Pair of Linear Equations in Two Variables — CBSE Class 10 Maths Textbook Solutions

(i) Let the number of girls be $x$x and the number of boys be $y$y.

Total students: $x + y = 10$x + y = 10 ...(1)
Girls 4 more than boys: $x - y = 4$x - y = 4 ...(2)

Two points on line (1): $x+y=10$x+y=10 passes through $(0,10)$(0,10) and $(10,0)$(10,0).
Two points on line (2): $x-y=4$x-y=4 passes through $(0,-4)$(0,-4) and $(4,0)$(4,0).

Plotting both lines on the same axes, they intersect at the point $(7, 3)$(7, 3).

So $x = 7$x = 7, $y = 3$y = 3, i.e. the number of girls is 7 and the number of boys is 3.

Check: $7+3=10$7+3=10 ✓ and $7-3=4$7-3=4 ✓.

(ii) Let the cost of one pencil be ₹$x$x and the cost of one pen be ₹$y$y.

$5x + 7y = 50$5x + 7y = 50 ...(1)
$7x + 5y = 46$7x + 5y = 46 ...(2)

Two points on line (1): it passes through $(10, 0)$(10, 0) and $(3, 5)$(3, 5).
Two points on line (2): it passes through $(8, -2)$(8, -2) and $(3, 5)$(3, 5).

Plotting both lines, they intersect at the point $(3, 5)$(3, 5).

So $x = 3$x = 3, $y = 5$y = 5, i.e. the cost of one pencil is ₹3 and the cost of one pen is ₹5.

Check: $5(3)+7(5)=15+35=50$5(3)+7(5)=15+35=50 ✓ and $7(3)+5(5)=21+25=46$7(3)+5(5)=21+25=46 ✓.

(i) $a_1=5,\ b_1=-4,\ c_1=8$a_1=5, b_1=-4, c_1=8 and $a_2=7,\ b_2=6,\ c_2=-9$a_2=7, b_2=6, c_2=-9.

$\dfrac{a_1}{a_2}=\dfrac{5}{7}$a_1/a_2=5/7, $\dfrac{b_1}{b_2}=\dfrac{-4}{6}=-\dfrac{2}{3}$b_1/b_2=-4/6=-2/3.

Since $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$a_1/a_2_1/b_2, the lines intersect at a point (the pair has a unique solution).

(ii) $a_1=9,\ b_1=3,\ c_1=12$a_1=9, b_1=3, c_1=12 and $a_2=18,\ b_2=6,\ c_2=24$a_2=18, b_2=6, c_2=24.

$\dfrac{a_1}{a_2}=\dfrac{9}{18}=\dfrac{1}{2}$a_1/a_2=9/18=1/2, $\dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac{1}{2}$b_1/b_2=3/6=1/2, $\dfrac{c_1}{c_2}=\dfrac{12}{24}=\dfrac{1}{2}$c_1/c_2=12/24=1/2.

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$a_1/a_2=b_1/b_2=c_1/c_2, the lines are coincident (infinitely many solutions).

(iii) $a_1=6,\ b_1=-3,\ c_1=10$a_1=6, b_1=-3, c_1=10 and $a_2=2,\ b_2=-1,\ c_2=9$a_2=2, b_2=-1, c_2=9.

$\dfrac{a_1}{a_2}=\dfrac{6}{2}=3$a_1/a_2=6/2=3, $\dfrac{b_1}{b_2}=\dfrac{-3}{-1}=3$b_1/b_2=-3/-1=3, $\dfrac{c_1}{c_2}=\dfrac{10}{9}$c_1/c_2=10/9.

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$a_1/a_2=b_1/b_2_1/c_2, the lines are parallel (no solution).

(i) $\dfrac{a_1}{a_2}=\dfrac{3}{2}$a_1/a_2=3/2, $\dfrac{b_1}{b_2}=\dfrac{2}{-3}=-\dfrac{2}{3}$b_1/b_2=2/-3=-2/3. Since $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$a_1/a_2_1/b_2, the pair is consistent (unique solution).

(ii) $\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}$a_1/a_2=2/4=1/2, $\dfrac{b_1}{b_2}=\dfrac{-3}{-6}=\dfrac{1}{2}$b_1/b_2=-3/-6=1/2, $\dfrac{c_1}{c_2}=\dfrac{8}{9}$c_1/c_2=8/9. Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$a_1/a_2=b_1/b_2_1/c_2, the pair is inconsistent (no solution).

