\displaystyle \text{Solve each pair of equations given below using the substitution}
\displaystyle \text{method:}

\displaystyle \textbf{Question 1: }\quad x+y=12,\qquad x-y=2
\displaystyle \text{Answer:}
\displaystyle x+y=12\qquad\text{...(i)}
\displaystyle x-y=2\qquad\text{...(ii)}
\displaystyle \text{From (ii),}
\displaystyle x=y+2
\displaystyle \text{Substituting }x=y+2\text{ in (i),}
\displaystyle y+2+y=12
\displaystyle \Rightarrow 2y+2=12
\displaystyle \Rightarrow 2y=10
\displaystyle \Rightarrow y=5
\displaystyle \text{Substituting }y=5\text{ in }x=y+2,
\displaystyle x=5+2=7
\displaystyle \therefore x=7,\qquad y=5.
\displaystyle \text{Hence, the solution is }(7,5).
\displaystyle \\

\displaystyle \textbf{Question 2: }\quad 5x+3y=24,\qquad 3x-y=20
\displaystyle \text{Answer:}
\displaystyle 5x+3y=24\qquad\text{...(i)}
\displaystyle 3x-y=20\qquad\text{...(ii)}
\displaystyle \text{From (ii),}
\displaystyle y=3x-20
\displaystyle \text{Substituting }y=3x-20\text{ in (i),}
\displaystyle 5x+3(3x-20)=24
\displaystyle \Rightarrow 5x+9x-60=24
\displaystyle \Rightarrow 14x-60=24
\displaystyle \Rightarrow 14x=84
\displaystyle \Rightarrow x=6
\displaystyle \text{Substituting }x=6\text{ in }y=3x-20,
\displaystyle y=3(6)-20=18-20=-2
\displaystyle \therefore x=6,\qquad y=-2.
\displaystyle \text{Hence, the solution is }(6,-2).
\displaystyle \\

\displaystyle \textbf{Question 3: }\quad x-3y=2,\qquad 2x+7y=30
\displaystyle \text{Answer:}
\displaystyle x-3y=2\qquad\text{...(i)}
\displaystyle 2x+7y=30\qquad\text{...(ii)}
\displaystyle \text{From (i),}
\displaystyle x=3y+2
\displaystyle \text{Substituting }x=3y+2\text{ in (ii),}
\displaystyle 2(3y+2)+7y=30
\displaystyle \Rightarrow 6y+4+7y=30
\displaystyle \Rightarrow 13y+4=30
\displaystyle \Rightarrow 13y=26
\displaystyle \Rightarrow y=2
\displaystyle \text{Substituting }y=2\text{ in }x=3y+2,
\displaystyle x=3(2)+2=8
\displaystyle \therefore x=8,\qquad y=2.
\displaystyle \text{Hence, the solution is }(8,2).
\displaystyle \\

\displaystyle \textbf{Question 4: }\quad x+4y=5,\qquad 4x+y=50
\displaystyle \text{Answer:}
\displaystyle x+4y=5\qquad\text{...(i)}
\displaystyle 4x+y=50\qquad\text{...(ii)}
\displaystyle \text{From (i),}
\displaystyle x=5-4y
\displaystyle \text{Substituting }x=5-4y\text{ in (ii),}
\displaystyle 4(5-4y)+y=50
\displaystyle \Rightarrow 20-16y+y=50
\displaystyle \Rightarrow 20-15y=50
\displaystyle \Rightarrow -15y=30
\displaystyle \Rightarrow y=-2
\displaystyle \text{Substituting }y=-2\text{ in }x=5-4y,
\displaystyle x=5-4(-2)=5+8=13
\displaystyle \therefore x=13,\qquad y=-2.
\displaystyle \text{Hence, the solution is }(13,-2).
\displaystyle \\

\displaystyle \textbf{Question 5: }\quad 2x+3y=6,\qquad 3x+5y=15
\displaystyle \text{Answer:}
\displaystyle 2x+3y=6\qquad\text{...(i)}
\displaystyle 3x+5y=15\qquad\text{...(ii)}
\displaystyle \text{From (i),}
\displaystyle 2x=6-3y
\displaystyle \Rightarrow x=3-\frac{3}{2}y
\displaystyle \text{Substituting }x=3-\frac{3}{2}y\text{ in (ii),}
\displaystyle 3\left(3-\frac{3}{2}y\right)+5y=15
\displaystyle \Rightarrow 9-\frac{9}{2}y+5y=15
\displaystyle \Rightarrow 9+\frac{1}{2}y=15
\displaystyle \Rightarrow \frac{1}{2}y=6
\displaystyle \Rightarrow y=12
\displaystyle \text{Substituting }y=12\text{ in }x=3-\frac{3}{2}y,
\displaystyle x=3-\frac{3}{2}(12)=3-18=-15
\displaystyle \therefore x=-15,\qquad y=12.
\displaystyle \text{Hence, the solution is }(-15,12).
\displaystyle \\

