \)
\( \therefore \begin{vmatrix} \frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2} \end{vmatrix} = \frac{-4}{3} \)
Determinants are useful mathematical tools for solving systems of linear equations.
In simple words: To find the value of a \( 2 \times 2 \) determinant, we cross-multiply the numbers (top-left with bottom-right, and top-right with bottom-left) and subtract the second product from the first.

🎯 Exam Tip: Be extremely careful with negative signs when cross-multiplying. Double-check calculations like \( 0 - (-21) = 21 \) to avoid silly sign errors.

Question 3. Solve the following simultaneous equations using Cramer’s rule.
(i) \( 3x - 4y = 10 \); \( 4x + 3y = 5 \)
(ii) \( 4x + 3y - 4 = 0 \); \( 6x = 8 - 5y \)
(iii) \( x + 2y = -1 \); \( 2x - 3y = 12 \)
(iv) \( 6x - 4y = -12 \); \( 8x - 3y = -2 \)
(v) \( 4m + 6n = 54 \); \( 3m + 2n = 28 \)
(vi) \( 2x + 3y = 2 \); \( x - \frac{y}{2} = \frac{1}{2} \)
Answer:
(i) The given simultaneous equations are:
\( 3x - 4y = 10 \) ...(i)
\( 4x + 3y = 5 \) ...(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:
\( a_1 = 3, b_1 = -4, c_1 = 10 \)
\( a_2 = 4, b_2 = 3, c_2 = 5 \)
\( D = \begin{vmatrix} 3 & -4 \\ 4 & 3 \end{vmatrix} = (3 \times 3) - (-4 \times 4) = 9 - (-16) = 9 + 16 = 25 \)
\( D_x = \begin{vmatrix} 10 & -4 \\ 5 & 3 \end{vmatrix} = (10 \times 3) - (-4 \times 5) = 30 - (-20) = 30 + 20 = 50 \)
\( D_y = \begin{vmatrix} 3 & 10 \\ 4 & 5 \end{vmatrix} = (3 \times 5) - (10 \times 4) = 15 - 40 = -25 \)
By Cramer's rule,
\( x = \frac{D_x}{D} = \frac{50}{25} = 2 \)
\( y = \frac{D_y}{D} = \frac{-25}{25} = -1 \)
\( \therefore (x, y) = (2, -1) \) is the solution of the given simultaneous equations. This systematic method is highly reliable and reduces the chance of algebraic errors compared to elimination.
In simple words: Cramer's rule is a neat way to solve two equations by using determinants. We calculate three determinants (\( D \), \( D_x \), and \( D_y \)) and then divide to find the values of \( x \) and \( y \).

🎯 Exam Tip: Always write down the general form \( a_1x + b_1y = c_1 \) and clearly list the values of \( a_1, b_1, c_1, a_2, b_2, c_2 \) before calculating the determinants to prevent substitution mistakes.

Question 1. Solve the following simultaneous equations using Cramer's rule:
(i) \( 3x - 4y = 10 \); \( 4x + 3y = 5 \)
(ii) \( 4x + 3y - 4 = 0 \); \( 6x = 8 - 5y \)
Answer:
(i) \( 3x - 4y = 10 \); \( 4x + 3y = 5 \)
Comparing the given equations with \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), we get
\( a_1 = 3, b_1 = -4, c_1 = 10 \) and
\( a_2 = 4, b_2 = 3, c_2 = 5 \)
\( \therefore D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 3 & -4 \\ 4 & 3 \end{vmatrix} = (3 \times 3) - (-4 \times 4) \)
\( = 9 - (-16) \)
\( = 9 + 16 = 25 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} 10 & -4 \\ 5 & 3 \end{vmatrix} = (10 \times 3) - (-4 \times 5) \)
\( = 30 - (-20) \)
\( = 30 + 20 = 50 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 3 & 10 \\ 4 & 5 \end{vmatrix} = (3 \times 5) - (10 \times 4) \)
\( = 15 - 40 = -25 \)
\( \therefore \) By Cramer's rule, we get
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \dots x = \frac{50}{25} \) and \( y = \frac{-25}{25} \)
\( \therefore x = 2 \) and \( y = -1 \)
\( \therefore (x, y) = (2, -1) \) is the solution of the given simultaneous equations.