(iii) $\dfrac{a_1}{a_2}=\dfrac{3/2}{9}=\dfrac{1}{6}$a_1/a_2=3/2/9=1/6, $\dfrac{b_1}{b_2}=\dfrac{5/3}{-10}=-\dfrac{1}{6}$b_1/b_2=5/3/-10=-1/6. Since $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$a_1/a_2_1/b_2, the pair is consistent (unique solution).

(iv) $\dfrac{a_1}{a_2}=\dfrac{5}{-10}=-\dfrac{1}{2}$a_1/a_2=5/-10=-1/2, $\dfrac{b_1}{b_2}=\dfrac{-3}{6}=-\dfrac{1}{2}$b_1/b_2=-3/6=-1/2, $\dfrac{c_1}{c_2}=\dfrac{11}{-22}=-\dfrac{1}{2}$c_1/c_2=11/-22=-1/2. Since all three ratios are equal, the pair is consistent (coincident lines, infinitely many solutions).

(v) $\dfrac{a_1}{a_2}=\dfrac{4/3}{2}=\dfrac{2}{3}$a_1/a_2=4/3/2=2/3, $\dfrac{b_1}{b_2}=\dfrac{2}{3}$b_1/b_2=2/3, $\dfrac{c_1}{c_2}=\dfrac{8}{12}=\dfrac{2}{3}$c_1/c_2=8/12=2/3. Since all three ratios are equal, the pair is consistent (coincident lines, infinitely many solutions).

(i) $\dfrac{a_1}{a_2}=\dfrac{1}{2}$a_1/a_2=1/2, $\dfrac{b_1}{b_2}=\dfrac{1}{2}$b_1/b_2=1/2, $\dfrac{c_1}{c_2}=\dfrac{-5}{-10}=\dfrac{1}{2}$c_1/c_2=-5/-10=1/2. All three ratios are equal, so the pair is consistent with infinitely many solutions — the two equations represent the same line. Every point satisfying $y = 5-x$y = 5-x is a solution (e.g. $(0,5), (1,4), (2,3),\dots$(0,5), (1,4), (2,3),).

(ii) $\dfrac{a_1}{a_2}=\dfrac{1}{3}$a_1/a_2=1/3, $\dfrac{b_1}{b_2}=\dfrac{-1}{-3}=\dfrac{1}{3}$b_1/b_2=-1/-3=1/3, $\dfrac{c_1}{c_2}=\dfrac{-8}{-16}=\dfrac{1}{2}$c_1/c_2=-8/-16=1/2. Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$a_1/a_2=b_1/b_2_1/c_2, the pair is inconsistent (parallel lines, no solution).

(iii) $\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}$a_1/a_2=2/4=1/2, $\dfrac{b_1}{b_2}=\dfrac{1}{-2}=-\dfrac{1}{2}$b_1/b_2=1/-2=-1/2. Since $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$a_1/a_2_1/b_2, the pair is consistent with a unique solution. To find it: from $2x+y=6$2x+y=6, $y=6-2x$y=6-2x. Substitute in $4x-2y=4$4x-2y=4: $4x-2(6-2x)=4 \Rightarrow 4x-12+4x=4 \Rightarrow 8x=16 \Rightarrow x=2$4x-2(6-2x)=4 4x-12+4x=4 8x=16 x=2, so $y=6-4=2$y=6-4=2. The lines intersect at $(2,2)$(2,2), the required solution.

(iv) $\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}$a_1/a_2=2/4=1/2, $\dfrac{b_1}{b_2}=\dfrac{-2}{-4}=\dfrac{1}{2}$b_1/b_2=-2/-4=1/2, $\dfrac{c_1}{c_2}=\dfrac{-2}{-5}=\dfrac{2}{5}$c_1/c_2=-2/-5=2/5. Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$a_1/a_2=b_1/b_2_1/c_2, the pair is inconsistent (parallel lines, no solution).

Let the width of the garden be $x$x m and the length be $y$y m.

Length is 4 m more than width: $y = x + 4$y = x + 4 ...(1)

Half the perimeter is 36 m, so the perimeter is 72 m: $2(x+y) = 72 \Rightarrow x + y = 36$2(x+y) = 72 x + y = 36 ...(2)

Substituting (1) in (2): $x + (x+4) = 36 \Rightarrow 2x + 4 = 36 \Rightarrow 2x = 32 \Rightarrow x = 16$x + (x+4) = 36 2x + 4 = 36 2x = 32 x = 16.