\displaystyle \textbf{Question 6: }\quad 5x-7y=9,\qquad 2x+5y=12
\displaystyle \text{Answer:}
\displaystyle 5x-7y=9\qquad\text{...(i)}
\displaystyle 2x+5y=12\qquad\text{...(ii)}
\displaystyle \text{From (i),}
\displaystyle 5x=7y+9
\displaystyle \Rightarrow x=\frac{7}{5}y+\frac{9}{5}
\displaystyle \text{Substituting }x=\frac{7}{5}y+\frac{9}{5}\text{ in (ii),}
\displaystyle 2\left(\frac{7}{5}y+\frac{9}{5}\right)+5y=12
\displaystyle \Rightarrow \frac{14}{5}y+\frac{18}{5}+5y=12
\displaystyle \Rightarrow \frac{39}{5}y+\frac{18}{5}=12
\displaystyle \Rightarrow \frac{39}{5}y=\frac{42}{5}
\displaystyle \Rightarrow y=\frac{14}{13}
\displaystyle \text{Substituting }y=\frac{14}{13}\text{ in } x=\frac{7}{5}y+\frac{9}{5},
\displaystyle x=\frac{7}{5}\left(\frac{14}{13}\right)+\frac{9}{5}
\displaystyle =\frac{98}{65}+\frac{117}{65} =\frac{215}{65}=\frac{43}{13}
\displaystyle \therefore x=\frac{43}{13},\qquad y=\frac{14}{13}.
\displaystyle \text{Hence, the solution is } \left(\frac{43}{13},\frac{14}{13}\right).
\displaystyle \\

\displaystyle \text{Solve each pair of equations given below using the elimination}
\displaystyle \text{method:}

\displaystyle \textbf{Question 7: }\quad x+2y=39,\qquad 2x-3y=1
\displaystyle \text{Answer:}
\displaystyle x+2y=39\qquad\text{...(i)}
\displaystyle 2x-3y=1\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }2,\text{ we get}
\displaystyle 2x+4y=78\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle (2x+4y)-(2x-3y)=78-1
\displaystyle \Rightarrow 2x+4y-2x+3y=77
\displaystyle \Rightarrow 7y=77
\displaystyle \Rightarrow y=11
\displaystyle \text{Substituting }y=11\text{ in equation (i),}
\displaystyle x+2(11)=39
\displaystyle \Rightarrow x=39-22=17
\displaystyle \therefore x=17,\qquad y=11.
\displaystyle \text{Hence, the solution is }(17,11).
\displaystyle \\

\displaystyle \textbf{Question 8: }\quad 3x-y=5,\qquad 5x-2y=4
\displaystyle \text{Answer:}
\displaystyle 3x-y=5\qquad\text{...(i)}
\displaystyle 5x-2y=4\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }2,\text{ we get}
\displaystyle 6x-2y=10\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle (6x-2y)-(5x-2y)=10-4
\displaystyle \Rightarrow 6x-2y-5x+2y=6
\displaystyle \Rightarrow x=6
\displaystyle \text{Substituting }x=6\text{ in equation (i),}
\displaystyle 3(6)-y=5
\displaystyle \Rightarrow 18-y=5
\displaystyle \Rightarrow y=13
\displaystyle \therefore x=6,\qquad y=13.
\displaystyle \text{Hence, the solution is }(6,13).
\displaystyle \\

\displaystyle \textbf{Question 9: }\quad 14x-3y=54,\qquad 21x-8y=95
\displaystyle \text{Answer:}
\displaystyle 14x-3y=54\qquad\text{...(i)}
\displaystyle 21x-8y=95\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }3,\text{ we get}
\displaystyle 42x-9y=162\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle 42x-16y=190\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (42x-9y)-(42x-16y)=162-190
\displaystyle \Rightarrow 42x-9y-42x+16y=-28
\displaystyle \Rightarrow 7y=-28
\displaystyle \Rightarrow y=-4
\displaystyle \text{Substituting }y=-4\text{ in equation (i),}
\displaystyle 14x-3(-4)=54
\displaystyle \Rightarrow 14x+12=54
\displaystyle \Rightarrow 14x=42
\displaystyle \Rightarrow x=3
\displaystyle \therefore x=3,\qquad y=-4.
\displaystyle \text{Hence, the solution is }(3,-4).
\displaystyle \\

