\displaystyle \textbf{Question 1: }\text{Solve the following pairs of simultaneous equations using the substitution}
\displaystyle \text{method:}
\displaystyle \text{(i) }2x+3y=12\text{ and }3x+2y=18
\displaystyle \text{(ii) }\frac{b}{a}x+\frac{a}{b}y=a^2+b^2\text{ and }x+y=2ab,\quad a\ne0,\ b\ne0
\displaystyle \text{(iii) }\frac{x}{6}=y-6\text{ and }\frac{3x}{4}=1+y
\displaystyle \text{(iv) }\frac{x}{2}+\frac{2y}{3}=-1\text{ and }x-\frac{y}{3}=3
\displaystyle \text{(v) }9-(x-4)=y+7\text{ and }2(x+y)=4-3y
\displaystyle \text{Answer:}

\displaystyle \text{(i) The given equations are}
\displaystyle 2x+3y=12\qquad\ldots\text{(1)}
\displaystyle 3x+2y=18\qquad\ldots\text{(2)}
\displaystyle \text{From equation (1),}
\displaystyle 2x=12-3y
\displaystyle \Rightarrow x=6-\frac{3y}{2}
\displaystyle \text{Substituting }x=6-\frac{3y}{2}\text{ in equation (2),}
\displaystyle 3\left(6-\frac{3y}{2}\right)+2y=18
\displaystyle \Rightarrow 18-\frac{9y}{2}+2y=18
\displaystyle \Rightarrow 18-\frac{5y}{2}=18
\displaystyle \Rightarrow -\frac{5y}{2}=0
\displaystyle \Rightarrow y=0
\displaystyle \text{Substituting }y=0\text{ in }x=6-\frac{3y}{2},
\displaystyle x=6-\frac{3}{2}(0)=6
\displaystyle \therefore \text{The solution is }x=6,\ y=0.
\displaystyle \\

\displaystyle \text{(ii) The given equations are}
\displaystyle \frac{b}{a}x+\frac{a}{b}y=a^2+b^2\qquad\ldots\text{(1)}
\displaystyle x+y=2ab\qquad\ldots\text{(2)}
\displaystyle \text{From equation (2),}
\displaystyle x=2ab-y
\displaystyle \text{Substituting }x=2ab-y\text{ in equation (1),}
\displaystyle \frac{b}{a}(2ab-y)+\frac{a}{b}y=a^2+b^2
\displaystyle \Rightarrow 2b^2-\frac{b}{a}y+\frac{a}{b}y=a^2+b^2
\displaystyle \Rightarrow \left(\frac{a}{b}-\frac{b}{a}\right)y=a^2-b^2
\displaystyle \Rightarrow \frac{a^2-b^2}{ab}y=a^2-b^2
\displaystyle \text{When }a^2\ne b^2,
\displaystyle \frac{y}{ab}=1
\displaystyle \Rightarrow y=ab
\displaystyle \text{Substituting }y=ab\text{ in }x=2ab-y,
\displaystyle x=2ab-ab=ab
\displaystyle \therefore \text{When }a^2\ne b^2,\text{ the unique solution is }x=ab,\ y=ab.
\displaystyle \text{When }a^2=b^2,\text{ the two equations are dependent and have infinitely many solutions.}
\displaystyle \\

\displaystyle \text{(iii) The given equations are}
\displaystyle \frac{x}{6}=y-6\qquad\ldots\text{(1)}
\displaystyle \frac{3x}{4}=1+y\qquad\ldots\text{(2)}
\displaystyle \text{From equation (1),}
\displaystyle y=\frac{x}{6}+6
\displaystyle \text{Substituting }y=\frac{x}{6}+6\text{ in equation (2),}
\displaystyle \frac{3x}{4}=1+\frac{x}{6}+6
\displaystyle \Rightarrow \frac{3x}{4}-\frac{x}{6}=7
\displaystyle \Rightarrow \left(\frac{9-2}{12}\right)x=7
\displaystyle \Rightarrow \frac{7x}{12}=7
\displaystyle \Rightarrow x=12
\displaystyle \text{Substituting }x=12\text{ in }y=\frac{x}{6}+6,
\displaystyle y=\frac{12}{6}+6=8
\displaystyle \therefore \text{The solution is }x=12,\ y=8.
\displaystyle \\

