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Problem Set 1 Algebra 10th Standard Maths Part 1 Chapter 1 Linear Equations in Two Variables Solutions Maharashtra Board

Problem Set 1 Solutions

Question 1: Choose correct alternative for each of the following questions.

  1. To draw graph of $4x+5y=19$ find $y$ when $x=1$
    • (a) 4
    • (b) 3
    • (c) 2
    • (d) -3
    Answer: (b)
  2. For simultaneous equations in variables $x$ and $y$, $D_x=49$, $D_y=-63$, $D=7$ then what is $x$?
    • (a) 7
    • (b) -7
    • (c) $\frac{1}{7}$
    • (d) $\frac{-1}{7}$
    Answer: (a)
  3. Find the value of $|\begin{matrix}5&3\\ -7&-4\end{matrix}|$.
    • (a) -1
    • (b) -41
    • (c) 41
    • (d) 1
    Answer: (d)
  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. $ax+by=c$ 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)

Question 2. Complete the following table to draw the graph of $2x-6y=3$.

$x$ -5 $\frac{3}{2}$
$y$ $\frac{-13}{6}$ 0
$(x,y)$ $(-5, \frac{-13}{6})$ $(\frac{3}{2}, 0)$

Question 3. Solve the following simultaneous equations graphically.

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

The given simultaneous equations are $2x+3y=12 \Rightarrow 3y=12-2x \Rightarrow y=\frac{12-2x}{3}$ and $x-y=1 \Rightarrow y=x-1$.

Table for $2x+3y=12$:

$x$ 0 6 3 -3
$y$ 4 0 2 6
$(x,y)$ (0,4) (6,0) (3,2) (-3,6)

Table for $x-y=1$:

$x$ 0 2 4 5
$y$ -1 1 3 4
$(x,y)$ (0,-1) (2,1) (4,3) (5,4)
Graph 1.1

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 $x-y=1$.

ii. $x-3y=1$; $3x-2y+4=0$

The given simultaneous equations are $x-3y=1 \Rightarrow 3y=x-1 \Rightarrow y=\frac{x-1}{3}$ and $3x-2y+4=0 \Rightarrow 2y=3x+4 \Rightarrow y=\frac{3x+4}{2}$.

Table for $x-3y=1$:

$x$ 4 -2 -5 1
$y$ 1 -1 -2 0
$(x,y)$ (4,1) (-2,-1) (-5,-2) (1,0)

Table for $3x-2y+4=0$:

$x$ 0 -2 2 -4
$y$ 2 -1 5 -4
$(x,y)$ (0,2) (-2,-1) (2,5) (-4,-4)
Graph 1.2

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 $3x-2y+4=0$.

iii. $5x-6y+30=0$; $5x+4y-20=0$

The given simultaneous equations are $5x-6y+30=0 \Rightarrow 6y=5x+30 \Rightarrow y=\frac{5x+30}{6}$ and $5x+4y-20=0 \Rightarrow 4y=20-5x \Rightarrow y=\frac{20-5x}{4}$.

Table for $5x-6y+30=0$:

$x$ 0 -6 6 3
$y$ 5 0 10 7.5
$(x,y)$ (0,5) (-6,0) (6,10) (3,7.5)

Table for $5x+4y-20=0$:

$x$ 0 4 -4 2
$y$ 5 0 10 2.5
$(x,y)$ (0,5) (4,0) (-4,10) (2,2.5)
Graph 1.3

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 $5x+4y-20=0$.

iv. $3x-y-2=0$; $2x+y=8$

The given simultaneous equations are $3x-y-2=0 \Rightarrow y=3x-2$ and $2x+y=8 \Rightarrow y=8-2x$.

Table for $3x-y-2=0$:

$x$ 0 1 3 -1
$y$ -2 1 7 -5
$(x,y)$ (0,-2) (1, 1) (3,7) (-1,-5)

Table for $2x+y=8$:

$x$ 0 4 1 3
$y$ 8 0 6 2
$(x,y)$ (0,8) (4,0) (1,6) (3,2)
Graph 1.4

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 $2x+y=8$.

v. $3x+y=10$; $x-y=2$

The given simultaneous equations are $3x+y=10 \Rightarrow y=10-3x$ and $x-y=2 \Rightarrow y=x-2$.