(ii) \( 4x + 3y - 4 = 0 \); \( 6x = 8 - 5y \)
The given simultaneous equations are
\( 4x + 3y - 4 = 0 \)
\( \therefore 4x + 3y = 4 \) ...(i)
\( 6x = 8 - 5y \)
\( \therefore 6x + 5y = 8 \) ...(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
\( a_1 = 4, b_1 = 3, c_1 = 4 \) and
\( a_2 = 6, b_2 = 5, c_2 = 8 \)
\( \therefore D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 4 & 3 \\ 6 & 5 \end{vmatrix} = (4 \times 5) - (3 \times 6) \)
\( = 20 - 18 = 2 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} 4 & 3 \\ 8 & 5 \end{vmatrix} = (4 \times 5) - (3 \times 8) \)
\( = 20 - 24 = -4 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 4 & 4 \\ 6 & 8 \end{vmatrix} = (4 \times 8) - (4 \times 6) \)
\( = 32 - 24 = 8 \)
\( \dots \) By Cramer's rule, we get
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \therefore x = \frac{-4}{2} \) and \( y = \frac{8}{2} \)
\( \therefore x = -2 \) and \( y = 4 \)
\( \therefore (x, y) = (-2, 4) \) is the solution of the given simultaneous equations.
In simple words: To solve these equations using Cramer's rule, we first write them in standard form, find the determinants \( D \), \( D_x \), and \( D_y \), and then divide to find the values of \( x \) and \( y \).

🎯 Exam Tip: Always double-check the signs when calculating determinants, especially when multiplying negative numbers, as a single sign error will affect the entire solution.

Question 3(ii). Solve the following simultaneous equations using Cramer's rule: \( 4x + 3y = 4 \); \( 6x + 5y = 8 \)
Answer:
\( D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 4 & 3 \\ 6 & 5 \end{vmatrix} = (4 \times 5) - (3 \times 6) \)
\( = 20 - 18 = 2 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} 4 & 3 \\ 8 & 5 \end{vmatrix} = (4 \times 5) - (3 \times 8) \)
\( = 20 - 24 = -4 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 4 & 4 \\ 6 & 8 \end{vmatrix} = (4 \times 8) - (4 \times 6) \)
\( = 32 - 24 = 8 \)
By Cramer's rule, we get
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)

\( \implies x = \frac{-4}{2} \) and \( y = \frac{8}{2} \)

\( \implies x = -2 \) and \( y = 4 \)
\( \therefore (x, y) = (-2, 4) \) is the solution of the given simultaneous equations. An extra step of verification can confirm that these values satisfy both original equations.
In simple words: We find the values of D, Dx, and Dy using determinants, and then divide Dx and Dy by D to get the values of x and y.

🎯 Exam Tip: Always write the formula \( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \) clearly before substituting the values to ensure you get full step marks.

Question 3(iii). Solve the following simultaneous equations using Cramer's rule: \( x + 2y = -1 \); \( 2x - 3y = 12 \)
Answer:
The given simultaneous equations are:
\( x + 2y = -1 \) ...(i)
\( 2x - 3y = 12 \) ...(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:
\( a_1 = 1, b_1 = 2, c_1 = -1 \) and
\( a_2 = 2, b_2 = -3, c_2 = 12 \)
Now, let's find the determinants:
\( D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix} = (1 \times -3) - (2 \times 2) = -3 - 4 = -7 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -1 & 2 \\ 12 & -3 \end{vmatrix} = (-1 \times -3) - (2 \times 12) = 3 - 24 = -21 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 1 & -1 \\ 2 & 12 \end{vmatrix} = (1 \times 12) - (-1 \times 2) = 12 + 2 = 14 \)
By Cramer's rule, we get:
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)

\( \implies x = \frac{-21}{-7} \) and \( y = \frac{14}{-7} \)

\( \implies x = 3 \) and \( y = -2 \)
\( \therefore (x, y) = (3, -2) \) is the solution of the given simultaneous equations. This method is highly efficient for solving linear equations with integer coefficients.
In simple words: We compare the equations to the standard form to find the coefficients, calculate the three determinants, and then divide to find the final values of x and y.

🎯 Exam Tip: Be extremely careful with negative signs when calculating determinants, as a single sign error will lead to incorrect values for x and y.