\displaystyle \textbf{Question 10: }\quad 7x-6y=37,\qquad 5x+4y=43
\displaystyle \text{Answer:}
\displaystyle 7x-6y=37\qquad\text{...(i)}
\displaystyle 5x+4y=43\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }5,\text{ we get}
\displaystyle 35x-30y=185\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }7,\text{ we get}
\displaystyle 35x+28y=301\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (35x-30y)-(35x+28y)=185-301
\displaystyle \Rightarrow 35x-30y-35x-28y=-116
\displaystyle \Rightarrow -58y=-116
\displaystyle \Rightarrow y=2
\displaystyle \text{Substituting }y=2\text{ in equation (i),}
\displaystyle 7x-6(2)=37
\displaystyle \Rightarrow 7x=49
\displaystyle \Rightarrow x=7
\displaystyle \therefore x=7,\qquad y=2.
\displaystyle \text{Hence, the solution is }(7,2).
\displaystyle \\

\displaystyle \textbf{Question 11: }\quad 10x+3y=36,\qquad 15x-14y=17
\displaystyle \text{Answer:}
\displaystyle 10x+3y=36\qquad\text{...(i)}
\displaystyle 15x-14y=17\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }3,\text{ we get}
\displaystyle 30x+9y=108\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle 30x-28y=34\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (30x+9y)-(30x-28y)=108-34
\displaystyle \Rightarrow 30x+9y-30x+28y=74
\displaystyle \Rightarrow 37y=74
\displaystyle \Rightarrow y=2
\displaystyle \text{Substituting }y=2\text{ in equation (i),}
\displaystyle 10x+3(2)=36
\displaystyle \Rightarrow 10x=30
\displaystyle \Rightarrow x=3
\displaystyle \therefore x=3,\qquad y=2.
\displaystyle \text{Hence, the solution is }(3,2).
\displaystyle \\

\displaystyle \textbf{Question 12: }\quad 4x+3y=14,\qquad 9x-5y=55
\displaystyle \text{Answer:}
\displaystyle 4x+3y=14\qquad\text{...(i)}
\displaystyle 9x-5y=55\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }9,\text{ we get}
\displaystyle 36x+27y=126\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }4,\text{ we get}
\displaystyle 36x-20y=220\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (36x+27y)-(36x-20y)=126-220
\displaystyle \Rightarrow 36x+27y-36x+20y=-94
\displaystyle \Rightarrow 47y=-94
\displaystyle \Rightarrow y=-2
\displaystyle \text{Substituting }y=-2\text{ in equation (i),}
\displaystyle 4x+3(-2)=14
\displaystyle \Rightarrow 4x-6=14
\displaystyle \Rightarrow 4x=20
\displaystyle \Rightarrow x=5
\displaystyle \therefore x=5,\qquad y=-2.
\displaystyle \text{Hence, the solution is }(5,-2).
\displaystyle \\

\displaystyle \textbf{Question 13: }\quad 4x-3y=11,\qquad 2x-5y=-5
\displaystyle \text{Answer:}
\displaystyle 4x-3y=11\qquad\text{...(i)}
\displaystyle 2x-5y=-5\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle 4x-10y=-10\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (i),}
\displaystyle (4x-3y)-(4x-10y)=11-(-10)
\displaystyle \Rightarrow 4x-3y-4x+10y=21
\displaystyle \Rightarrow 7y=21
\displaystyle \Rightarrow y=3
\displaystyle \text{Substituting }y=3\text{ in equation (i),}
\displaystyle 4x-3(3)=11
\displaystyle \Rightarrow 4x=20
\displaystyle \Rightarrow x=5
\displaystyle \therefore x=5,\qquad y=3.
\displaystyle \text{Hence, the solution is }(5,3).
\displaystyle \\

\displaystyle \text{Solve the following simultaneous equations:}

\displaystyle \textbf{Question 14: }\quad 11x-8y=46,\qquad 2x+7y=-17
\displaystyle \text{Answer:}
\displaystyle 11x-8y=46\qquad\text{...(i)}
\displaystyle 2x+7y=-17\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }2,\text{ we get}
\displaystyle 22x-16y=92\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }11,\text{ we get}
\displaystyle 22x+77y=-187\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (22x-16y)-(22x+77y)=92-(-187)
\displaystyle \Rightarrow 22x-16y-22x-77y=279
\displaystyle \Rightarrow -93y=279
\displaystyle \Rightarrow y=-3
\displaystyle \text{Substituting }y=-3\text{ in equation (i),}
\displaystyle 11x-8(-3)=46
\displaystyle \Rightarrow 11x+24=46
\displaystyle \Rightarrow 11x=22
\displaystyle \Rightarrow x=2
\displaystyle \therefore x=2,\qquad y=-3.
\displaystyle \text{Hence, the solution is }(2,-3).
\displaystyle \\