\displaystyle \text{(iv) The given equations are}
\displaystyle \frac{x}{2}+\frac{2y}{3}=-1\qquad\ldots\text{(1)}
\displaystyle x-\frac{y}{3}=3\qquad\ldots\text{(2)}
\displaystyle \text{From equation (2),}
\displaystyle x=3+\frac{y}{3}
\displaystyle \text{Substituting }x=3+\frac{y}{3}\text{ in equation (1),}
\displaystyle \frac{1}{2}\left(3+\frac{y}{3}\right)+\frac{2y}{3}=-1
\displaystyle \Rightarrow \frac{3}{2}+\frac{y}{6}+\frac{2y}{3}=-1
\displaystyle \Rightarrow \frac{y}{6}+\frac{4y}{6}=-1-\frac{3}{2}
\displaystyle \Rightarrow \frac{5y}{6}=-\frac{5}{2}
\displaystyle \Rightarrow y=-3
\displaystyle \text{Substituting }y=-3\text{ in }x=3+\frac{y}{3},
\displaystyle x=3+\frac{-3}{3}=2
\displaystyle \therefore \text{The solution is }x=2,\ y=-3.
\displaystyle \\

\displaystyle \text{(v) The given equations are}
\displaystyle 9-(x-4)=y+7\qquad\ldots\text{(1)}
\displaystyle 2(x+y)=4-3y\qquad\ldots\text{(2)}
\displaystyle \text{Simplifying equation (1),}
\displaystyle 9-x+4=y+7
\displaystyle \Rightarrow x+y=6\qquad\ldots\text{(3)}
\displaystyle \text{Simplifying equation (2),}
\displaystyle 2x+2y=4-3y
\displaystyle \Rightarrow 2x+5y=4\qquad\ldots\text{(4)}
\displaystyle \text{From equation (3),}
\displaystyle x=6-y
\displaystyle \text{Substituting }x=6-y\text{ in equation (4),}
\displaystyle 2(6-y)+5y=4
\displaystyle \Rightarrow 12-2y+5y=4
\displaystyle \Rightarrow 3y=-8
\displaystyle \Rightarrow y=-\frac{8}{3}
\displaystyle \text{Substituting }y=-\frac{8}{3}\text{ in }x=6-y,
\displaystyle x=6-\left(-\frac{8}{3}\right)=\frac{18+8}{3}=\frac{26}{3}
\displaystyle \therefore \text{The solution is }x=\frac{26}{3},\ y=-\frac{8}{3}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following systems of equations using the method of elimination:}
\displaystyle \text{(i) }\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}\text{ and }\frac{3}{x}+\frac{2}{y}=0
\displaystyle \text{(ii) }8v-3u=5uv\text{ and }6v-5u=-2uv
\displaystyle \text{(iii) }\frac{1}{2(2x+3y)}+\frac{12}{7(3x-2y)}=\frac{1}{2}\text{ and }
\displaystyle \frac{7}{2x+3y}+\frac{4}{3x-2y}=2
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle 12u+4v=1\qquad\ldots\text{(1)}
\displaystyle 3u+2v=0\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (2) by }4,
\displaystyle 12u+8v=0\qquad\ldots\text{(3)}
\displaystyle \text{Subtracting equation (3) from equation (1),}
\displaystyle 12u+4v=1
\displaystyle \underline{(-)\hspace{0.5cm}12u+8v=0}
\displaystyle -4v=1
\displaystyle \Rightarrow v=-\frac{1}{4}
\displaystyle \text{Substituting }v=-\frac{1}{4}\text{ in equation (2),}
\displaystyle 3u+2\left(-\frac{1}{4}\right)=0
\displaystyle \Rightarrow 3u=\frac{1}{2}
\displaystyle \Rightarrow u=\frac{1}{6}
\displaystyle \therefore x=\frac{1}{u}=6\text{ and }y=\frac{1}{v}=-4.
\displaystyle \\