Then $y = 16 + 4 = 20$y = 16 + 4 = 20.

So the width of the garden is 16 m and the length is 20 m.

Check: half perimeter $= x+y = 16+20 = 36$= x+y = 16+20 = 36 m ✓.

Here $a_1=2,\ b_1=3,\ c_1=-8$a_1=2, b_1=3, c_1=-8. Any second equation with coefficients $a_2, b_2, c_2$a_2, b_2, c_2 needs to satisfy the condition below for each case (one possible answer is given in each case).

(i) Intersecting lines — need $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$a_1/a_2_1/b_2. One example: $3x + 2y - 7 = 0$3x + 2y - 7 = 0. Here $\dfrac{a_1}{a_2}=\dfrac{2}{3}$a_1/a_2=2/3 and $\dfrac{b_1}{b_2}=\dfrac{3}{2}$b_1/b_2=3/2, which are unequal, so the lines intersect at a point.

(ii) Parallel lines — need $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$a_1/a_2=b_1/b_2_1/c_2. One example: $2x + 3y - 12 = 0$2x + 3y - 12 = 0. Here $\dfrac{a_1}{a_2}=1$a_1/a_2=1, $\dfrac{b_1}{b_2}=1$b_1/b_2=1, but $\dfrac{c_1}{c_2}=\dfrac{-8}{-12}=\dfrac{2}{3}\neq 1$c_1/c_2=-8/-12=2/3≠ 1, so the lines are parallel.

(iii) Coincident lines — need $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$a_1/a_2=b_1/b_2=c_1/c_2. One example: $4x + 6y - 16 = 0$4x + 6y - 16 = 0 (simply double every term of the original equation). Here all three ratios equal $\dfrac{1}{2}$1/2, so the lines coincide.

Line 1: $x - y + 1 = 0 \Rightarrow y = x+1$x - y + 1 = 0 y = x+1.
$x$x-intercept (put $y=0$y=0): $x = -1$x = -1, giving the point $(-1, 0)$(-1, 0).
Another point: $x=2 \Rightarrow y=3$x=2 y=3, giving $(2,3)$(2,3).

Line 2: $3x + 2y - 12 = 0$3x + 2y - 12 = 0.
$x$x-intercept (put $y=0$y=0): $3x=12 \Rightarrow x=4$3x=12 x=4, giving the point $(4, 0)$(4, 0).
Another point: $x=2 \Rightarrow 6+2y=12 \Rightarrow y=3$x=2 6+2y=12 y=3, giving $(2,3)$(2,3).

Intersection of the two lines: solve $x-y=-1$x-y=-1 and $3x+2y=12$3x+2y=12 together. From the first, $y=x+1$y=x+1. Substitute: $3x+2(x+1)=12 \Rightarrow 5x+2=12 \Rightarrow 5x=10 \Rightarrow x=2$3x+2(x+1)=12 5x+2=12 5x=10 x=2, so $y=3$y=3. The lines meet at $(2, 3)$(2, 3).

The triangle is formed by the two lines and the $x$x-axis, so its three vertices are the two $x$x-intercepts and the point of intersection:

$(-1, 0), \quad (4, 0), \quad (2, 3)$(-1, 0), (4, 0), (2, 3)

Plot these three points and shade the triangle bounded by the segment of the $x$x-axis from $(-1,0)$(-1,0) to $(4,0)$(4,0) and the two line segments meeting at $(2,3)$(2,3).

(i) From $x+y=14$x+y=14: $x = 14-y$x = 14-y. Substitute in $x-y=4$x-y=4: $(14-y)-y=4 \Rightarrow 14-2y=4 \Rightarrow 2y=10 \Rightarrow y=5$(14-y)-y=4 14-2y=4 2y=10 y=5. Then $x=14-5=9$x=14-5=9.

Solution: $x=9,\ y=5$x=9, y=5.

(ii) From $s-t=3$s-t=3: $s=t+3$s=t+3. Substitute in $\dfrac{s}{3}+\dfrac{t}{2}=6$s/3+t/2=6: $\dfrac{t+3}{3}+\dfrac{t}{2}=6$t+3/3+t/2=6. Multiply throughout by 6: $2(t+3)+3t=36 \Rightarrow 2t+6+3t=36 \Rightarrow 5t=30 \Rightarrow t=6$2(t+3)+3t=36 2t+6+3t=36 5t=30 t=6. Then $s=6+3=9$s=6+3=9.