\displaystyle \textbf{Question 15: }\quad 5x-3y=13,\qquad 3x-2y=5
\displaystyle \text{Answer:}
\displaystyle 5x-3y=13\qquad\text{...(i)}
\displaystyle 3x-2y=5\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }3,\text{ we get}
\displaystyle 15x-9y=39\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }5,\text{ we get}
\displaystyle 15x-10y=25\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (15x-9y)-(15x-10y)=39-25
\displaystyle \Rightarrow 15x-9y-15x+10y=14
\displaystyle \Rightarrow y=14
\displaystyle \text{Substituting }y=14\text{ in equation (i),}
\displaystyle 5x-3(14)=13
\displaystyle \Rightarrow 5x-42=13
\displaystyle \Rightarrow 5x=55
\displaystyle \Rightarrow x=11
\displaystyle \therefore x=11,\qquad y=14.
\displaystyle \text{Hence, the solution is }(11,14).
\displaystyle \\

\displaystyle \textbf{Question 16: }\quad 5a+4b=22,\qquad 4a-5b=23
\displaystyle \text{Answer:}
\displaystyle 5a+4b=22\qquad\text{...(i)}
\displaystyle 4a-5b=23\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }4,\text{ we get}
\displaystyle 20a+16b=88\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }5,\text{ we get}
\displaystyle 20a-25b=115\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (20a+16b)-(20a-25b)=88-115
\displaystyle \Rightarrow 20a+16b-20a+25b=-27
\displaystyle \Rightarrow 41b=-27
\displaystyle \Rightarrow b=-\frac{27}{41}
\displaystyle \text{Substituting }b=-\frac{27}{41}\text{ in equation (i),}
\displaystyle 5a+4\left(-\frac{27}{41}\right)=22
\displaystyle \Rightarrow 5a-\frac{108}{41}=22
\displaystyle \Rightarrow 5a=\frac{1010}{41}
\displaystyle \Rightarrow a=\frac{202}{41}
\displaystyle \therefore a=\frac{202}{41},\qquad b=-\frac{27}{41}.
\displaystyle \text{Hence, the solution is } \left(\frac{202}{41},-\frac{27}{41}\right).
\displaystyle \\

\displaystyle \textbf{Question 17: }\quad 8x+3y=0,\qquad 3x+5(y+3)=-16
\displaystyle \text{Answer:}
\displaystyle 8x+3y=0\qquad\text{...(i)}
\displaystyle 3x+5(y+3)=-16\qquad\text{...(ii)}
\displaystyle \text{Simplifying equation (ii),}
\displaystyle 3x+5y+15=-16
\displaystyle \Rightarrow 3x+5y=-31\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (i) by }3,\text{ we get}
\displaystyle 24x+9y=0\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }8,\text{ we get}
\displaystyle 24x+40y=-248\qquad\text{...(v)}
\displaystyle \text{Subtracting equation (iv) from equation (v),}
\displaystyle (24x+40y)-(24x+9y)=-248-0
\displaystyle \Rightarrow 31y=-248
\displaystyle \Rightarrow y=-8
\displaystyle \text{Substituting }y=-8\text{ in equation (i),}
\displaystyle 8x+3(-8)=0
\displaystyle \Rightarrow 8x=24
\displaystyle \Rightarrow x=3
\displaystyle \therefore x=3,\qquad y=-8.
\displaystyle \text{Hence, the solution is }(3,-8).
\displaystyle \\

\displaystyle \textbf{Question 18: }\quad 8y-5z=7,\qquad 3y=4(z-2)
\displaystyle \text{Answer:}
\displaystyle 8y-5z=7\qquad\text{...(i)}
\displaystyle 3y=4(z-2)\qquad\text{...(ii)}
\displaystyle \text{Simplifying equation (ii),}
\displaystyle 3y=4z-8
\displaystyle \Rightarrow 3y-4z=-8\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (i) by }3,\text{ we get}
\displaystyle 24y-15z=21\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }8,\text{ we get}
\displaystyle 24y-32z=-64\qquad\text{...(v)}
\displaystyle \text{Subtracting equation (v) from equation (iv),}
\displaystyle (24y-15z)-(24y-32z)=21-(-64)
\displaystyle \Rightarrow 24y-15z-24y+32z=85
\displaystyle \Rightarrow 17z=85
\displaystyle \Rightarrow z=5
\displaystyle \text{Substituting }z=5\text{ in equation (i),}
\displaystyle 8y-5(5)=7
\displaystyle \Rightarrow 8y=32
\displaystyle \Rightarrow y=4
\displaystyle \therefore y=4,\qquad z=5.
\displaystyle \text{Hence, the solution is }(y,z)=(4,5).
\displaystyle \\