Table for $3x+y=10$:

$x$ 2 3 4 5
$y$ 4 1 -2 -5
$(x,y)$ (2,4) (3,1) (4,-2) (5,-5)

Table for $x-y=2$:

$x$ 0 2 4 5
$y$ -2 0 2 3
$(x,y)$ (0,-2) (2,0) (4,2) (5,3)
Graph 1.5

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 $x-y=2$.

Question 4. Find the values of each of the following determinants.

i. $|\begin{matrix}4&3\\ 2&7\end{matrix}|$

$=(4\times 7)-(3\times 2) = 28-6 = 22$

ii. $|\begin{matrix}5&-2\\ -3&1\end{matrix}|$

$=(5\times 1)-(-2\times -3) = 5-6 = -1$

iii. $|\begin{matrix}3&-1\\ 1&4\end{matrix}|$

$=(3\times 4)-(-1\times 1) = 12-(-1) = 12+1 = 13$

Question 5. Solve the following equations by Cramer's method.

i. $6x-3y=-10$; $3x+5y-8=0$

  • Equations: $6x-3y=-10$ and $3x+5y=8$.
  • Comparing: $a_1=6, b_1=-3, c_1=-10$ and $a_2=3, b_2=5, c_2=8$.
  • $D = |\begin{matrix}6&-3\\ 3&5\end{matrix}| = (6\times 5)-(-3\times 3) = 30+9 = 39 \neq 0$.
  • $D_x = |\begin{matrix}-10&-3\\ 8&5\end{matrix}| = (-10\times 5)-(-3\times 8) = -50+24 = -26$.
  • $D_y = |\begin{matrix}6&-10\\ 3&8\end{matrix}| = (6\times 8)-(-10\times 3) = 48+30 = 78$.
  • By Cramer's rule: $x = \frac{D_x}{D} = \frac{-26}{39} = \frac{-2}{3}$ and $y = \frac{D_y}{D} = \frac{78}{39} = 2$.
  • $(x,y) = (\frac{-2}{3}, 2)$ is the solution.

ii. $4m-2n=-4$; $4m+3n=16$

  • Equations: $4m-2n=-4$ and $4m+3n=16$.
  • Comparing: $a_1=4, b_1=-2, c_1=-4$ and $a_2=4, b_2=3, c_2=16$.
  • $D = |\begin{matrix}4&-2\\ 4&3\end{matrix}| = (4\times 3)-(-2\times 4) = 12+8 = 20 \neq 0$.
  • $D_m = |\begin{matrix}-4&-2\\ 16&3\end{matrix}| = (-4\times 3)-(-2\times 16) = -12+32 = 20$.
  • $D_n = |\begin{matrix}4&-4\\ 4&16\end{matrix}| = (4\times 16)-(-4\times 4) = 64+16 = 80$.
  • By Cramer's rule: $m = \frac{D_m}{D} = \frac{20}{20} = 1$ and $n = \frac{D_n}{D} = \frac{80}{20} = 4$.
  • $(m,n) = (1, 4)$ is the solution.

iii. $3x-2y=\frac{5}{2}$; $\frac{1}{3}x+3y=-\frac{4}{3}$

  • Equations: $3x-2y=\frac{5}{2}$ and $x+9y=-4$ (Multiplying second equation by 3).
  • Comparing: $a_1=3, b_1=-2, c_1=\frac{5}{2}$ and $a_2=1, b_2=9, c_2=-4$.
  • $D = |\begin{matrix}3&-2\\ 1&9\end{matrix}| = (3\times 9)-(-2\times 1) = 27+2 = 29 \neq 0$.
  • $D_x = |\begin{matrix}\frac{5}{2}&-2\\ -4&9\end{matrix}| = (\frac{5}{2}\times 9)-(-2\times -4) = \frac{45}{2}-8 = \frac{29}{2}$.
  • $D_y = |\begin{matrix}3&\frac{5}{2}\\ 1&-4\end{matrix}| = (3\times -4)-(\frac{5}{2}\times 1) = -12-\frac{5}{2} = -\frac{29}{2}$.
  • By Cramer's rule: $x = \frac{\frac{29}{2}}{29} = \frac{1}{2}$ and $y = \frac{-\frac{29}{2}}{29} = -\frac{1}{2}$.
  • $(x,y) = (\frac{1}{2}, -\frac{1}{2})$ is the solution.

