Practice Set 1.1 Algebra 10th Standard Maths Part 1 Chapter 1 Linear Equations in Two Variables Solutions Maharashtra Board

Practice Set 1.1 Solutions

Question 1

Complete the following activity to solve the simultaneous equations.

$5x+3y=9$ ... (i)

$2x-3y=12$ ... (ii)

Add equations (i) and (ii).

$5x+3y=9$ ... (1)

$2x-3y=12$ ... (2)

$7x = \boxed{21}$

$x = \frac{21}{7}$

$x = 3$

Place $x=3$ in equation (1).

$5(3) + 3y = 9$

$3y = 9 - \boxed{15}$

$3y = \boxed{-6}$

$y = \frac{\boxed{-6}}{3}$

$y = \boxed{-2}$

Solution is $(x, y) = (\boxed{3}, \boxed{-2})$.

Question 2

Solve the following simultaneous equations.

i. $3a+5b=26$; $a+5b=22$

$3a+5b=26$ ... (i)
$a+5b=22$ ... (ii)
Subtracting equation (ii) from (i), we get $2a=4$, therefore $a=2$.
Substituting $a=2$ in equation (ii) yields $2+5b=22$.
$5b=20$, so $b=4$.
$(a,b)=(2,4)$ is the solution.

ii. $x+7y=10$; $3x-2y=7$

$x+7y=10 \Rightarrow x=10-7y$ ... (i)
$3x-2y=7$ ... (ii)
Substituting $x=10-7y$ in equation (ii) yields $3(10-7y)-2y=7$.
$30-23y=7 \Rightarrow -23y=-23 \Rightarrow y=1$.
Substituting $y=1$ in equation (i) gives $x=10-7(1)=3$.
$(x,y)=(3,1)$ is the solution.

iii. $2x-3y=9$; $2x+y=13$

$2x-3y=9$ ... (i)
$2x+y=13$ ... (ii)
Subtracting equation (ii) from (i) yields $-4y=-4$, therefore $y=1$.
Substituting $y=1$ in equation (ii) yields $2x+1=13 \Rightarrow 2x=12 \Rightarrow x=6$.
$(x,y)=(6,1)$ is the solution.

iv. $5m-3n=19$; $m-6n=-7$

$5m-3n=19$ ... (i)
$m-6n=-7 \Rightarrow m=6n-7$ ... (ii)
Substituting $m=6n-7$ in equation (i) yields $5(6n-7)-3n=19$.
$30n-35-3n=19 \Rightarrow 27n=54 \Rightarrow n=2$.
Substituting $n=2$ in equation (ii) gives $m=6(2)-7=5$.
$(m,n)=(5,2)$ is the solution.

v. $5x+2y=-3$; $x+5y=4$

$5x+2y=-3$ ... (i)
$x+5y=4 \Rightarrow x=4-5y$ ... (ii)
Substituting $x=4-5y$ in equation (i) yields $5(4-5y)+2y=-3$.
$20-23y=-3 \Rightarrow -23y=-23 \Rightarrow y=1$.
Substituting $y=1$ in equation (ii) gives $x=4-5(1)=-1$.
$(x,y)=(-1,1)$ is the solution.

vi. $\frac{1}{3}x+y=\frac{10}{3}$; $2x+\frac{1}{4}y=\frac{11}{4}$

Multiplying the first equation by 3 yields $x+3y=10 \Rightarrow x=10-3y$ ... (i).
Multiplying the second equation by 4 yields $8x+y=11$ ... (ii).
Substituting $x=10-3y$ in equation (ii) gives $8(10-3y)+y=11$.
$80-23y=11 \Rightarrow -23y=-69 \Rightarrow y=3$.
Substituting $y=3$ in equation (i) gives $x=10-3(3)=1$.
$(x,y)=(1,3)$ is the solution.

vii. $99x+101y=499$; $101x+99y=501$

$99x+101y=499$ ... (i)
$101x+99y=501$ ... (ii)
Adding equations (i) and (ii) gives $200x+200y=1000 \Rightarrow x+y=5$ ... (iii).
Subtracting equation (ii) from (i) gives $-2x+2y=-2 \Rightarrow x-y=1$ ... (iv).
Adding (iii) and (iv) yields $2x=6 \Rightarrow x=3$.
Substituting $x=3$ in equation (iii) gives $3+y=5 \Rightarrow y=2$.
$(x,y)=(3,2)$ is the solution.

viii. $49x-57y=172$; $57x-49y=252$

$49x-57y=172$ ... (i)
$57x-49y=252$ ... (ii)
Adding equations (i) and (ii) gives $106x-106y=424 \Rightarrow x-y=4$ ... (iii).
Subtracting equation (ii) from (i) gives $-8x-8y=-80 \Rightarrow x+y=10$ ... (iv).
Adding (iii) and (iv) yields $2x=14 \Rightarrow x=7$.
Substituting $x=7$ in equation (iv) gives $7+y=10 \Rightarrow y=3$.
$(x,y)=(7,3)$ is the solution.

Complete the following table (Textbook pg. no. 1)

No.	Equation	Is the equation a linear equation in 2 variables	Reason
1	$4m+3n=12$	Yes	Two variables each with degree 1
2	$3x^{2}-7y=13$	No	The degree of variable x is 2
3	$\sqrt{2}x-\sqrt{5}y=16$	Yes	Two variables each with degree 1
4	$0x+6y-3=0$	No	Only one variable y
5	$0.3x+0y-36=0$	No	Only one variable x
6	$\frac{4}{x}+\frac{5}{y}=4$	No	The degree of variables is -1
7	$4xy-5y-8=0$	No	The degree of xy is 2

Question 1 (Textbook pg. no. 3)

Solve: $3x+2y=29$; $5x-y=18$

$3x+2y=29$ ... (i)

$5x-y=18$ ... (ii)

Let's solve the equations by eliminating 'y'.

Multiplying equation (ii) by 2, we get:

$5x \times \boxed{2} - y \times \boxed{2} = 18 \times \boxed{2}$

$10x - 2y = \boxed{36}$ ... (iii)

Add equations (i) and (iii):

$3x+2y=29$

$+\boxed{10x}-\boxed{2y}=\boxed{36}$

$\boxed{13x}=\boxed{65}$

$x=\frac{65}{13}=\boxed{5}$

Substituting $x=5$ in equation (i):

$3x+2y=29$

$3 \times \boxed{5} + 2y = 29$

$\boxed{15} + 2y = 29$

$2y = 29 - \boxed{15}$

$2y = \boxed{14}$

$y = \frac{14}{2} = \boxed{7}$

$(x,y)=(\boxed{5},\boxed{7})$ is the solution.