Pair of Linear Equations in Two Variables — Exercise 3.2

(i) x+y=14; x−y=4

From the first, x=14−y. Substitute into the second:

\[14-y-y=4\Rightarrow y=5,\quad x=9\]

(ii) s−t=3; s/3+t/2=6

Use s=t+3 and multiply the second equation by 6:

\[2(t+3)+3t=36\Rightarrow5t=30\]

(iii) 3x−y=3; 9x−3y=9

The first gives y=3x−3. Substituting into the second gives 9x−3(3x−3)=9, or 9=9. No new restriction remains.

(iv) 0.2x+0.3y=1.3; 0.4x+0.5y=2.3

Multiply both equations by 10. Then x=(13−3y)/2.

\[4\left(\frac{13-3y}{2}\right)+5y=23\]

\[26-6y+5y=23\Rightarrow y=3,\quad x=2\]

(v) √2x+√3y=0; √3x−√8y=0

The first gives \(x=-\sqrt3y/\sqrt2\). Insert it into the second and use √8=2√2:

\[-\frac3{\sqrt2}y-2\sqrt2y=0\]

\[-\frac7{\sqrt2}y=0\Rightarrow y=0,\quad x=0\]

(vi) 3x/2−5y/3=−2; x/3+y/2=13/6

Multiplying both equations by 6 gives 9x−10y=−12 and 2x+3y=13. Use x=(13−3y)/2:

\[9\left(\frac{13-3y}{2}\right)-10y=-12\]

\[117-47y=-24\Rightarrow y=3,\quad x=2\]

Express x from the first equation and substitute:

\[x=\frac{11-3y}{2}\]

\[11-3y-4y=-24\Rightarrow -7y=-35\]

\[y=5,\qquad x=\frac{11-15}{2}=-2\]

Now use the extra relation y=mx+3:

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

(i) Difference 26; larger number is triple

Let the smaller number be x and the larger y. Then y−x=26 and y=3x.

\[3x-x=26\Rightarrow x=13,\quad y=39\]

(ii) Supplementary angles differing by 18°

Let the smaller angle be x° and the larger y°. Then x+y=180 and y=x+18.

\[x+(x+18)=180\Rightarrow x=81,\quad y=99\]

(iii) Bats and balls

Let a bat cost ₹b and a ball ₹c. Then 7b+6c=3800 and 3b+5c=1750. Use b=(1750−5c)/3 in the first:

\[7(1750-5c)+18c=11400\]

\[12250-17c=11400\Rightarrow c=50\]

\[b=\frac{1750-250}{3}=500\]

(iv) Fixed and per-kilometre taxi fare

Let the fixed charge be ₹f and the rate ₹r per kilometre. Then f+10r=105 and f+15r=155. Use f=105−10r:

\[105-10r+15r=155\Rightarrow r=10\]

\[f=105-100=5\]

For 25 km, fare=5+25×10=255.

(v) Fraction after adding 2 or 3 to both terms

Let the original numerator be x and denominator y. The two conditions give:

\[\frac{x+2}{y+2}=\frac9{11}\Rightarrow11x-9y=-4\]

\[\frac{x+3}{y+3}=\frac56\Rightarrow6x-5y=-3\]

From the first, x=(9y−4)/11. Substitute:

\[6(9y-4)-55y=-33\Rightarrow y=9\]

\[x=\frac{81-4}{11}=7\]

(vi) Jacob and his son

Let their present ages be J and S. Five years later J+5=3(S+5), so J=3S+10. Five years earlier J−5=7(S−5), so J=7S−30.

\[3S+10=7S-30\Rightarrow S=10\]

\[J=3(10)+10=40\]