iv. $7x+3y=15$; $12y-5x=39$

  • Equations: $7x+3y=15$ and $-5x+12y=39$.
  • Comparing: $a_1=7, b_1=3, c_1=15$ and $a_2=-5, b_2=12, c_2=39$.
  • $D = |\begin{matrix}7&3\\ -5&12\end{matrix}| = (7\times 12)-(3\times -5) = 84+15 = 99 \neq 0$.
  • $D_x = |\begin{matrix}15&3\\ 39&12\end{matrix}| = (15\times 12)-(3\times 39) = 180-117 = 63$.
  • $D_y = |\begin{matrix}7&15\\ -5&39\end{matrix}| = (7\times 39)-(15\times -5) = 273+75 = 348$.
  • By Cramer's rule: $x = \frac{63}{99} = \frac{7}{11}$ and $y = \frac{348}{99} = \frac{116}{33}$.
  • $(x,y) = (\frac{7}{11}, \frac{116}{33})$ is the solution.

v. $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$

  • From $\frac{x+y-8}{2}=\frac{x+2y-14}{3}$: $3(x+y-8)=2(x+2y-14) \Rightarrow 3x+3y-24=2x+4y-28 \Rightarrow x-y=-4$ ... (i).
  • From $\frac{x+y-8}{2}=\frac{3x-y}{4}$: $4(x+y-8)=2(3x-y) \Rightarrow 4x+4y-32=6x-2y \Rightarrow -2x+6y=32 \Rightarrow x-3y=-16$ ... (ii) [Dividing by -2].
  • Comparing: $a_1=1, b_1=-1, c_1=-4$ and $a_2=1, b_2=-3, c_2=-16$.
  • $D = |\begin{matrix}1&-1\\ 1&-3\end{matrix}| = (1\times -3)-(-1\times 1) = -3+1 = -2 \neq 0$.
  • $D_x = |\begin{matrix}-4&-1\\ -16&-3\end{matrix}| = (-4\times -3)-(-1\times -16) = 12-16 = -4$.
  • $D_y = |\begin{matrix}1&-4\\ 1&-16\end{matrix}| = (1\times -16)-(-4\times 1) = -16+4 = -12$.
  • By Cramer's rule: $x = \frac{-4}{-2} = 2$ and $y = \frac{-12}{-2} = 6$.
  • $(x,y) = (2, 6)$ is the solution.

Question 6. Solve the following simultaneous equations:

i. $\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}$; $\frac{3}{x}+\frac{2}{y}=0$

  • Let $\frac{1}{x}=p$ and $\frac{1}{y}=q$.
  • Equations become: $2p+\frac{2}{3}q=\frac{1}{6} \Rightarrow 12p+4q=1$ and $3p+2q=0 \Rightarrow 6p+4q=0$ (Multiplying by 2).
  • Alternatively: $6p+2q=\frac{1}{2}$ (Multiplying first eq by 3) ... (iii) and $3p+2q=0$ ... (iv).
  • Subtracting (iv) from (iii): $3p = \frac{1}{2} \Rightarrow p=\frac{1}{6}$.
  • Substituting $p=\frac{1}{6}$ in (iv): $3(\frac{1}{6})+2q=0 \Rightarrow \frac{1}{2}+2q=0 \Rightarrow 2q=-\frac{1}{2} \Rightarrow q=-\frac{1}{4}$.
  • Resubstituting: $\frac{1}{x}=\frac{1}{6} \Rightarrow x=6$ and $\frac{1}{y}=-\frac{1}{4} \Rightarrow y=-4$.
  • $(x,y) = (6, -4)$ is the solution.

ii. $\frac{7}{2x+1}+\frac{13}{y+2}=27$; $\frac{13}{2x+1}+\frac{7}{y+2}=33$

  • Let $\frac{1}{2x+1}=p$ and $\frac{1}{y+2}=q$.
  • Equations become: $7p+13q=27$ ... (iii) and $13p+7q=33$ ... (iv).
  • Adding (iii) and (iv): $20p+20q=60 \Rightarrow p+q=3$ ... (v).
  • Subtracting (iv) from (iii): $-6p+6q=-6 \Rightarrow p-q=1$ ... (vi).
  • Adding (v) and (vi): $2p=4 \Rightarrow p=2$.
  • Substituting $p=2$ in (v): $2+q=3 \Rightarrow q=1$.
  • Resubstituting: $\frac{1}{2x+1}=2 \Rightarrow 2(2x+1)=1 \Rightarrow 4x+2=1 \Rightarrow 4x=-1 \Rightarrow x=-\frac{1}{4}$.
  • $\frac{1}{y+2}=1 \Rightarrow y+2=1 \Rightarrow y=-1$.
  • $(x,y) = (-\frac{1}{4}, -1)$ is the solution.