\displaystyle \textbf{Question 19: }\quad 2(a-3)+3(b-5)=0,\qquad 5(a-1)+4(b-4)=0
\displaystyle \text{Answer:}
\displaystyle 2(a-3)+3(b-5)=0\qquad\text{...(i)}
\displaystyle 5(a-1)+4(b-4)=0\qquad\text{...(ii)}
\displaystyle \text{Simplifying equation (i),}
\displaystyle 2a-6+3b-15=0
\displaystyle \Rightarrow 2a+3b=21\qquad\text{...(iii)}
\displaystyle \text{Simplifying equation (ii),}
\displaystyle 5a-5+4b-16=0
\displaystyle \Rightarrow 5a+4b=21\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }5,\text{ we get}
\displaystyle 10a+15b=105\qquad\text{...(v)}
\displaystyle \text{Multiplying equation (iv) by }2,\text{ we get}
\displaystyle 10a+8b=42\qquad\text{...(vi)}
\displaystyle \text{Subtracting equation (vi) from equation (v),}
\displaystyle (10a+15b)-(10a+8b)=105-42
\displaystyle \Rightarrow 7b=63
\displaystyle \Rightarrow b=9
\displaystyle \text{Substituting }b=9\text{ in equation (iii),}
\displaystyle 2a+3(9)=21
\displaystyle \Rightarrow 2a=-6
\displaystyle \Rightarrow a=-3
\displaystyle \therefore a=-3,\qquad b=9.
\displaystyle \text{Hence, the solution is }(a,b)=(-3,9).
\displaystyle \\

\displaystyle \textbf{Question 20: }\quad 4(3x-y)=9x+5,\qquad 3(2x+3y)=13(x+y-5)
\displaystyle \text{Answer:}
\displaystyle 4(3x-y)=9x+5\qquad\text{...(i)}
\displaystyle 3(2x+3y)=13(x+y-5)\qquad\text{...(ii)}
\displaystyle \text{Simplifying equation (i),}
\displaystyle 12x-4y=9x+5
\displaystyle \Rightarrow 3x-4y=5\qquad\text{...(iii)}
\displaystyle \text{Simplifying equation (ii),}
\displaystyle 6x+9y=13x+13y-65
\displaystyle \Rightarrow 7x+4y=65\qquad\text{...(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle (3x-4y)+(7x+4y)=5+65
\displaystyle \Rightarrow 10x=70
\displaystyle \Rightarrow x=7
\displaystyle \text{Substituting }x=7\text{ in equation (iii),}
\displaystyle 3(7)-4y=5
\displaystyle \Rightarrow 21-4y=5
\displaystyle \Rightarrow 4y=16
\displaystyle \Rightarrow y=4
\displaystyle \therefore x=7,\qquad y=4.
\displaystyle \text{Hence, the solution is }(7,4).
\displaystyle \\

\displaystyle \textbf{Question 21: }\quad \frac{x}{2}-\frac{y}{3}=3,\qquad 4x-3y=22
\displaystyle \text{Answer:}
\displaystyle \frac{x}{2}-\frac{y}{3}=3\qquad\text{...(i)}
\displaystyle 4x-3y=22\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }6,\text{ we get}
\displaystyle 3x-2y=18\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,\text{ we get}
\displaystyle 12x-9y=66\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }4,\text{ we get}
\displaystyle 12x-8y=72\qquad\text{...(v)}
\displaystyle \text{Subtracting equation (v) from equation (iv),}
\displaystyle (12x-9y)-(12x-8y)=66-72
\displaystyle \Rightarrow -y=-6
\displaystyle \Rightarrow y=6
\displaystyle \text{Substituting }y=6\text{ in equation (ii),}
\displaystyle 4x-3(6)=22
\displaystyle \Rightarrow 4x-18=22
\displaystyle \Rightarrow 4x=40
\displaystyle \Rightarrow x=10
\displaystyle \therefore x=10,\qquad y=6.
\displaystyle \text{Hence, the solution is }(10,6).
\displaystyle \\

\displaystyle \textbf{Question 22: }\quad \frac{x}{3}-\frac{y}{4}=0,\qquad \frac{2x}{3}+\frac{3y}{4}=5
\displaystyle \text{Answer:}
\displaystyle \frac{x}{3}-\frac{y}{4}=0\qquad\text{...(i)}
\displaystyle \frac{2x}{3}+\frac{3y}{4}=5\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }12,\text{ we get}
\displaystyle 4x-3y=0\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }12,\text{ we get}
\displaystyle 8x+9y=60\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }2,\text{ we get}
\displaystyle 8x-6y=0\qquad\text{...(v)}
\displaystyle \text{Subtracting equation (iv) from equation (v),}
\displaystyle (8x-6y)-(8x+9y)=0-60
\displaystyle \Rightarrow -15y=-60
\displaystyle \Rightarrow y=4
\displaystyle \text{Substituting }y=4\text{ in equation (iii),}
\displaystyle 4x-3(4)=0
\displaystyle \Rightarrow 4x=12
\displaystyle \Rightarrow x=3
\displaystyle \therefore x=3,\qquad y=4.
\displaystyle \text{Hence, the solution is }(3,4).
\displaystyle \\

