Linear Equations in Two Variables: Solutions for Practice Set 1.3 Algebra 10th of Maharashtra State Board Solutions. 


Practice Set 1.3

Q.1) Fill in the blanks with correct number  

 \begin{vmatrix}3 & 4 \\2 & 5\\ \end{vmatrix}= 3 × □ - □ × 4 = □ - 8 = □

Solution:-

 \begin{vmatrix}3 & 4 \\2 & 5\\ \end{vmatrix}= 3 × 5 - 2 × 4 = 15 - 8 = 7

Q.2) Find the values of following determinant

 \begin{vmatrix}-1 & 2 \\7 & 4\\ \end{vmatrix}

Solution:-

 \begin{vmatrix}-1 & 2 \\7 & 4\\ \end{vmatrix}= ([-1]× 4) - (2 × 7) = -4 - 14 = -18

Q.3) Find the values of following determinant

 \begin{vmatrix}5 & 3 \\-7 & 0\\ \end{vmatrix}

Solution:-

 \begin{vmatrix}5 & 3 \\-7 & 0\\ \end{vmatrix}= (5 × 0) − (3 × [− 7]) = 0 - (-21) = 0 + 21 = 21

Q.4) Find the values of following determinant

\begin{vmatrix}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\\ \end{vmatrix}

Solution:-

\begin{vmatrix}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\\ \end{vmatrix}

$$=\frac{7}{3}×\frac{1}{2}−\frac{5}{3}×\frac{3}{2}=\frac{7}{6}-\frac{15}{6}=-\frac{8}{6}=-\frac{4}{3}$$

Q.5) Solve the following simultaneous equations using cramer's rule.

3x − 4y = 10; 4x + 3y = 5

Solution:-

Given Equations, 

3x − 4y = 10

4x + 3y = 5

$$D=\begin{vmatrix}3 & − 4 \\ 4& 3\\ \end{vmatrix}$$

= 3 × 3 −  ( − 4) × 4 = 9 + 16 = 25

$$Dx=\begin{vmatrix}10 & − 4 \\ 5& 3\\ \end{vmatrix}$$ = 10 × 3 − ( − 4) × 5 = 30 + 20 = 50 $$Dy=\begin{vmatrix}3 & 10 \\ 4& 5\\ \end{vmatrix}$$ = 3 × 5 − 10 × 4 = 15 − 40 = −25 
 By using Cramer's rule, $$x = \frac{Dx}{D}=\frac{50}{25}=2$$
$$y = \frac{Dy}{D}=\frac{−25}{25}=−1$$
(x,y) = (2,−1) is the solution.

Q.6) Solve the following simultaneous equations using cramer's rule.

4x + 3y − 4 = 0; 6x = 8 − 5y 

Solution:-

4x + 3y − 4 = 0

6x = 8 − 5y 

Write the given equation in the form of  ax + by = c

4x + 3y =  4

6x + 5y= 8 

$$D=\begin{vmatrix}4 & 3 \\ 6& 5\\ \end{vmatrix}$$

= 4 × 5 − 3 × 6 = 20 − 18 = 2

$$Dx=\begin{vmatrix}4 & 3 \\ 8& 5\\ \end{vmatrix}$$ = 4 × 5 − 3 × 8 = 20 − 24 = −4 $$Dy=\begin{vmatrix}4 & 4 \\ 6& 8\\ \end{vmatrix}$$ = 4 × 8 − 4 × 6 = 32 − 24 = 8

 By using Cramer's rule, $$x = \frac{Dx}{D}=\frac{−4}{2}=−2$$
$$y = \frac{Dy}{D}=\frac{8}{2}=4$$
(x,y) = (−2,2) is the solution.