iii. $\frac{148}{x}+\frac{231}{y}=\frac{527}{xy}$; $\frac{231}{x}+\frac{148}{y}=\frac{610}{xy}$

  • Multiplying equations by $xy$:
  • $148y+231x=527 \Rightarrow 231x+148y=527$ ... (i).
  • $231y+148x=610 \Rightarrow 148x+231y=610$ ... (ii).
  • Adding (i) and (ii): $379x+379y=1137 \Rightarrow x+y=3$ ... (iii).
  • Subtracting (ii) from (i): $83x-83y=-83 \Rightarrow x-y=-1$ ... (iv).
  • Adding (iii) and (iv): $2x=2 \Rightarrow x=1$.
  • Substituting $x=1$ in (iii): $1+y=3 \Rightarrow y=2$.
  • $(x,y) = (1, 2)$ is the solution.

iv. $\frac{7x-2y}{xy}=5$; $\frac{8x+7y}{xy}=15$

  • Separating terms: $\frac{7x}{xy}-\frac{2y}{xy}=5 \Rightarrow \frac{7}{y}-\frac{2}{x}=5$ ... (i).
  • $\frac{8x}{xy}+\frac{7y}{xy}=15 \Rightarrow \frac{8}{y}+\frac{7}{x}=15$ ... (ii).
  • Let $\frac{1}{x}=p$ and $\frac{1}{y}=q$. Equations become: $-2p+7q=5$ ... (iii) and $7p+8q=15$ ... (iv).
  • Multiplying (iii) by 7: $-14p+49q=35$ ... (v).
  • Multiplying (iv) by 2: $14p+16q=30$ ... (vi).
  • Adding (v) and (vi): $65q=65 \Rightarrow q=1$.
  • Substituting $q=1$ in (iv): $7p+8(1)=15 \Rightarrow 7p=7 \Rightarrow p=1$.
  • Resubstituting: $\frac{1}{x}=1 \Rightarrow x=1$ and $\frac{1}{y}=1 \Rightarrow y=1$.
  • $(x,y) = (1, 1)$ is the solution.

v. $\frac{1}{2(3x+4y)}+\frac{1}{5(2x-3y)}=\frac{1}{4}$; $\frac{5}{(3x+4y)}-\frac{2}{(2x-3y)}=-\frac{3}{2}$

  • Let $\frac{1}{3x+4y}=p$ and $\frac{1}{2x-3y}=q$.
  • Equations become: $\frac{1}{2}p+\frac{1}{5}q=\frac{1}{4} \Rightarrow 10p+4q=5$ ... (iii) [Multiplying by 20].
  • $5p-2q=-\frac{3}{2} \Rightarrow 10p-4q=-3$ ... (iv) [Multiplying by 2].
  • Adding (iii) and (iv): $20p=2 \Rightarrow p=\frac{1}{10}$.
  • Substituting $p=\frac{1}{10}$ in (iii): $10(\frac{1}{10})+4q=5 \Rightarrow 1+4q=5 \Rightarrow 4q=4 \Rightarrow q=1$.
  • Resubstituting: $\frac{1}{3x+4y}=\frac{1}{10} \Rightarrow 3x+4y=10$ ... (v).
  • $\frac{1}{2x-3y}=1 \Rightarrow 2x-3y=1$ ... (vi).
  • Multiplying (v) by 3: $9x+12y=30$ ... (vii).
  • Multiplying (vi) by 4: $8x-12y=4$ ... (viii).
  • Adding (vii) and (viii): $17x=34 \Rightarrow x=2$.
  • Substituting $x=2$ in (v): $3(2)+4y=10 \Rightarrow 6+4y=10 \Rightarrow 4y=4 \Rightarrow y=1$.
  • $(x,y) = (2, 1)$ is the solution.

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.

  • Let the digit in unit's place be $x$ and that in the ten's place be $y$.
  • The number $= 10y+x$. The number obtained by interchanging the digits is $10x+y$.
  • According to the first condition: $(10y+x) + (10x+y) = 143 \Rightarrow 11x+11y=143 \Rightarrow x+y=13$ ... (i).
  • According to the second condition: $x = y+3 \Rightarrow x-y=3$ ... (ii).
  • Adding equations (i) and (ii): $2x=16 \Rightarrow x=8$.
  • Putting $x=8$ in equation (i): $8+y=13 \Rightarrow y=5$.
  • The original number is $10y+x = 10(5)+8 = 50+8 = 58$.

ii. Kantabai bought $1\frac{1}{2}$ kg tea and 5 kg sugar from a shop. She paid 50 as return fare for rickshaw. Total expense was 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 880 for that. Find the rate of sugar and tea per kg.

