Home » » 10th Standard Maths 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

10th Standard Maths 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Practice Set 1.3 Solutions

Question 1

Fill in the blanks with correct number.

$|\begin{matrix}3&2\\ 4&5\end{matrix}| = 3 \times \boxed{5} - \boxed{2} \times 4$

$= 15 - 8$

$= \boxed{7}$

Question 2

Find the values of following determinants.

i. $|\begin{matrix}-1&7\\ 2&4\end{matrix}|$

$|\begin{matrix}-1&7\\ 2&4\end{matrix}| = (-1 \times 4) - (7 \times 2)$

$= -4 - 14$

$= -18$

ii. $|\begin{matrix}5&3\\ -7&0\end{matrix}|$

$|\begin{matrix}5&3\\ -7&0\end{matrix}| = (5 \times 0) - (3 \times -7)$

$= 0 - (-21)$

$= 21$

iii. $|\begin{matrix}\frac{7}{3}&\frac{5}{3}\\ \frac{3}{2}&\frac{1}{2}\end{matrix}|$

$|\begin{matrix}\frac{7}{3}&\frac{5}{3}\\ \frac{3}{2}&\frac{1}{2}\end{matrix}| = (\frac{7}{3} \times \frac{1}{2}) - (\frac{5}{3} \times \frac{3}{2})$

$= \frac{7}{6} - \frac{15}{6}$

$= \frac{7-15}{6}$

$= \frac{-8}{6}$

$= \frac{-4}{3}$

Question 3

Solve the following simultaneous equations using Cramer's rule.

i. $3x-4y=10$; $4x+3y=5$

  • The given equations are $3x-4y=10$ and $4x+3y=5$.
  • Comparing with $a_{1}x+b_{1}y=c_{1}$ and $a_{2}x+b_{2}y=c_{2}$, we get $a_{1}=3, b_{1}=-4, c_{1}=10$ and $a_{2}=4, b_{2}=3, c_{2}=5$.
  • $D = |\begin{matrix}3&-4\\ 4&3\end{matrix}| = (3 \times 3) - (-4 \times 4) = 9 + 16 = 25 \neq 0$.
  • $D_{x} = |\begin{matrix}10&-4\\ 5&3\end{matrix}| = (10 \times 3) - (-4 \times 5) = 30 + 20 = 50$.
  • $D_{y} = |\begin{matrix}3&10\\ 4&5\end{matrix}| = (3 \times 5) - (10 \times 4) = 15 - 40 = -25$.
  • By Cramer's rule: $x = \frac{D_{x}}{D} = \frac{50}{25} = 2$ and $y = \frac{D_{y}}{D} = \frac{-25}{25} = -1$.
  • $(x,y) = (2,-1)$ is the solution.

ii. $4x+3y-4=0$; $6x=8-5y$

  • The given equations can be written as $4x+3y=4$ and $6x+5y=8$.
  • Comparing, we get $a_{1}=4, b_{1}=3, c_{1}=4$ and $a_{2}=6, b_{2}=5, c_{2}=8$.
  • $D = |\begin{matrix}4&3\\ 6&5\end{matrix}| = (4 \times 5) - (3 \times 6) = 20 - 18 = 2 \neq 0$.
  • $D_{x} = |\begin{matrix}4&3\\ 8&5\end{matrix}| = (4 \times 5) - (3 \times 8) = 20 - 24 = -4$.
  • $D_{y} = |\begin{matrix}4&4\\ 6&8\end{matrix}| = (4 \times 8) - (4 \times 6) = 32 - 24 = 8$.
  • By Cramer's rule: $x = \frac{-4}{2} = -2$ and $y = \frac{8}{2} = 4$.
  • $(x,y) = (-2,4)$ is the solution.

iii. $x+2y=-1$; $2x-3y=12$

  • The given equations are $x+2y=-1$ and $2x-3y=12$.
  • Comparing, we get $a_{1}=1, b_{1}=2, c_{1}=-1$ and $a_{2}=2, b_{2}=-3, c_{2}=12$.
  • $D = |\begin{matrix}1&2\\ 2&-3\end{matrix}| = (1 \times -3) - (2 \times 2) = -3 - 4 = -7 \neq 0$.
  • $D_{x} = |\begin{matrix}-1&2\\ 12&-3\end{matrix}| = (-1 \times -3) - (2 \times 12) = 3 - 24 = -21$.
  • $D_{y} = |\begin{matrix}1&-1\\ 2&12\end{matrix}| = (1 \times 12) - (-1 \times 2) = 12 + 2 = 14$.
  • By Cramer's rule: $x = \frac{-21}{-7} = 3$ and $y = \frac{14}{-7} = -2$.
  • $(x,y) = (3,-2)$ is the solution.

iv. $6x-4y=-12$; $8x-3y=-2$

  • Dividing the first equation by 2, we get $3x-2y=-6$. The second equation is $8x-3y=-2$.
  • Comparing, we get $a_{1}=3, b_{1}=-2, c_{1}=-6$ and $a_{2}=8, b_{2}=-3, c_{2}=-2$.
  • $D = |\begin{matrix}3&-2\\ 8&-3\end{matrix}| = (3 \times -3) - (-2 \times 8) = -9 + 16 = 7 \neq 0$.
  • $D_{x} = |\begin{matrix}-6&-2\\ -2&-3\end{matrix}| = (-6 \times -3) - (-2 \times -2) = 18 - 4 = 14$.
  • $D_{y} = |\begin{matrix}3&-6\\ 8&-2\end{matrix}| = (3 \times -2) - (-6 \times 8) = -6 + 48 = 42$.
  • By Cramer's rule: $x = \frac{14}{7} = 2$ and $y = \frac{42}{7} = 6$.
  • $(x,y) = (2,6)$ is the solution.