Question 3. (iii) Solve the following simultaneous equations using Cramer's rule:
\( x + 2y = -1 \)
\( 2x - 3y = 12 \)
Answer:
The given simultaneous equations are:
\( x + 2y = -1 \) ...(i)
\( 2x - 3y = 12 \) ...(ii)
Comparing these equations with \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), we get:
\( a_1 = 1, b_1 = 2, c_1 = -1 \)
\( a_2 = 2, b_2 = -3, c_2 = 12 \)
Now, let us find the values of determinants \( D \), \( D_x \), and \( D_y \):
\( D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix} = (1 \times -3) - (2 \times 2) = -3 - 4 = -7 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -1 & 2 \\ 12 & -3 \end{vmatrix} = (-1 \times -3) - (2 \times 12) = 3 - 24 = -21 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 1 & -1 \\ 2 & 12 \end{vmatrix} = (1 \times 12) - (-1 \times 2) = 12 - (-2) = 12 + 2 = 14 \)
By Cramer's rule, we get:
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \implies x = \frac{-21}{-7} \) and \( y = \frac{14}{-7} \)
\( \implies x = 3 \) and \( y = -2 \)
\( \therefore (x, y) = (3, -2) \) is the solution of the given simultaneous equations. We can easily verify this solution by substituting these values back into the original equations.
In simple words: We find the determinant values \( D \), \( D_x \), and \( D_y \) using the coefficients of the equations, and then divide \( D_x \) and \( D_y \) by \( D \) to get the values of \( x \) and \( y \).

🎯 Exam Tip: Always double-check the signs when calculating determinants, as a single sign error will lead to incorrect values for both variables.

Question 3. (iv) Solve the following simultaneous equations using Cramer's rule:
\( 6x - 4y = -12 \)
\( 8x - 3y = -2 \)
Answer:
The given simultaneous equations are:
\( 6x - 4y = -12 \)
Dividing both sides by 2, we get:
\( \implies 3x - 2y = -6 \) ...(i)
\( 8x - 3y = -2 \) ...(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:
\( a_1 = 3, b_1 = -2, c_1 = -6 \) and
\( a_2 = 8, b_2 = -3, c_2 = -2 \)
Now, let us find the values of determinants \( D \), \( D_x \), and \( D_y \):
\( D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 3 & -2 \\ 8 & -3 \end{vmatrix} = (3 \times -3) - (-2 \times 8) = -9 - (-16) = -9 + 16 = 7 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -6 & -2 \\ -2 & -3 \end{vmatrix} = (-6 \times -3) - (-2 \times -2) = 18 - 4 = 14 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 3 & -6 \\ 8 & -2 \end{vmatrix} = (3 \times -2) - (-6 \times 8) = -6 - (-48) = -6 + 48 = 42 \)
By Cramer's rule, we get:
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \implies x = \frac{14}{7} \) and \( y = \frac{42}{7} \)
\( \implies x = 2 \) and \( y = 6 \)
\( \therefore (x, y) = (2, 6) \) is the solution of the given simultaneous equations. This systematic method of determinants is highly reliable for solving linear equations.
In simple words: We simplify the first equation by dividing by 2, identify the coefficients, calculate the determinants, and use Cramer's formulas to find the values of \( x \) and \( y \).

🎯 Exam Tip: Simplifying equations by dividing by a common factor makes the determinant calculations much easier and reduces the chance of arithmetic mistakes.

Question 3(iv). Solve the following simultaneous equations using Cramer's rule:
\( 6x - 4y = -12 \)
\( 8x - 3y = -2 \)
Answer:
The given simultaneous equations are:
\( 6x - 4y = -12 \)
Dividing both sides by 2, we get:
\( 3x - 2y = -6 \) ---(i)
\( 8x - 3y = -2 \) ---(ii)
Comparing these equations with \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), we get:
\( a_1 = 3, b_1 = -2, c_1 = -6 \)
\( a_2 = 8, b_2 = -3, c_2 = -2 \)

Now, find the determinants:
\( D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 3 & -2 \\ 8 & -3 \end{vmatrix} = (3 \times -3) - (-2 \times 8) = -9 - (-16) = -9 + 16 = 7 \neq 0 \)
\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -6 & -2 \\ -2 & -3 \end{vmatrix} = (-6 \times -3) - (-2 \times -2) = 18 - 4 = 14 \)
\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 3 & -6 \\ 8 & -2 \end{vmatrix} = (3 \times -2) - (-6 \times 8) = -6 - (-48) = -6 + 48 = 42 \)

By Cramer's rule, we get:
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)

\( \implies x = \frac{14}{7} \) and \( y = \frac{42}{7} \)

\( \implies x = 2 \) and \( y = 6 \)

\( \therefore (x, y) = (2, 6) \) is the solution of the given simultaneous equations.
In simple words: To solve these equations, we first simplify them and find three determinant values (D, Dx, and Dy). Dividing Dx and Dy by D gives us the final values of x and y.