\displaystyle \textbf{Question 23: }\quad \frac{x}{3}-\frac{5y}{6}=3,\qquad \frac{3x}{4}-\frac{5y}{2}=8
\displaystyle \text{Answer:}
\displaystyle \frac{x}{3}-\frac{5y}{6}=3\qquad\text{...(i)}
\displaystyle \frac{3x}{4}-\frac{5y}{2}=8\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }6,\text{ we get}
\displaystyle 2x-5y=18\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }4,\text{ we get}
\displaystyle 3x-10y=32\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }3,\text{ we get}
\displaystyle 6x-15y=54\qquad\text{...(v)}
\displaystyle \text{Multiplying equation (iv) by }2,\text{ we get}
\displaystyle 6x-20y=64\qquad\text{...(vi)}
\displaystyle \text{Subtracting equation (vi) from equation (v),}
\displaystyle (6x-15y)-(6x-20y)=54-64
\displaystyle \Rightarrow 5y=-10
\displaystyle \Rightarrow y=-2
\displaystyle \text{Substituting }y=-2\text{ in equation (iii),}
\displaystyle 2x-5(-2)=18
\displaystyle \Rightarrow 2x+10=18
\displaystyle \Rightarrow 2x=8
\displaystyle \Rightarrow x=4
\displaystyle \therefore x=4,\qquad y=-2.
\displaystyle \text{Hence, the solution is }(4,-2).
\displaystyle \\

\displaystyle \textbf{Question 24: }\quad \frac{3a+5}{4}=\frac{2b-1}{6},\qquad \frac{4a}{3}+\frac{b}{6}=-1
\displaystyle \text{Answer:}
\displaystyle \frac{3a+5}{4}=\frac{2b-1}{6}\qquad\text{...(i)}
\displaystyle \frac{4a}{3}+\frac{b}{6}=-1\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }12,\text{ we get}
\displaystyle 3(3a+5)=2(2b-1)
\displaystyle \Rightarrow 9a+15=4b-2
\displaystyle \Rightarrow 9a-4b=-17\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }6,\text{ we get}
\displaystyle 8a+b=-6\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iv) by }4,\text{ we get}
\displaystyle 32a+4b=-24\qquad\text{...(v)}
\displaystyle \text{Adding equations (iii) and (v),}
\displaystyle (9a-4b)+(32a+4b)=-17-24
\displaystyle \Rightarrow 41a=-41
\displaystyle \Rightarrow a=-1
\displaystyle \text{Substituting }a=-1\text{ in equation (iv),}
\displaystyle 8(-1)+b=-6
\displaystyle \Rightarrow -8+b=-6
\displaystyle \Rightarrow b=2
\displaystyle \therefore a=-1,\qquad b=2.
\displaystyle \text{Hence, the solution is }(a,b)=(-1,2).
\displaystyle \\

\displaystyle \textbf{Question 25: }\quad 23x+31y=7,\qquad 31x+23y=47
\displaystyle \text{Answer:}
\displaystyle 23x+31y=7\qquad\text{...(i)}
\displaystyle 31x+23y=47\qquad\text{...(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 54x+54y=54
\displaystyle \Rightarrow x+y=1\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (iii) by }23,\text{ we get}
\displaystyle 23x+23y=23\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (i),}
\displaystyle (23x+31y)-(23x+23y)=7-23
\displaystyle \Rightarrow 8y=-16
\displaystyle \Rightarrow y=-2
\displaystyle \text{Substituting }y=-2\text{ in equation (iii),}
\displaystyle x+(-2)=1
\displaystyle \Rightarrow x=3
\displaystyle \therefore x=3,\qquad y=-2.
\displaystyle \text{Hence, the solution is }(3,-2).
\displaystyle \\

\displaystyle \textbf{Question 26: }\quad 97x-78y=59,\qquad 78x-97y=116
\displaystyle \text{Answer:}
\displaystyle 97x-78y=59\qquad\text{...(i)}
\displaystyle 78x-97y=116\qquad\text{...(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 175x-175y=175
\displaystyle \Rightarrow x-y=1\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (iii) by }97,\text{ we get}
\displaystyle 97x-97y=97\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (i),}
\displaystyle (97x-78y)-(97x-97y)=59-97
\displaystyle \Rightarrow 19y=-38
\displaystyle \Rightarrow y=-2
\displaystyle \text{Substituting }y=-2\text{ in equation (iii),}
\displaystyle x-(-2)=1
\displaystyle \Rightarrow x+2=1
\displaystyle \Rightarrow x=-1
\displaystyle \therefore x=-1,\qquad y=-2.
\displaystyle \text{Hence, the solution is }(-1,-2).
\displaystyle \\