v. $4m+6n=54$; $3m+2n=28$

  • Dividing the first equation by 2, we get $2m+3n=27$. The second equation is $3m+2n=28$.
  • Comparing, we get $a_{1}=2, b_{1}=3, c_{1}=27$ and $a_{2}=3, b_{2}=2, c_{2}=28$.
  • $D = |\begin{matrix}2&3\\ 3&2\end{matrix}| = (2 \times 2) - (3 \times 3) = 4 - 9 = -5 \neq 0$.
  • $D_{m} = |\begin{matrix}27&3\\ 28&2\end{matrix}| = (27 \times 2) - (3 \times 28) = 54 - 84 = -30$.
  • $D_{n} = |\begin{matrix}2&27\\ 3&28\end{matrix}| = (2 \times 28) - (27 \times 3) = 56 - 81 = -25$.
  • By Cramer's rule: $m = \frac{-30}{-5} = 6$ and $n = \frac{-25}{-5} = 5$.
  • $(m,n) = (6,5)$ is the solution.

vi. $2x+3y=2$; $x-\frac{y}{2}=\frac{1}{2}$

  • Multiplying the second equation by 2 gives $2x-y=1$. The first equation is $2x+3y=2$.
  • Comparing, we get $a_{1}=2, b_{1}=3, c_{1}=2$ and $a_{2}=2, b_{2}=-1, c_{2}=1$.
  • $D = |\begin{matrix}2&3\\ 2&-1\end{matrix}| = (2 \times -1) - (3 \times 2) = -2 - 6 = -8 \neq 0$.
  • $D_{x} = |\begin{matrix}2&3\\ 1&-1\end{matrix}| = (2 \times -1) - (3 \times 1) = -2 - 3 = -5$.
  • $D_{y} = |\begin{matrix}2&2\\ 2&1\end{matrix}| = (2 \times 1) - (2 \times 2) = 2 - 4 = -2$.
  • By Cramer's rule: $x = \frac{-5}{-8} = \frac{5}{8}$ and $y = \frac{-2}{-8} = \frac{1}{4}$.
  • $(x,y) = (\frac{5}{8},\frac{1}{4})$ is the solution.

Question 1 (Textbook pg. no. 14)

To solve the simultaneous equations by determinant method, fill in the blanks: $y+2x-19=0$; $2x-3y+3=0$

Write the given equations in the form $ax+by=c$:

$2x+y=19$

$2x-3y=-3$

$D = |\begin{matrix}\boxed{2}&\boxed{1}\\ 2&-3\end{matrix}| = [\boxed{2} \times (-3)] - [2 \times (\boxed{1})] = \boxed{-6} - \boxed{2} = \boxed{-8}$

$D_{x} = |\begin{matrix}19&\boxed{1}\\ \boxed{-3}&-3\end{matrix}| = [19 \times (\boxed{-3})] - [(\boxed{-3}) \times (\boxed{1})] = \boxed{-57} - \boxed{-3} = \boxed{-54}$

$D_{y} = |\begin{matrix}\boxed{2}&19\\ 2&\boxed{-3}\end{matrix}| = [(\boxed{2}) \times (\boxed{-3})] - [(\boxed{2}) \times (\boxed{19})] = \boxed{-6} - \boxed{38} = \boxed{-44}$

By Cramer's rule, we get:

$x = \frac{D_{x}}{D} = \frac{\boxed{-54}}{\boxed{-8}} = \frac{\boxed{27}}{\boxed{4}}$ and $y = \frac{D_{y}}{D} = \frac{\boxed{-44}}{\boxed{-8}} = \frac{\boxed{11}}{\boxed{2}}$

$(x,y) = (\frac{\boxed{27}}{\boxed{4}}, \frac{\boxed{11}}{\boxed{2}})$ is the solution of the given simultaneous equations.

Question 2 (Textbook pg. no. 15)

Complete the following activity.

Equations: $3x-2y=3$ and $2x+y=16$

$D = |\begin{matrix}3&-2\\ 2&1\end{matrix}| = \boxed{7}$

$D_{x} = |\begin{matrix}3&-2\\ 16&1\end{matrix}| = \boxed{35}$

$D_{y} = |\begin{matrix}3&3\\ 2&16\end{matrix}| = \boxed{42}$

$x = \frac{D_{x}}{D} = \frac{\boxed{35}}{\boxed{7}} = \boxed{5}$

$y = \frac{D_{y}}{D} = \frac{\boxed{42}}{\boxed{7}} = \boxed{6}$

$(x,y) = (\boxed{5}, \boxed{6})$ is the solution.

Question 3 (Textbook pg. no. 16)

What is the nature of solution if $D=0$?

If $D=0$ i.e. $a_{1}b_{2}-b_{1}a_{2}=0$, then the two simultaneous equations do not have a unique solution.

Examples:

  • $2x-4y=8$ and $x-2y=4$: Here, $a_{1}b_{2}-b_{1}a_{2}=(2)(-2)-(-4)(1) = -4+4=0$. Graphically, we can check that these two lines coincide and hence will have infinite solutions.
  • $2x-y=-1$ and $2x-y=-4$: Here, $a_{1}b_{2}-b_{1}a_{2}=(2)(-1)-(-1)(2) = -2+2=0$. Graphically, we can check that these two lines are parallel and hence they do not have a solution.

Question 4 (Textbook pg. no. 16)

What can you say about lines if common solution is not possible?

If the common solution is not possible, then the lines will either coincide or will be parallel to each other.

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