  • Let the rate of tea be $x$ per kg and that of sugar be $y$ per kg.
  • First condition: $1\frac{1}{2}x + 5y + 50 = 700 \Rightarrow \frac{3}{2}x + 5y = 650 \Rightarrow 3x+10y=1300$ ... (i) [Multiplying by 2].
  • Second condition: $2x+7y=880$ ... (ii).
  • Multiplying (i) by 2: $6x+20y=2600$ ... (iii).
  • Multiplying (ii) by 3: $6x+21y=2640$ ... (iv).
  • Subtracting (iii) from (iv): $y=40$.
  • Substituting $y=40$ in (i): $3x+10(40)=1300 \Rightarrow 3x+400=1300 \Rightarrow 3x=900 \Rightarrow x=300$.
  • The rate of tea is 300 per kg and that of sugar is 40 per kg.

iii. To find number of notes that Anushka had, complete the following activity.

  • Anushka had $x$ notes of 100 and $y$ notes of 50.
  • First condition: $100x+50y=2500 \Rightarrow 2x+y=50$ ... (i) [Dividing by 50].
  • Second condition: $100y+50x=2000 \Rightarrow 2y+x=40 \Rightarrow x+2y=40 \Rightarrow 2x+4y=80$ ... (ii) [Dividing by 50, then multiplying by 2].
  • Subtracting (i) from (ii): $3y=30 \Rightarrow y=10$.
  • Substituting $y=10$ in (i): $2x+10=50 \Rightarrow 2x=40 \Rightarrow x=20$.
  • Anushka had 20 notes of 100 and 10 notes of 50.

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.

  • Let present ages of Manish and Savita be $x$ and $y$ respectively.
  • First condition: $x+y=31$ ... (i).
  • 3 years ago: Manish's age $=(x-3)$, Savita's age $=(y-3)$.
  • Second condition: $(x-3)=4(y-3) \Rightarrow x-3=4y-12 \Rightarrow x-4y=-9$ ... (ii).
  • Subtracting (ii) from (i): $5y=40 \Rightarrow y=8$.
  • Substituting $y=8$ in (i): $x+8=31 \Rightarrow x=23$.
  • Present ages of Manish and Savita are 23 years and 8 years respectively.

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 720. Find daily wages of skilled and unskilled workers.

  • Let the daily wages of skilled workers be $x$ and unskilled workers be $y$.
  • First condition: $\frac{x}{y}=\frac{5}{3} \Rightarrow 3x=5y \Rightarrow 3x-5y=0$ ... (i).
  • Second condition: $x+y=720 \Rightarrow 5x+5y=3600$ ... (ii) [Multiplying by 5].
  • Adding (i) and (ii): $8x=3600 \Rightarrow x=450$.
  • Substituting $x=450$ in (i): $3(450)-5y=0 \Rightarrow 1350=5y \Rightarrow y=270$.
  • The daily wages of skilled workers is 450 and that of unskilled workers is 270.

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.

  • Let speeds of Hamid and Joseph be $x$ km/hr and $y$ km/hr respectively.
  • Distance travelled by Hamid in 20 minutes ($\frac{1}{3}$ hrs) $= \frac{1}{3}x$ km.
  • Distance travelled by Joseph in $\frac{1}{3}$ hrs $= \frac{1}{3}y$ km.
  • First condition: $\frac{1}{3}x+\frac{1}{3}y=30 \Rightarrow x+y=90$ ... (i) [Multiplying by 3].
  • Distance travelled by Hamid in 3 hrs $= 3x$ km. Distance travelled by Joseph in 3 hrs $= 3y$ km.
  • Second condition (Hamid catches Joseph, meaning Hamid covers the 30km gap plus the distance Joseph travels): $3x-3y=30 \Rightarrow x-y=10$ ... (ii) [Dividing by 3].
  • Adding (i) and (ii): $2x=100 \Rightarrow x=50$.
  • Substituting $x=50$ in (i): $50+y=90 \Rightarrow y=40$.
  • The speeds of Hamid and Joseph are 50 km/hr and 40 km/hr respectively.
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