\displaystyle \text{(ii) If }u=0,\text{ then }v=0,\text{ and if }v=0,\text{ then }u=0.
\displaystyle \therefore (u,v)=(0,0)\text{ is one solution.}
\displaystyle \text{For }u\ne0,\ v\ne0,\text{ divide each equation by }uv.
\displaystyle \frac{8}{u}-\frac{3}{v}=5\qquad\ldots\text{(1)}
\displaystyle \frac{6}{u}-\frac{5}{v}=-2\qquad\ldots\text{(2)}
\displaystyle \text{Let }\frac{1}{u}=x\text{ and }\frac{1}{v}=y.
\displaystyle \text{Then}
\displaystyle 8x-3y=5\qquad\ldots\text{(3)}
\displaystyle 6x-5y=-2\qquad\ldots\text{(4)}
\displaystyle \text{Multiplying equation (3) by }3\text{ and equation (4) by }4,
\displaystyle 24x-9y=15\qquad\ldots\text{(5)}
\displaystyle 24x-20y=-8\qquad\ldots\text{(6)}
\displaystyle \text{Subtracting equation (6) from equation (5),}
\displaystyle 24x-9y=15
\displaystyle \underline{(-)\hspace{0.5cm}24x-20y=-8}
\displaystyle 11y=23
\displaystyle \Rightarrow y=\frac{23}{11}
\displaystyle \text{Substituting }y=\frac{23}{11}\text{ in equation (3),}
\displaystyle 8x-3\left(\frac{23}{11}\right)=5
\displaystyle \Rightarrow x=\frac{31}{22}
\displaystyle \therefore u=\frac{1}{x}=\frac{22}{31}\text{ and }v=\frac{1}{y}=\frac{11}{23}.
\displaystyle \text{Hence the solutions are }(u,v)=(0,0)\text{ and }\left(\frac{22}{31},\frac{11}{23}\right).
\displaystyle \\

\displaystyle \text{(iii) Let }\frac{1}{2x+3y}=u\text{ and }\frac{1}{3x-2y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle \frac{u}{2}+\frac{12v}{7}=\frac{1}{2}
\displaystyle \Rightarrow 7u+24v=7\qquad\ldots\text{(1)}
\displaystyle 7u+4v=2\qquad\ldots\text{(2)}
\displaystyle \text{Subtracting equation (2) from equation (1),}
\displaystyle 7u+24v=7
\displaystyle \underline{(-)\hspace{0.5cm}7u+4v=2}
\displaystyle 20v=5
\displaystyle \Rightarrow v=\frac{1}{4}
\displaystyle \text{Substituting }v=\frac{1}{4}\text{ in equation (1),}
\displaystyle 7u+24\left(\frac{1}{4}\right)=7
\displaystyle \Rightarrow 7u=1
\displaystyle \Rightarrow u=\frac{1}{7}
\displaystyle \therefore 2x+3y=7\qquad\ldots\text{(3)}
\displaystyle \therefore 3x-2y=4\qquad\ldots\text{(4)}
\displaystyle \text{Multiplying equation (4) by }2\text{ and equation (3) by }3,
\displaystyle 6x+9y=21\qquad\ldots\text{(5)}
\displaystyle 6x-4y=8\qquad\ldots\text{(6)}
\displaystyle \text{Subtracting equation (6) from equation (5),}
\displaystyle 6x+9y=21
\displaystyle \underline{(-)\hspace{0.5cm}6x-4y=8}
\displaystyle 13y=13
\displaystyle \Rightarrow y=1
\displaystyle \text{Substituting }y=1\text{ in equation (5),}
\displaystyle 6x+9=21
\displaystyle \Rightarrow x=2
\displaystyle \therefore \text{The solution is }x=2,\ y=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following systems of equations using the method of elimination:}
\displaystyle \text{(iv) }0.4x+0.3y=1.7\text{ and }0.7x-0.2y=0.8