Solution: $s=9,\ t=6$s=9, t=6.

(iii) From $3x-y=3$3x-y=3: $y=3x-3$y=3x-3. Substitute in $9x-3y=9$9x-3y=9: $9x-3(3x-3)=9 \Rightarrow 9x-9x+9=9 \Rightarrow 9=9$9x-3(3x-3)=9 9x-9x+9=9 9=9, which is true for every value of $x$x.

This means the two equations represent the same line ($9x-3y=9$9x-3y=9 is just $3(3x-y=3)$3(3x-y=3)), so the pair has infinitely many solutions, given by $y = 3x-3$y = 3x-3 for any real value of $x$x.

(iv) From $0.2x+0.3y=1.3$0.2x+0.3y=1.3: $x = \dfrac{1.3-0.3y}{0.2}=6.5-1.5y$x = 1.3-0.3y/0.2=6.5-1.5y. Substitute in $0.4x+0.5y=2.3$0.4x+0.5y=2.3: $0.4(6.5-1.5y)+0.5y=2.3 \Rightarrow 2.6-0.6y+0.5y=2.3 \Rightarrow 2.6-0.1y=2.3 \Rightarrow 0.1y=0.3 \Rightarrow y=3$0.4(6.5-1.5y)+0.5y=2.3 2.6-0.6y+0.5y=2.3 2.6-0.1y=2.3 0.1y=0.3 y=3. Then $x=6.5-1.5(3)=6.5-4.5=2$x=6.5-1.5(3)=6.5-4.5=2.

Solution: $x=2,\ y=3$x=2, y=3.

(v) From $\sqrt{2}x+\sqrt{3}y=0$√2x+√3y=0: $x = -\dfrac{\sqrt{3}}{\sqrt{2}}y$x = -√3/√2y. Substitute in $\sqrt{3}x-\sqrt{8}y=0$√3x-√8y=0 (note $\sqrt{8}=2\sqrt{2}$√8=2√2):

$\sqrt{3}\left(-\dfrac{\sqrt{3}}{\sqrt{2}}y\right) - 2\sqrt{2}\,y = 0 \Rightarrow -\dfrac{3y}{\sqrt{2}} - 2\sqrt{2}y = 0$√3(-√3/√2y) - 2√2\,y = 0 -3y/√2 - 2√2y = 0

Multiply by $\sqrt{2}$√2: $-3y - 4y = 0 \Rightarrow -7y=0 \Rightarrow y=0$-3y - 4y = 0 -7y=0 y=0. Then $x=0$x=0.

Solution: $x=0,\ y=0$x=0, y=0.

(vi) Multiply $\dfrac{3x}{2}-\dfrac{5y}{3}=-2$3x/2-5y/3=-2 by 6: $9x-10y=-12$9x-10y=-12 ...(1)
Multiply $\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{13}{6}$x/3+y/2=13/6 by 6: $2x+3y=13$2x+3y=13 ...(2)

From (2): $x = \dfrac{13-3y}{2}$x = 13-3y/2. Substitute in (1): $9\left(\dfrac{13-3y}{2}\right)-10y=-12$9(13-3y/2)-10y=-12. Multiply by 2: $9(13-3y)-20y=-24 \Rightarrow 117-27y-20y=-24 \Rightarrow 117-47y=-24 \Rightarrow -47y=-141 \Rightarrow y=3$9(13-3y)-20y=-24 117-27y-20y=-24 117-47y=-24 -47y=-141 y=3. Then $x=\dfrac{13-9}{2}=2$x=13-9/2=2.

Solution: $x=2,\ y=3$x=2, y=3.

Subtracting the second equation from the first eliminates $x$x:

$(2x+3y) - (2x-4y) = 11-(-24) \Rightarrow 7y = 35 \Rightarrow y = 5$(2x+3y) - (2x-4y) = 11-(-24) 7y = 35 y = 5

Substitute $y=5$y=5 in $2x+3y=11$2x+3y=11: $2x+15=11 \Rightarrow 2x=-4 \Rightarrow x=-2$2x+15=11 2x=-4 x=-2.

Solution: $x=-2,\ y=5$x=-2, y=5.