🎯 Exam Tip: Always double-check the signs when calculating determinants, especially when multiplying negative numbers like \( -2 \times -3 \).

Question 3(v). Solve the following simultaneous equations using Cramer's rule:
\( 4m + 6n = 54 \)
\( 3m + 2n = 28 \)
Answer:
The given simultaneous equations are:
\( 4m + 6n = 54 \)
Dividing both sides by 2, we get:
\( 2m + 3n = 27 \) ---(i)
\( 3m + 2n = 28 \) ---(ii)
Comparing these equations with \( a_1m + b_1n = c_1 \) and \( a_2m + b_2n = c_2 \), we get:
\( a_1 = 2, b_1 = 3, c_1 = 27 \)
\( a_2 = 3, b_2 = 2, c_2 = 28 \)

Now, find the determinants:
\( D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = (2 \times 2) - (3 \times 3) = 4 - 9 = -5 \neq 0 \)
\( D_m = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} 27 & 3 \\ 28 & 2 \end{vmatrix} = (27 \times 2) - (3 \times 28) = 54 - 84 = -30 \)
\( D_n = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 2 & 27 \\ 3 & 28 \end{vmatrix} = (2 \times 28) - (27 \times 3) = 56 - 81 = -25 \)

By Cramer's rule, we get:
\( m = \frac{D_m}{D} \) and \( n = \frac{D_n}{D} \)

\( \implies m = \frac{-30}{-5} \) and \( n = \frac{-25}{-5} \)

\( \implies m = 6 \) and \( n = 5 \)

\( \therefore (m, n) = (6, 5) \) is the solution of the given simultaneous equations.
In simple words: We simplify the first equation by dividing by 2, then calculate the determinants D, Dm, and Dn. Dividing Dm and Dn by D gives us the values of m and n.

🎯 Exam Tip: Remember to simplify equations first if all terms have a common divisor, as this makes determinant calculations much easier and less prone to errors.

Question 3. (v) Solve the simultaneous equations: \( 4m + 6n = 54 \), \( 3m + 2n = 28 \) using Cramer's rule.
Answer: The first equation can be simplified by dividing both sides by 2, giving \( 2m + 3n = 27 \). This systematic method of determinants is highly efficient for solving linear equations.
\( \therefore D = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = (2 \times 2) - (3 \times 3) = 4 - 9 = -5 \neq 0 \)
\( D_m = \begin{vmatrix} 27 & 3 \\ 28 & 2 \end{vmatrix} = (27 \times 2) - (3 \times 28) = 54 - 84 = -30 \)
\( D_n = \begin{vmatrix} 2 & 27 \\ 3 & 28 \end{vmatrix} = (2 \times 28) - (27 \times 3) = 56 - 81 = -25 \)
By Cramer's rule, we get:
\( m = \frac{D_m}{D} \) and \( n = \frac{D_n}{D} \)
\( \implies m = \frac{-30}{-5} \) and \( n = \frac{-25}{-5} \)
\( \implies m = 6 \) and \( n = 5 \)
\( \implies (m, n) = (6, 5) \) is the solution of the given simultaneous equations.
In simple words: We find the values of \( m \) and \( n \) by dividing their respective determinants (\( D_m \) and \( D_n \)) by the main determinant (\( D \)). This gives us the final answer \( (6, 5) \).

🎯 Exam Tip: Always double-check the signs when calculating determinants, as a single negative sign error will lead to incorrect values for the variables.