\displaystyle \text{(v) }\frac{x}{2}+y=0.8\text{ and }\frac{7}{x+\frac{y}{2}}=10

\displaystyle \text{(vi) }\frac{x}{3}+\frac{y}{4}=11\text{ and }\frac{5x}{6}-\frac{y}{3}=-7
\displaystyle \text{Answer:}
\displaystyle \text{(iv) The given equations are}
\displaystyle 0.4x+0.3y=1.7\qquad\ldots\text{(1)}
\displaystyle 0.7x-0.2y=0.8\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }20\text{ and equation (2) by }30,
\displaystyle 8x+6y=34\qquad\ldots\text{(3)}
\displaystyle 21x-6y=24\qquad\ldots\text{(4)}
\displaystyle \text{Adding equations (3) and (4),}
\displaystyle 8x+6y=34
\displaystyle \underline{(+)\hspace{0.5cm}21x-6y=24}
\displaystyle 29x=58
\displaystyle \Rightarrow x=2
\displaystyle \text{Substituting }x=2\text{ in equation (3),}
\displaystyle 8(2)+6y=34
\displaystyle \Rightarrow 16+6y=34
\displaystyle \Rightarrow 6y=18
\displaystyle \Rightarrow y=3
\displaystyle \therefore \text{The solution is }x=2,\ y=3.
\displaystyle \\

\displaystyle \text{(v) The given equations are}
\displaystyle \frac{x}{2}+y=0.8
\displaystyle \Rightarrow x+2y=1.6\qquad\ldots\text{(1)}
\displaystyle \frac{7}{x+\frac{y}{2}}=10
\displaystyle \Rightarrow 7=10\left(x+\frac{y}{2}\right)
\displaystyle \Rightarrow 7=10x+5y
\displaystyle \Rightarrow 2x+y=1.4\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }2,
\displaystyle 2x+4y=3.2\qquad\ldots\text{(3)}
\displaystyle 2x+y=1.4\qquad\ldots\text{(2)}
\displaystyle \text{Subtracting equation (2) from equation (3),}
\displaystyle 2x+4y=3.2
\displaystyle \underline{(-)\hspace{0.5cm}2x+y=1.4}
\displaystyle 3y=1.8
\displaystyle \Rightarrow y=0.6
\displaystyle \text{Substituting }y=0.6\text{ in equation (1),}
\displaystyle x+2(0.6)=1.6
\displaystyle \Rightarrow x=1.6-1.2=0.4
\displaystyle \therefore \text{The solution is }x=0.4,\ y=0.6.
\displaystyle \\