Now substitute this solution into $y = mx+3$y = mx+3: $5 = m(-2)+3 \Rightarrow 5-3=-2m \Rightarrow 2=-2m \Rightarrow m=-1$5 = m(-2)+3 5-3=-2m 2=-2m m=-1.

So $m = -1$m = -1.

(i) Let the two numbers be $x$x and $y$y with $x>y$x>y.

$x - y = 26$x - y = 26 ...(1)
$x = 3y$x = 3y ...(2)

Substitute (2) in (1): $3y - y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13$3y - y = 26 2y = 26 y = 13. Then $x = 3(13)=39$x = 3(13)=39.

The numbers are 39 and 13.

(ii) Let the two supplementary angles be $x$x (larger) and $y$y (smaller).

$x + y = 180$x + y = 180 ...(1) (supplementary angles)
$x - y = 18$x - y = 18 ...(2)

From (2): $x = y+18$x = y+18. Substitute in (1): $(y+18)+y=180 \Rightarrow 2y=162 \Rightarrow y=81$(y+18)+y=180 2y=162 y=81. Then $x=81+18=99$x=81+18=99.

The angles are $99°$99° and $81°$81°.

(iii) Let the cost of one bat be ₹$x$x and one ball be ₹$y$y.

$7x + 6y = 3800$7x + 6y = 3800 ...(1)
$3x + 5y = 1750$3x + 5y = 1750 ...(2)

From (2): $x = \dfrac{1750-5y}{3}$x = 1750-5y/3. Substitute in (1): $7\left(\dfrac{1750-5y}{3}\right)+6y=3800$7(1750-5y/3)+6y=3800. Multiply by 3: $7(1750-5y)+18y=11400 \Rightarrow 12250-35y+18y=11400 \Rightarrow 12250-17y=11400 \Rightarrow 17y=850 \Rightarrow y=50$7(1750-5y)+18y=11400 12250-35y+18y=11400 12250-17y=11400 17y=850 y=50. Then $x=\dfrac{1750-250}{3}=\dfrac{1500}{3}=500$x=1750-250/3=1500/3=500.

Cost of one bat = ₹500, cost of one ball = ₹50.

(iv) Let the fixed charge be ₹$x$x and the charge per km be ₹$y$y.

$x + 10y = 105$x + 10y = 105 ...(1)
$x + 15y = 155$x + 15y = 155 ...(2)

Subtracting (1) from (2): $5y = 50 \Rightarrow y = 10$5y = 50 y = 10. Then from (1), $x = 105-100=5$x = 105-100=5.

Fixed charge = ₹5, charge per km = ₹10.

Charge for 25 km $= x + 25y = 5 + 250 = 255$= x + 25y = 5 + 250 = 255, i.e. ₹255.

(v) Let the fraction be $\dfrac{x}{y}$x/y.

$\dfrac{x+2}{y+2}=\dfrac{9}{11} \Rightarrow 11(x+2)=9(y+2) \Rightarrow 11x+22=9y+18 \Rightarrow 11x-9y+4=0$x+2/y+2=9/11 11(x+2)=9(y+2) 11x+22=9y+18 11x-9y+4=0 ...(1)

$\dfrac{x+3}{y+3}=\dfrac{5}{6} \Rightarrow 6(x+3)=5(y+3) \Rightarrow 6x+18=5y+15 \Rightarrow 6x-5y+3=0$x+3/y+3=5/6 6(x+3)=5(y+3) 6x+18=5y+15 6x-5y+3=0 ...(2)

From (2): $x = \dfrac{5y-3}{6}$x = 5y-3/6. Substitute in (1): $11\left(\dfrac{5y-3}{6}\right)-9y+4=0$11(5y-3/6)-9y+4=0. Multiply by 6: $11(5y-3)-54y+24=0 \Rightarrow 55y-33-54y+24=0 \Rightarrow y-9=0 \Rightarrow y=9$11(5y-3)-54y+24=0 55y-33-54y+24=0 y-9=0 y=9. Then $x=\dfrac{5(9)-3}{6}=\dfrac{42}{6}=7$x=5(9)-3/6=42/6=7.

The fraction is $\dfrac{7}{9}$7/9.

(vi) Let the present ages of Jacob and his son be $x$x and $y$y years.