\displaystyle \textbf{Question 27: }\quad \frac{1}{x}+\frac{1}{y}=7,\qquad \frac{1}{x}-\frac{1}{y}=1
\displaystyle \text{Answer:}
\displaystyle \frac{1}{x}+\frac{1}{y}=7\qquad\text{...(i)}
\displaystyle \frac{1}{x}-\frac{1}{y}=1\qquad\text{...(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle \frac{2}{x}=8
\displaystyle \Rightarrow 2=8x
\displaystyle \Rightarrow x=\frac{1}{4}
\displaystyle \text{Substituting }x=\frac{1}{4}\text{ in equation (i),}
\displaystyle \frac{1}{\frac{1}{4}}+\frac{1}{y}=7
\displaystyle \Rightarrow 4+\frac{1}{y}=7
\displaystyle \Rightarrow \frac{1}{y}=3
\displaystyle \Rightarrow y=\frac{1}{3}
\displaystyle \therefore x=\frac{1}{4},\qquad y=\frac{1}{3}.
\displaystyle \text{Hence, the solution is } \left(\frac{1}{4},\frac{1}{3}\right).
\displaystyle \\

\displaystyle \textbf{Question 28: }\quad \frac{2}{x}+\frac{10}{y}=3,\qquad \frac{8}{x}-\frac{15}{y}=1
\displaystyle \text{Answer:}
\displaystyle \frac{2}{x}+\frac{10}{y}=3\qquad\text{...(i)}
\displaystyle \frac{8}{x}-\frac{15}{y}=1\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }4,\text{ we get}
\displaystyle \frac{8}{x}+\frac{40}{y}=12\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle \left(\frac{8}{x}+\frac{40}{y}\right)- \left(\frac{8}{x}-\frac{15}{y}\right)=12-1
\displaystyle \Rightarrow \frac{55}{y}=11
\displaystyle \Rightarrow 55=11y
\displaystyle \Rightarrow y=5
\displaystyle \text{Substituting }y=5\text{ in equation (ii),}
\displaystyle \frac{8}{x}-\frac{15}{5}=1
\displaystyle \Rightarrow \frac{8}{x}-3=1
\displaystyle \Rightarrow \frac{8}{x}=4
\displaystyle \Rightarrow x=2
\displaystyle \therefore x=2,\qquad y=5.
\displaystyle \text{Hence, the solution is }(2,5).
\displaystyle \\

\displaystyle \textbf{Question 29: }\quad 2x+\frac{3}{y}=20,\qquad 4x-\frac{9}{y}=10
\displaystyle \text{Answer:}
\displaystyle 2x+\frac{3}{y}=20\qquad\text{...(i)}
\displaystyle 4x-\frac{9}{y}=10\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }2,\text{ we get}
\displaystyle 4x+\frac{6}{y}=40\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle \left(4x+\frac{6}{y}\right)- \left(4x-\frac{9}{y}\right)=40-10
\displaystyle \Rightarrow \frac{15}{y}=30
\displaystyle \Rightarrow 15=30y
\displaystyle \Rightarrow y=\frac{1}{2}
\displaystyle \text{Substituting }y=\frac{1}{2}\text{ in equation (i),}
\displaystyle 2x+\frac{3}{\frac{1}{2}}=20
\displaystyle \Rightarrow 2x+6=20
\displaystyle \Rightarrow 2x=14
\displaystyle \Rightarrow x=7
\displaystyle \therefore x=7,\qquad y=\frac{1}{2}.
\displaystyle \text{Hence, the solution is } \left(7,\frac{1}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 30: }\quad \frac{6}{x}-4y=9,\qquad \frac{4}{x}-y=1
\displaystyle \text{Answer:}
\displaystyle \frac{6}{x}-4y=9\qquad\text{...(i)}
\displaystyle \frac{4}{x}-y=1\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (ii) by }4,\text{ we get}
\displaystyle \frac{16}{x}-4y=4\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (i),}
\displaystyle \left(\frac{6}{x}-4y\right)- \left(\frac{16}{x}-4y\right)=9-4
\displaystyle \Rightarrow -\frac{10}{x}=5
\displaystyle \Rightarrow -10=5x
\displaystyle \Rightarrow x=-2
\displaystyle \text{Substituting }x=-2\text{ in equation (ii),}
\displaystyle \frac{4}{-2}-y=1
\displaystyle \Rightarrow -2-y=1
\displaystyle \Rightarrow y=-3
\displaystyle \therefore x=-2,\qquad y=-3.
\displaystyle \text{Hence, the solution is }(-2,-3).
\displaystyle \\