\displaystyle \text{(vi) The given equations are}
\displaystyle \frac{x}{3}+\frac{y}{4}=11
\displaystyle \Rightarrow 4x+3y=132\qquad\ldots\text{(1)}
\displaystyle \frac{5x}{6}-\frac{y}{3}=-7
\displaystyle \Rightarrow 5x-2y=-42\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }2\text{ and equation (2) by }3,
\displaystyle 8x+6y=264\qquad\ldots\text{(3)}
\displaystyle 15x-6y=-126\qquad\ldots\text{(4)}
\displaystyle \text{Adding equations (3) and (4),}
\displaystyle 8x+6y=264
\displaystyle \underline{(+)\hspace{0.5cm}15x-6y=-126}
\displaystyle 23x=138
\displaystyle \Rightarrow x=6
\displaystyle \text{Substituting }x=6\text{ in equation (1),}
\displaystyle 4(6)+3y=132
\displaystyle \Rightarrow 24+3y=132
\displaystyle \Rightarrow 3y=108
\displaystyle \Rightarrow y=36
\displaystyle \therefore \text{The solution is }x=6,\ y=36.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following systems of equations using the method of elimination:}
\displaystyle \text{(vii) }3x-\frac{y+7}{11}+2=10\text{ and }2y+\frac{x+11}{7}=10
\displaystyle \text{(viii) }\frac{1}{2x}-\frac{1}{y}=-1\text{ and }\frac{1}{x}+\frac{1}{2y}=8
\displaystyle \text{(ix) }\frac{6}{x+y}=\frac{7}{x-y}+3\text{ and }\frac{1}{2(x+y)}=\frac{1}{3(x-y)}
\displaystyle \text{Answer:}
\displaystyle \text{(vii) The given equations are}
\displaystyle 3x-\frac{y+7}{11}+2=10
\displaystyle \Rightarrow 33x-y=95\qquad\ldots\text{(1)}
\displaystyle 2y+\frac{x+11}{7}=10
\displaystyle \Rightarrow x+14y=59\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }14,
\displaystyle 462x-14y=1330\qquad\ldots\text{(3)}
\displaystyle x+14y=59\qquad\ldots\text{(2)}
\displaystyle \text{Adding equations (3) and (2),}
\displaystyle 462x-14y=1330
\displaystyle \underline{(+)\hspace{0.5cm}x+14y=59}
\displaystyle 463x=1389
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (1),}
\displaystyle 33(3)-y=95
\displaystyle \Rightarrow 99-y=95
\displaystyle \Rightarrow y=4
\displaystyle \therefore \text{The solution is }x=3,\ y=4.
\displaystyle \\

\displaystyle \text{(viii) Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle \frac{u}{2}-v=-1
\displaystyle \Rightarrow u-2v=-2\qquad\ldots\text{(1)}
\displaystyle u+\frac{v}{2}=8
\displaystyle \Rightarrow 2u+v=16\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }2,
\displaystyle 2u-4v=-4\qquad\ldots\text{(3)}
\displaystyle \text{Subtracting equation (2) from equation (3),}
\displaystyle 2u-4v=-4
\displaystyle \underline{(-)\hspace{0.5cm}2u+v=16}
\displaystyle -5v=-20
\displaystyle \Rightarrow v=4
\displaystyle \text{Substituting }v=4\text{ in equation (1),}
\displaystyle u-2(4)=-2
\displaystyle \Rightarrow u=6
\displaystyle \therefore x=\frac{1}{u}=\frac{1}{6}\text{ and }y=\frac{1}{v}=\frac{1}{4}.
\displaystyle \therefore \text{The solution is }x=\frac{1}{6},\ y=\frac{1}{4}.
\displaystyle \\