Question 3. (vi) Solve the simultaneous equations: \( 2x + 3y = 2 \), \( x - \frac{y}{2} = \frac{1}{2} \) using Cramer's rule.
Answer: The given simultaneous equations are:
\( 2x + 3y = 2 \) ...(i)
\( x - \frac{y}{2} = \frac{1}{2} \)

\( \implies 2x - y = 1 \) ...(ii) [Multiplying both sides by 2]
Equations (i) and (ii) are in \( ax + by = c \) form. Converting fractional coefficients into integers makes the determinant calculations much simpler.
Comparing the given equations with \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), we get:
\( a_1 = 2, b_1 = 3, c_1 = 2 \) and \( a_2 = 2, b_2 = -1, c_2 = 1 \)

\( \implies D = \begin{vmatrix} 2 & 3 \\ 2 & -1 \end{vmatrix} = (2 \times -1) - (3 \times 2) = -2 - 6 = -8 \)

\( \implies D_x = \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = (2 \times -1) - (3 \times 1) = -2 - 3 = -5 \)

\( \implies D_y = \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} = (2 \times 1) - (2 \times 2) = 2 - 4 = -2 \)
By Cramer's rule, we get:
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)

\( \implies x = \frac{-5}{-8} = \frac{5}{8} \) and \( y = \frac{-2}{-8} = \frac{1}{4} \)

\( \implies (x, y) = \left(\frac{5}{8}, \frac{1}{4}\right) \) is the solution of the given simultaneous equations.
In simple words: First, we multiply the second equation by 2 to clear the fractions. Then, we find the determinants \( D \), \( D_x \), and \( D_y \) to calculate the values of \( x \) and \( y \).

🎯 Exam Tip: When equations contain fractions, always multiply the entire equation by the denominator first to simplify the calculations and avoid errors with fractions.

Question. Solve the simultaneous equations \( 2x + 3y = 2 \) and \( 2x - y = 1 \) using Cramer's rule.
Answer:
Comparing the given equations with \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), we get:
\( a_1 = 2, b_1 = 3, c_1 = 2 \)
\( a_2 = 2, b_2 = -1, c_2 = 1 \)

\( D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ 2 & -1 \end{vmatrix} = (2 \times -1) - (3 \times 2) = -2 - 6 = -8 \neq 0 \)

\( D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = (2 \times -1) - (3 \times 1) = -2 - 3 = -5 \)

\( D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} = (2 \times 1) - (2 \times 2) = 2 - 4 = -2 \)

By Cramer's rule, we get:
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)
\( \therefore x = \frac{-5}{-8} \) and \( y = \frac{-2}{-8} \)
\( \therefore x = \frac{5}{8} \) and \( y = \frac{1}{4} \)
\( \therefore (x, y) = \left(\frac{5}{8}, \frac{1}{4}\right) \) is the solution of the given simultaneous equations. This method provides a very systematic way of solving linear equations using determinants.
In simple words: To solve these equations, we first find the main determinant D, and then the specific determinants Dx and Dy. Dividing Dx and Dy by D gives us the values of x and y.

🎯 Exam Tip: Always double-check the signs when calculating the determinants, especially when multiplying negative numbers, to avoid simple calculation errors.

Question 1. To solve the simultaneous equations by determinant method, fill in the blanks, \( y + 2x - 19 = 0 \); \( 2x - 3y + 3 = 0 \)
Answer:
Write the given equations in the form \( ax + by = c \):
\( 2x + y = 19 \)
\( 2x - 3y = -3 \)

Now, find the values of determinants \( D \), \( D_x \), and \( D_y \):
\( D = \begin{vmatrix} 2 & 1 \\ 2 & -3 \end{vmatrix} = 2 \times (-3) - 1 \times 2 = -6 - 2 = -8 \)

\( D_x = \begin{vmatrix} 19 & 1 \\ -3 & -3 \end{vmatrix} = 19 \times (-3) - 1 \times (-3) = -57 + 3 = -54 \)

\( D_y = \begin{vmatrix} 2 & 19 \\ 2 & -3 \end{vmatrix} = 2 \times (-3) - 19 \times 2 = -6 - 38 = -44 \)

By Cramer's rule:
\( x = \frac{D_x}{D} = \frac{-54}{-8} = \frac{27}{4} \)
\( y = \frac{D_y}{D} = \frac{-44}{-8} = \frac{11}{2} \)

Therefore, the solution is \( (x, y) = \left(\frac{27}{4}, \frac{11}{2}\right) \). Converting the equations to standard form first is a crucial step to ensure the coefficients are aligned correctly.
In simple words: First, rearrange the equations so that x and y are on the left and the numbers are on the right. Then, calculate the determinants to find the values of x and y.

🎯 Exam Tip: Remember to write the equations in the standard form \( ax + by = c \) before finding the determinants, otherwise your signs for the constant terms will be incorrect.