Five years hence: $x+5 = 3(y+5) \Rightarrow x+5=3y+15 \Rightarrow x-3y-10=0$x+5 = 3(y+5) x+5=3y+15 x-3y-10=0 ...(1)
Five years ago: $x-5 = 7(y-5) \Rightarrow x-5=7y-35 \Rightarrow x-7y+30=0$x-5 = 7(y-5) x-5=7y-35 x-7y+30=0 ...(2)

From (1): $x = 3y+10$x = 3y+10. Substitute in (2): $(3y+10)-7y+30=0 \Rightarrow -4y+40=0 \Rightarrow y=10$(3y+10)-7y+30=0 -4y+40=0 y=10. Then $x=3(10)+10=40$x=3(10)+10=40.

Jacob's present age is 40 years, and his son's present age is 10 years.

(i) $x+y=5$x+y=5, $2x-3y=4$2x-3y=4

Elimination: Multiply the first equation by 3: $3x+3y=15$3x+3y=15. Add to the second equation: $3x+3y+2x-3y=15+4 \Rightarrow 5x=19 \Rightarrow x=\dfrac{19}{5}$3x+3y+2x-3y=15+4 5x=19 x=19/5. Then from $x+y=5$x+y=5: $y=5-\dfrac{19}{5}=\dfrac{6}{5}$y=5-19/5=6/5.

Substitution (check): From $x+y=5$x+y=5, $y=5-x$y=5-x. Substitute in $2x-3y=4$2x-3y=4: $2x-3(5-x)=4 \Rightarrow 2x-15+3x=4 \Rightarrow 5x=19 \Rightarrow x=\dfrac{19}{5}$2x-3(5-x)=4 2x-15+3x=4 5x=19 x=19/5, so $y=\dfrac{6}{5}$y=6/5 — the same answer.

Solution: $x=\dfrac{19}{5},\ y=\dfrac{6}{5}$x=19/5, y=6/5.

(ii) $3x+4y=10$3x+4y=10, $2x-2y=2$2x-2y=2

Elimination: Multiply the second equation by 2: $4x-4y=4$4x-4y=4. Add to the first equation: $3x+4y+4x-4y=10+4 \Rightarrow 7x=14 \Rightarrow x=2$3x+4y+4x-4y=10+4 7x=14 x=2. From $2x-2y=2$2x-2y=2: $4-2y=2 \Rightarrow y=1$4-2y=2 y=1.

Substitution (check): From $2x-2y=2$2x-2y=2, $x=y+1$x=y+1. Substitute in $3x+4y=10$3x+4y=10: $3(y+1)+4y=10 \Rightarrow 7y+3=10 \Rightarrow y=1$3(y+1)+4y=10 7y+3=10 y=1, so $x=2$x=2 — the same answer.

Solution: $x=2,\ y=1$x=2, y=1.

(iii) $3x-5y=4$3x-5y=4, $9x-2y=7$9x-2y=7 (rewriting $9x=2y+7$9x=2y+7)

Elimination: Multiply the first equation by 3: $9x-15y=12$9x-15y=12. Subtract the second equation: $(9x-15y)-(9x-2y)=12-7 \Rightarrow -13y=5 \Rightarrow y=-\dfrac{5}{13}$(9x-15y)-(9x-2y)=12-7 -13y=5 y=-5/13. From $3x-5y=4$3x-5y=4: $3x=4+5\left(-\dfrac{5}{13}\right)=4-\dfrac{25}{13}=\dfrac{27}{13} \Rightarrow x=\dfrac{9}{13}$3x=4+5(-5/13)=4-25/13=27/13 x=9/13.

Substitution (check): From $3x-5y=4$3x-5y=4, $x=\dfrac{4+5y}{3}$x=4+5y/3. Substitute in $9x-2y=7$9x-2y=7: $3(4+5y)-2y=7 \Rightarrow 12+15y-2y=7 \Rightarrow 13y=-5 \Rightarrow y=-\dfrac{5}{13}$3(4+5y)-2y=7 12+15y-2y=7 13y=-5 y=-5/13, giving the same $x=\dfrac{9}{13}$x=9/13.

Solution: $x=\dfrac{9}{13},\ y=-\dfrac{5}{13}$x=9/13, y=-5/13.

(iv) $\dfrac{x}{2}+\dfrac{2y}{3}=-1$x/2+2y/3=-1, $x-\dfrac{y}{3}=3$x-y/3=3

Clear fractions: multiply the first by 6: $3x+4y=-6$3x+4y=-6 ...(1); multiply the second by 3: $3x-y=9$3x-y=9 ...(2)

Elimination: Subtract (2) from (1): $(3x+4y)-(3x-y)=-6-9 \Rightarrow 5y=-15 \Rightarrow y=-3$(3x+4y)-(3x-y)=-6-9 5y=-15 y=-3. From (2): $3x-(-3)=9 \Rightarrow 3x=6 \Rightarrow x=2$3x-(-3)=9 3x=6 x=2.