\displaystyle \text{(ix) Let }\frac{1}{x+y}=u\text{ and }\frac{1}{x-y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle 6u=7v+3
\displaystyle \Rightarrow 6u-7v=3\qquad\ldots\text{(1)}
\displaystyle \frac{u}{2}=\frac{v}{3}
\displaystyle \Rightarrow 3u-2v=0\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (2) by }2,
\displaystyle 6u-4v=0\qquad\ldots\text{(3)}
\displaystyle \text{Subtracting equation (3) from equation (1),}
\displaystyle 6u-7v=3
\displaystyle \underline{(-)\hspace{0.5cm}6u-4v=0}
\displaystyle -3v=3
\displaystyle \Rightarrow v=-1
\displaystyle \text{Substituting }v=-1\text{ in equation (2),}
\displaystyle 3u-2(-1)=0
\displaystyle \Rightarrow 3u=-2
\displaystyle \Rightarrow u=-\frac{2}{3}
\displaystyle \therefore x+y=\frac{1}{u}=-\frac{3}{2}\qquad\ldots\text{(4)}
\displaystyle \therefore x-y=\frac{1}{v}=-1\qquad\ldots\text{(5)}
\displaystyle \text{Adding equations (4) and (5),}
\displaystyle 2x=-\frac{3}{2}-1=-\frac{5}{2}
\displaystyle \Rightarrow x=-\frac{5}{4}
\displaystyle \text{Substituting }x=-\frac{5}{4}\text{ in equation (5),}
\displaystyle -\frac{5}{4}-y=-1
\displaystyle \Rightarrow -y=\frac{1}{4}
\displaystyle \Rightarrow y=-\frac{1}{4}
\displaystyle \therefore \text{The solution is }x=-\frac{5}{4},\ y=-\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following systems of equations using the method of elimination:}
\displaystyle \text{(x) }\frac{5}{x+y}-\frac{2}{x-y}=-1\text{ and }\frac{15}{x+y}+\frac{7}{x-y}=10
\displaystyle \text{(xi) }\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=-\frac{3}{2}\text{ and}
\displaystyle \frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}
\displaystyle \text{(xii) }\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\text{ and}
\displaystyle \frac{5}{3x+2y}+\frac{1}{3x-2y}=2
\displaystyle \text{Answer:}
\displaystyle \text{(x) Let }\frac{1}{x+y}=u\text{ and }\frac{1}{x-y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle 5u-2v=-1\qquad\ldots\text{(1)}
\displaystyle 15u+7v=10\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }3,
\displaystyle 15u-6v=-3\qquad\ldots\text{(3)}
\displaystyle \text{Subtracting equation (2) from equation (3),}
\displaystyle 15u-6v=-3
\displaystyle \underline{(-)\hspace{0.5cm}15u+7v=10}
\displaystyle -13v=-13
\displaystyle \Rightarrow v=1
\displaystyle \text{Substituting }v=1\text{ in equation (2),}
\displaystyle 15u+7(1)=10
\displaystyle \Rightarrow 15u=3
\displaystyle \Rightarrow u=\frac{1}{5}
\displaystyle \therefore x+y=\frac{1}{u}=5\qquad\ldots\text{(4)}
\displaystyle \therefore x-y=\frac{1}{v}=1\qquad\ldots\text{(5)}
\displaystyle \text{Adding equations (4) and (5),}
\displaystyle 2x=6
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (5),}
\displaystyle 3-y=1
\displaystyle \Rightarrow y=2
\displaystyle \therefore \text{The solution is }x=3,\ y=2.
\displaystyle \\

\displaystyle \text{(xi) Let }\frac{1}{x+2y}=u\text{ and }\frac{1}{3x-2y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle \frac{u}{2}+\frac{5v}{3}=-\frac{3}{2}
\displaystyle \Rightarrow 3u+10v=-9\qquad\ldots\text{(1)}
\displaystyle \frac{5u}{4}-\frac{3v}{5}=\frac{61}{60}
\displaystyle \Rightarrow 75u-36v=61\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }25,
\displaystyle 75u+250v=-225\qquad\ldots\text{(3)}
\displaystyle \text{Subtracting equation (2) from equation (3),}
\displaystyle 75u+250v=-225
\displaystyle \underline{(-)\hspace{0.5cm}75u-36v=61}
\displaystyle 286v=-286
\displaystyle \Rightarrow v=-1
\displaystyle \text{Substituting }v=-1\text{ in equation (1),}
\displaystyle 3u+10(-1)=-9
\displaystyle \Rightarrow 3u=1
\displaystyle \Rightarrow u=\frac{1}{3}
\displaystyle \therefore x+2y=\frac{1}{u}=3\qquad\ldots\text{(4)}
\displaystyle \therefore 3x-2y=\frac{1}{v}=-1\qquad\ldots\text{(5)}
\displaystyle \text{Adding equations (4) and (5),}
\displaystyle 4x=2
\displaystyle \Rightarrow x=\frac{1}{2}
\displaystyle \text{Substituting }x=\frac{1}{2}\text{ in equation (4),}
\displaystyle \frac{1}{2}+2y=3
\displaystyle \Rightarrow 2y=\frac{5}{2}
\displaystyle \Rightarrow y=\frac{5}{4}
\displaystyle \therefore \text{The solution is }x=\frac{1}{2},\ y=\frac{5}{4}.
\displaystyle \\