Q.7) Solve the following simultaneous equations using cramer's rule.

x + 2y = − 1; 2x − 3y = 12

Solution:-

x + 2y = − 1

2x − 3y = 12

$$D=\begin{vmatrix}1 & 2 \\ 2& − 3 \\ \end{vmatrix}$$

= 1 ×  (− 3) − 2 × 2 =  (− 3) − 4 =  − 7

$$Dx=\begin{vmatrix} − 1 & 2 \\ 12& − 3\\ \end{vmatrix}$$ =  (− 1) × ( − 3) − 2 × 12 = 3 − 24 = −21$$Dy=\begin{vmatrix}1 & − 1 \\ 2& 12\\ \end{vmatrix}$$  =1×12 − ( −1) × 2 = 12 − ( − 2) = 12 + 2 =14
 By using Cramer's rule, $$x = \frac{Dx}{D}=\frac{−21}{ − 7}=3$$
$$y = \frac{Dy}{D}=\frac{14}{ − 7}= −2$$
(x,y) = (3, − 2) is the solution.

Q.8) Solve the following simultaneous equations using cramer's rule.

6x − 4y = −12; 8x − 3y = −2

Solution:-

6x − 4y = −12

8x − 3y = −2

$$D=\begin{vmatrix}6 & −4 \\ 8& − 3 \\ \end{vmatrix}$$

= 6 ×  (− 3) − (−4) × 8 =  (− 18) − (−32) =  (−18) + 32 = 14

$$Dx=\begin{vmatrix} −12 & −4 \\ −2& − 3\\ \end{vmatrix}$$ =  (− 12) × ( − 3) − (−4) × (−2) = 36 − 8 = 28 $$Dy=\begin{vmatrix}6 & −12 \\ 8& −2\\ \end{vmatrix}$$ =6×(−2) − ( −12) × 8 = −12 − (−96) = −12 + 96 = 84
 By using Cramer's rule, $$x = \frac{Dx}{D}=\frac{28}{14}=2$$
$$y = \frac{Dy}{D}=\frac{84}{14}= 6$$
(x,y) = (2, 6) is the solution.

Q.9) Solve the following simultaneous equations using cramer's rule.

4m + 6n = 54; 3m + 2n = 28

Solution:-

4m + 6n = 54

3m + 2n = 28

$$D=\begin{vmatrix}4 & 6 \\ 3& 2 \\ \end{vmatrix}$$

= 4 × 2 − 6 × 3 =  8 − 18 = −10

$$Dm=\begin{vmatrix} 54 & 6 \\ 28& 2\\ \end{vmatrix}$$ =  54 × 2 − 6 × 28 = 108 − 168 = −60 $$Dn=\begin{vmatrix}4 & 54 \\ 3& 28\\ \end{vmatrix}$$ =4×28 − 54 × 3 = 112 − 162 = −50
 By using Cramer's rule, $$m = \frac{Dm}{D}=\frac{−60}{−10}=6$$
$$n = \frac{Dn}{D}=\frac{−50}{−10}= 5$$
(x,y) = (6, 5) is the solution.

Q.10) Solve the following simultaneous equations using cramer's rule.

$$2x + 3y = 2; x −\frac{y}{2}=\frac{1}{2}$$

Solution:-

$$2x + 3y = 2; x −\frac{y}{2}=\frac{1}{2}$$

$$D=\begin{vmatrix}2 & 3 \\ 1& \frac{-1}{2} \\ \end{vmatrix}$$ =  2 ×(-1/2)  − 3 × 1 =  −1 − 3 = −4

$$Dx=\begin{vmatrix} 2 &  3\\ \frac{1}{2}& \frac{− 1}{2}\\ \end{vmatrix}$$  =  2× (-1/2) − 3 ×1/2 = (-1/1)− (3/2) = (-2-3/2)= -5/2$$Dy=\begin{vmatrix}2 & 2 \\ 1& \frac{1}{2}\\ \end{vmatrix}$$ $$=2×\frac{1}{2} − 2 × 1 = 1 − 2 = −1$$ 

 By using Cramer's rule, $$x = \frac{Dx}{D}=\frac{-5/2}{-4}=\frac{-5}{2}×\frac{1}{-4}=\frac{-5}{-8}=\frac{5}{8}$$
$$y = \frac{Dy}{D}=\frac{-1}{-4}= \frac{1}{4}$$
$$(x,y) = \frac{5}{8},\frac{1}{4}$$