NCERT Exercise 3.2 — Pair of Linear Equations in Two Variables

Question

Question 1

substitution method

Solve by substitution.

$x=14-y\Rightarrow14-2y=4\Rightarrow(x,y)=(9,5)$
$s=t+3;\ 2s+3t=36\Rightarrow(s,t)=(9,6)$
The second equation is three times the first: infinitely many solutions.
$2x+3y=13,\ 4x+5y=23\Rightarrow(x,y)=(2,3)$
$x=\frac{8y}{3}\Rightarrow\frac{43y}{9}=0\Rightarrow(x,y)=(0,0)$
$9x-10y=-12,\ 2x+3y=13\Rightarrow(x,y)=(2,3)$

Question

substitution method

Solve 2x + 3y = 11 and 2x − 4y = −24; hence find m if y = mx + 3.

7y=35\Rightarrow y=5,\qquad x=-2

5=-2m+3\Rightarrow m=-1

Both original equations evaluate correctly.

x=-2,\quad y=5,\quad m=-1

Question

substitution method · word problems

Form equations and solve by substitution.

Two numbers differ by 26; one is three times the other.
Supplementary angles differ by 18°.
7 bats and 6 balls cost ₹3800; 3 bats and 5 balls cost ₹1750.
Taxi fares are ₹105 for 10 km and ₹155 for 15 km. Find fixed and per-km charges and the 25 km fare.
A fraction becomes 9/11 after adding 2 to numerator and denominator, and 5/6 after adding 3.
Five years hence Jacob is three times his son; five years ago he was seven times his son.

$x-y=26,\ x=3y\Rightarrow(x,y)=(39,13)$
$L+S=180,\ L-S=18\Rightarrow(L,S)=(99^\circ,81^\circ)$
$7b+6c=3800,\ 3b+5c=1750\Rightarrow(b,c)=(500,50)$
$f+10r=105,\ f+15r=155\Rightarrow(f,r)=(5,10)$
$f+25r=255$
$11x-9y=-4,\ 6x-5y=-3\Rightarrow(x,y)=(7,9)$
$J-3S=10,\ J-7S=-30\Rightarrow(J,S)=(40,10)$

Answers: 39 and 13; 99° and 81°; ₹500 and ₹50; ₹5 fixed, ₹10/km, ₹255; 7/9; 40 and 10 years.