\displaystyle \text{(xii) Let }\frac{1}{3x+2y}=u\text{ and }\frac{1}{3x-2y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle 2u+3v=\frac{17}{5}
\displaystyle \Rightarrow 10u+15v=17\qquad\ldots\text{(1)}
\displaystyle 5u+v=2\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (2) by }2,
\displaystyle 10u+2v=4\qquad\ldots\text{(3)}
\displaystyle \text{Subtracting equation (3) from equation (1),}
\displaystyle 10u+15v=17
\displaystyle \underline{(-)\hspace{0.5cm}10u+2v=4}
\displaystyle 13v=13
\displaystyle \Rightarrow v=1
\displaystyle \text{Substituting }v=1\text{ in equation (2),}
\displaystyle 5u+1=2
\displaystyle \Rightarrow 5u=1
\displaystyle \Rightarrow u=\frac{1}{5}
\displaystyle \therefore 3x+2y=\frac{1}{u}=5\qquad\ldots\text{(4)}
\displaystyle \therefore 3x-2y=\frac{1}{v}=1\qquad\ldots\text{(5)}
\displaystyle \text{Adding equations (4) and (5),}
\displaystyle 6x=6
\displaystyle \Rightarrow x=1
\displaystyle \text{Substituting }x=1\text{ in equation (4),}
\displaystyle 3(1)+2y=5
\displaystyle \Rightarrow 2y=2
\displaystyle \Rightarrow y=1
\displaystyle \therefore \text{The solution is }x=1,\ y=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following systems of equations using the method of elimination:}
\displaystyle \text{(xiii) }\frac{5}{x+1}-\frac{2}{y-1}=\frac{1}{2}\text{ and}
\displaystyle \frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}
\displaystyle \text{(xiv) }x-y+z=4,\quad x-2y-2z=9\text{ and }2x+y+3z=1
\displaystyle \text{(xv) }x-y+z=4,\quad x+y+z=2\text{ and }2x+y-3z=0
\displaystyle \text{Answer:}
\displaystyle \text{(xiii) Let }\frac{1}{x+1}=u\text{ and }\frac{1}{y-1}=v.
\displaystyle \text{Then the given equations become}
\displaystyle 10u-4v=1\qquad\ldots\text{(1)}
\displaystyle 20u+4v=5\qquad\ldots\text{(2)}
\displaystyle \text{Adding equations (1) and (2),}
\displaystyle 10u-4v=1
\displaystyle \underline{(+)\hspace{0.5cm}20u+4v=5}
\displaystyle 30u=6
\displaystyle \Rightarrow u=\frac{1}{5}
\displaystyle \text{Substituting }u=\frac{1}{5}\text{ in equation (1),}
\displaystyle 10\left(\frac{1}{5}\right)-4v=1
\displaystyle \Rightarrow 2-4v=1
\displaystyle \Rightarrow 4v=1
\displaystyle \Rightarrow v=\frac{1}{4}
\displaystyle \therefore x+1=\frac{1}{u}=5
\displaystyle \Rightarrow x=4
\displaystyle \therefore y-1=\frac{1}{v}=4
\displaystyle \Rightarrow y=5
\displaystyle \therefore \text{The solution is }x=4,\ y=5.
\displaystyle \\

