Pair of Linear Equations in Two Variables — Exercise 3.3

(i) x+y=5; 2x−3y=4

Elimination: Double the first equation to get 2x+2y=10. Subtract the second: 5y=6, so y=6/5; x=5−6/5=19/5.

Substitution: x=5−y gives 2(5−y)−3y=4, hence 10−5y=4 and the same values.

(ii) 3x+4y=10; 2x−2y=2

Elimination: Twice the first is 6x+8y=20; three times the second is 6x−6y=6. Subtract: 14y=14, giving y=1 and x=2.

Substitution: The second gives x=y+1. Then 3(y+1)+4y=10, so 7y=7.

(iii) 3x−5y=4; 9x−2y=7

Elimination: Triple the first: 9x−15y=12. Subtract this from the second: 13y=−5. Thus y=−5/13, and 3x=4+5y=27/13 gives x=9/13.

Substitution: x=(4+5y)/3 in the second gives 3(4+5y)−2y=7, so 13y=−5 again.

(iv) x/2+2y/3=−1; x−y/3=3

Clear denominators to get 3x+4y=−6 and 3x−y=9.

Elimination: Subtract the second equation from the first: 5y=−15, so y=−3 and x=2.

Substitution: x=3+y/3 in 3x+4y=−6 gives 9+y+4y=−6, hence y=−3.

(i) Fraction

Let numerator be x and denominator y. The conditions (x+1)/(y−1)=1 and x/(y+1)=1/2 give:

\[x-y=-2,\qquad2x-y=1\]

Subtract the first from the second: x=3. Then 3−y=−2 gives y=5.

(ii) Nuri and Sonu

Let their present ages be N and S. Five years ago: N−5=3(S−5), giving N−3S=−10. Ten years later: N+10=2(S+10), giving N−2S=10.

Subtract the first equation from the second: S=20. Then N=50.

(iii) Two-digit number

Let tens digit be x and units digit y. Then x+y=9. The number is 10x+y; its reversal is 10y+x.

\[9(10x+y)=2(10y+x)\]

\[88x-11y=0\Rightarrow8x-y=0\]

Add this to x+y=9: 9x=9, so x=1 and y=8.

(iv) ₹50 and ₹100 banknotes

Let x be the number of ₹50 notes and y the number of ₹100 notes. Then x+y=25 and 50x+100y=2000. Divide the second by 50: x+2y=40.

Subtract x+y=25 to get y=15, then x=10.

(v) Library charges

Let ₹f be the fixed charge for the first three days and ₹r the charge per additional day. Seven days has 4 extra days; five days has 2 extra days.

\[f+4r=27,\qquad f+2r=21\]

Subtract: 2r=6, so r=3. Then f=21−6=15.