Substitution (check): From (2), $x=\dfrac{9+y}{3}$x=9+y/3. Substitute in (1): $3\left(\dfrac{9+y}{3}\right)+4y=-6 \Rightarrow 9+y+4y=-6 \Rightarrow 5y=-15 \Rightarrow y=-3$3(9+y/3)+4y=-6 9+y+4y=-6 5y=-15 y=-3, giving the same $x=2$x=2.

Solution: $x=2,\ y=-3$x=2, y=-3.

(i) Let the fraction be $\dfrac{x}{y}$x/y.

$\dfrac{x+1}{y-1}=1 \Rightarrow x+1=y-1 \Rightarrow x-y+2=0$x+1/y-1=1 x+1=y-1 x-y+2=0 ...(1)
$\dfrac{x}{y+1}=\dfrac{1}{2} \Rightarrow 2x=y+1 \Rightarrow 2x-y-1=0$x/y+1=1/2 2x=y+1 2x-y-1=0 ...(2)

Subtract (1) from (2): $(2x-y-1)-(x-y+2)=0 \Rightarrow x-3=0 \Rightarrow x=3$(2x-y-1)-(x-y+2)=0 x-3=0 x=3. From (1): $y=x+2=5$y=x+2=5.

The fraction is $\dfrac{3}{5}$3/5.

(ii) Let the present ages of Nuri and Sonu be $x$x and $y$y years.

Five years ago: $x-5=3(y-5) \Rightarrow x-5=3y-15 \Rightarrow x-3y+10=0$x-5=3(y-5) x-5=3y-15 x-3y+10=0 ...(1)
Ten years later: $x+10=2(y+10) \Rightarrow x+10=2y+20 \Rightarrow x-2y-10=0$x+10=2(y+10) x+10=2y+20 x-2y-10=0 ...(2)

Subtract (1) from (2): $(x-2y-10)-(x-3y+10)=0 \Rightarrow y-20=0 \Rightarrow y=20$(x-2y-10)-(x-3y+10)=0 y-20=0 y=20. From (2): $x=2(20)+10=50$x=2(20)+10=50.

Nuri's present age is 50 years and Sonu's present age is 20 years.

(iii) Let the tens digit be $x$x and the units digit be $y$y, so the number is $10x+y$10x+y and the reversed number is $10y+x$10y+x.

$x+y=9$x+y=9 ...(1)
$9(10x+y)=2(10y+x) \Rightarrow 90x+9y=20y+2x \Rightarrow 88x-11y=0 \Rightarrow 8x-y=0$9(10x+y)=2(10y+x) 90x+9y=20y+2x 88x-11y=0 8x-y=0 ...(2)

From (2): $y=8x$y=8x. Substitute in (1): $x+8x=9 \Rightarrow 9x=9 \Rightarrow x=1$x+8x=9 9x=9 x=1, so $y=8$y=8.

The number is $10(1)+8=18$10(1)+8=18.

(iv) Let the number of ₹50 notes be $x$x and the number of ₹100 notes be $y$y.

$x+y=25$x+y=25 ...(1) (total notes)
$50x+100y=2000 \Rightarrow x+2y=40$50x+100y=2000 x+2y=40 ...(2)

Subtract (1) from (2): $(x+2y)-(x+y)=40-25 \Rightarrow y=15$(x+2y)-(x+y)=40-25 y=15. From (1): $x=25-15=10$x=25-15=10.

Meena received 10 notes of ₹50 and 15 notes of ₹100.

(v) Let the fixed charge for the first three days be ₹$x$x and the charge for each extra day be ₹$y$y.

Saritha kept the book 7 days, i.e. $7-3=4$7-3=4 extra days: $x+4y=27$x+4y=27 ...(1)
Susy kept the book 5 days, i.e. $5-3=2$5-3=2 extra days: $x+2y=21$x+2y=21 ...(2)

Subtract (2) from (1): $2y=6 \Rightarrow y=3$2y=6 y=3. From (2): $x=21-6=15$x=21-6=15.

Fixed charge = ₹15, additional charge per day = ₹3.