Question 1. Solve the simultaneous equations using Cramer's rule: \( 2x + y = 19 \) and \( 2x - 3y = -3 \).
Answer: Given equations:
\( 2x + y = 19 \)
\( 2x - 3y = -3 \)

We find the determinants \( D \), \( D_x \), and \( D_y \):
\( D = \begin{vmatrix} 2 & 1 \\ 2 & -3 \end{vmatrix} = [2 \times (-3)] - [2 \times (1)] = -6 - 2 = -8 \)
\( D_x = \begin{vmatrix} 19 & 1 \\ -3 & -3 \end{vmatrix} = [19 \times (-3)] - [(-3) \times (1)] = -57 - (-3) = -54 \)
\( D_y = \begin{vmatrix} 2 & 19 \\ 2 & -3 \end{vmatrix} = [(2) \times (-3)] - [(2) \times (19)] = -6 - 38 = -44 \)

By Cramer's rule, we get:
\( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \)

\( \implies x = \frac{-54}{-8} = \frac{27}{4} \)

\( \implies y = \frac{-44}{-8} = \frac{11}{2} \)

\( \therefore (x, y) = \left(\frac{27}{4}, \frac{11}{2}\right) \) is the solution of the given simultaneous equations.
In simple words: Cramer's rule uses determinants, which are simple cross-multiplication grids, to find the values of x and y that satisfy both equations.

🎯 Exam Tip: Always write down the formula for Cramer's rule clearly before substituting the values of the determinants to secure step-wise marks.

Question 2. Complete the following activity. (Textbook pg. no. 15)
Answer: Refer to the activity on page 15 of the textbook to fill in the boxes for solving simultaneous equations using determinants.
In simple words: This textbook activity guides you through the step-by-step process of solving linear equations using Cramer's rule.

🎯 Exam Tip: In activity-based questions, draw the boxes around your final numbers exactly as shown in the question paper to make it easy for the examiner to grade.

Solution Steps for Cramer's Rule

• Given Equations: \( 3x - 2y = 3 \) and \( 2x + y = 16 \)
• Find the values of determinants in the given equations:
  • \( D = \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = 7 \)
  • \( D_x = \begin{vmatrix} 3 & -2 \\ 16 & 1 \end{vmatrix} = 35 \)
  • \( D_y = \begin{vmatrix} 3 & 3 \\ 2 & 16 \end{vmatrix} = 42 \)
• Values according to Cramer's Rule:
  • \( x = \frac{D_x}{D} = \frac{35}{7} = 5 \)
  • \( y = \frac{D_y}{D} = \frac{42}{7} = 6 \)
• Therefore, \( (x, y) = (5, 6) \) is the solution.

Question 3. What is the nature of solution if D = 0? (Textbook pg. no. 16)
Answer: If \( D = 0 \), i.e. \( a_1b_2 - b_1a_2 = 0 \), then the two simultaneous equations do not have a unique solution. This condition indicates that the lines are either parallel or completely overlapping.
Examples:
i. \( 2x - 4y = 8 \) and \( x - 2y = 4 \)
Here, \( a_1b_2 - b_1a_2 = (2)(-2) - (-4)(1) = -4 + 4 = 0 \)
Graphically, we can check that these two lines coincide and hence will have infinite solutions.

ii. \( 2x - y = -1 \) and \( 2x - y = -4 \)
Here, \( a_1b_2 - b_1a_2 = (2)(-1) - (-1)(2) = -2 + 2 = 0 \)
Graphically, we can check that these two lines are parallel and hence they do not have a solution.
In simple words: When the determinant D is zero, it means the two lines are either lying directly on top of each other (giving infinite solutions) or running parallel to each other (giving no solution at all). In either case, you cannot find a single unique point where they cross.

🎯 Exam Tip: Remember that \( D = 0 \) always means there is no unique solution. Clearly state both cases—coinciding lines (infinite solutions) and parallel lines (no solution)—to secure full marks.

Question 4. What can you say about lines if common solution is not possible? (Textbook pg. no. 16)
Answer: If the common solution is not possible, then the lines will either coincide or will be parallel to each other. This occurs because they do not intersect at any single unique point.
In simple words: If two lines never cross each other, they are parallel. If they lie directly on top of each other, they are coinciding lines.

🎯 Exam Tip: Always remember that parallel lines have no points in common, which is why they have no common solution.