\displaystyle \textbf{Question 31: }\quad \frac{3}{2x}-\frac{5}{3y}=\frac{7}{6},\qquad \frac{4}{5x}+\frac{1}{y}=1
\displaystyle \text{Answer:}
\displaystyle \frac{3}{2x}-\frac{5}{3y}=\frac{7}{6} \qquad\text{...(i)}
\displaystyle \frac{4}{5x}+\frac{1}{y}=1 \qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (ii) by }\frac{5}{3}, \text{ we get}
\displaystyle \frac{4}{3x}+\frac{5}{3y}=\frac{5}{3} \qquad\text{...(iii)}
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle \frac{3}{2x}+\frac{4}{3x}=\frac{7}{6}+\frac{5}{3}
\displaystyle \Rightarrow \left(\frac{3}{2}+\frac{4}{3}\right)\frac{1}{x} =\frac{7}{6}+\frac{10}{6}
\displaystyle \Rightarrow \frac{17}{6}\cdot\frac{1}{x}=\frac{17}{6}
\displaystyle \Rightarrow x=1
\displaystyle \text{Substituting }x=1\text{ in equation (ii),}
\displaystyle \frac{4}{5}+\frac{1}{y}=1
\displaystyle \Rightarrow \frac{1}{y}=1-\frac{4}{5} =\frac{1}{5}
\displaystyle \Rightarrow y=5
\displaystyle \therefore x=1,\qquad y=5.
\displaystyle \text{Hence, the solution is }(1,5).
\displaystyle \\

\displaystyle \textbf{Question 32: }\quad \frac{3x+2}{2y+3}=\frac{1}{8},\qquad \frac{x+1}{3y-2}=\frac{1}{8}
\displaystyle \text{Answer:}
\displaystyle \frac{3x+2}{2y+3}=\frac{1}{8} \qquad\text{...(i)}
\displaystyle \frac{x+1}{3y-2}=\frac{1}{8} \qquad\text{...(ii)}
\displaystyle \text{Cross-multiplying equation (i),}
\displaystyle 8(3x+2)=2y+3
\displaystyle \Rightarrow 24x+16=2y+3
\displaystyle \Rightarrow 24x-2y=-13\qquad\text{...(iii)}
\displaystyle \text{Cross-multiplying equation (ii),}
\displaystyle 8(x+1)=3y-2
\displaystyle \Rightarrow 8x+8=3y-2
\displaystyle \Rightarrow 8x-3y=-10\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }3,\text{ we get}
\displaystyle 72x-6y=-39\qquad\text{...(v)}
\displaystyle \text{Multiplying equation (iv) by }2,\text{ we get}
\displaystyle 16x-6y=-20\qquad\text{...(vi)}
\displaystyle \text{Subtracting equation (vi) from equation (v),}
\displaystyle (72x-6y)-(16x-6y)=-39-(-20)
\displaystyle \Rightarrow 56x=-19
\displaystyle \Rightarrow x=-\frac{19}{56}
\displaystyle \text{Substituting }x=-\frac{19}{56} \text{ in equation (iv),}
\displaystyle 8\left(-\frac{19}{56}\right)-3y=-10
\displaystyle \Rightarrow -\frac{19}{7}-3y=-10
\displaystyle \Rightarrow -3y=-\frac{51}{7}
\displaystyle \Rightarrow y=\frac{17}{7}
\displaystyle \therefore x=-\frac{19}{56},\qquad y=\frac{17}{7}.
\displaystyle \text{Hence, the solution is } \left(-\frac{19}{56},\frac{17}{7}\right).
\displaystyle \\

\displaystyle \textbf{Question 33: }\quad \frac{2x+1}{5}-\frac{3x-y}{2}=y,\qquad \frac{3x+2}{2}+\frac{2-y}{3}=x-y
\displaystyle \text{Answer:}
\displaystyle \frac{2x+1}{5}-\frac{3x-y}{2}=y \qquad\text{...(i)}
\displaystyle \frac{3x+2}{2}+\frac{2-y}{3}=x-y \qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }10,\text{ we get}
\displaystyle 2(2x+1)-5(3x-y)=10y
\displaystyle \Rightarrow 4x+2-15x+5y=10y
\displaystyle \Rightarrow -11x-5y=-2
\displaystyle \Rightarrow 11x+5y=2\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }6,\text{ we get}
\displaystyle 3(3x+2)+2(2-y)=6(x-y)
\displaystyle \Rightarrow 9x+6+4-2y=6x-6y
\displaystyle \Rightarrow 3x+4y=-10\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iii) by }3,\text{ we get}
\displaystyle 33x+15y=6\qquad\text{...(v)}
\displaystyle \text{Multiplying equation (iv) by }11,\text{ we get}
\displaystyle 33x+44y=-110\qquad\text{...(vi)}
\displaystyle \text{Subtracting equation (vi) from equation (v),}
\displaystyle (33x+15y)-(33x+44y)=6-(-110)
\displaystyle \Rightarrow -29y=116
\displaystyle \Rightarrow y=-4
\displaystyle \text{Substituting }y=-4\text{ in equation (iii),}
\displaystyle 11x+5(-4)=2
\displaystyle \Rightarrow 11x-20=2
\displaystyle \Rightarrow 11x=22
\displaystyle \Rightarrow x=2
\displaystyle \therefore x=2,\qquad y=-4.
\displaystyle \text{Hence, the solution is }(2,-4).
\displaystyle \\

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