\displaystyle \text{(xiv) The given equations are}
\displaystyle x-y+z=4\qquad\ldots\text{(1)}
\displaystyle x-2y-2z=9\qquad\ldots\text{(2)}
\displaystyle 2x+y+3z=1\qquad\ldots\text{(3)}
\displaystyle \text{From equation (1),}
\displaystyle z=4-x+y
\displaystyle \text{Substituting }z=4-x+y\text{ in equation (2),}
\displaystyle x-2y-2(4-x+y)=9
\displaystyle \Rightarrow x-2y-8+2x-2y=9
\displaystyle \Rightarrow 3x-4y=17\qquad\ldots\text{(4)}
\displaystyle \text{Substituting }z=4-x+y\text{ in equation (3),}
\displaystyle 2x+y+3(4-x+y)=1
\displaystyle \Rightarrow 2x+y+12-3x+3y=1
\displaystyle \Rightarrow x-4y=11\qquad\ldots\text{(5)}
\displaystyle \text{Subtracting equation (5) from equation (4),}
\displaystyle 3x-4y=17
\displaystyle \underline{(-)\hspace{0.5cm}x-4y=11}
\displaystyle 2x=6
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (5),}
\displaystyle 3-4y=11
\displaystyle \Rightarrow -4y=8
\displaystyle \Rightarrow y=-2
\displaystyle \text{Substituting }x=3\text{ and }y=-2\text{ in }z=4-x+y,
\displaystyle z=4-3-2=-1
\displaystyle \therefore \text{The solution is }x=3,\ y=-2,\ z=-1.
\displaystyle \\

\displaystyle \text{(xv) The given equations are}
\displaystyle x-y+z=4\qquad\ldots\text{(1)}
\displaystyle x+y+z=2\qquad\ldots\text{(2)}
\displaystyle 2x+y-3z=0\qquad\ldots\text{(3)}
\displaystyle \text{From equation (1),}
\displaystyle y=x+z-4
\displaystyle \text{Substituting }y=x+z-4\text{ in equation (2),}
\displaystyle x+(x+z-4)+z=2
\displaystyle \Rightarrow 2x+2z=6\qquad\ldots\text{(4)}
\displaystyle \text{Substituting }y=x+z-4\text{ in equation (3),}
\displaystyle 2x+(x+z-4)-3z=0
\displaystyle \Rightarrow 3x-2z=4\qquad\ldots\text{(5)}
\displaystyle \text{Adding equations (4) and (5),}
\displaystyle 2x+2z=6
\displaystyle \underline{(+)\hspace{0.5cm}3x-2z=4}
\displaystyle 5x=10
\displaystyle \Rightarrow x=2
\displaystyle \text{Substituting }x=2\text{ in equation (4),}
\displaystyle 2(2)+2z=6
\displaystyle \Rightarrow 2z=2
\displaystyle \Rightarrow z=1
\displaystyle \text{Substituting }x=2\text{ and }z=1\text{ in }y=x+z-4,
\displaystyle y=2+1-4=-1
\displaystyle \therefore \text{The solution is }x=2,\ y=-1,\ z=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following systems of equations using the method of elimination:}
\displaystyle \text{(xvi) }\frac{10}{x+y}+\frac{2}{x-y}=4\text{ and }\frac{15}{x+y}-\frac{9}{x-y}=-2
\displaystyle \text{(xvii) }\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}\text{ and}
\displaystyle \frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8}
\displaystyle \text{Answer:}
\displaystyle \text{(xvi) Let }\frac{1}{x+y}=u\text{ and }\frac{1}{x-y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle 10u+2v=4\qquad\ldots\text{(1)}
\displaystyle 15u-9v=-2\qquad\ldots\text{(2)}
\displaystyle \text{Multiplying equation (1) by }3\text{ and equation (2) by }2,
\displaystyle 30u+6v=12\qquad\ldots\text{(3)}
\displaystyle 30u-18v=-4\qquad\ldots\text{(4)}
\displaystyle \text{Subtracting equation (4) from equation (3),}
\displaystyle 30u+6v=12
\displaystyle \underline{(-)\hspace{0.5cm}30u-18v=-4}
\displaystyle 24v=16
\displaystyle \Rightarrow v=\frac{2}{3}
\displaystyle \text{Substituting }v=\frac{2}{3}\text{ in equation (1),}
\displaystyle 10u+2\left(\frac{2}{3}\right)=4
\displaystyle \Rightarrow 10u=4-\frac{4}{3}=\frac{8}{3}
\displaystyle \Rightarrow u=\frac{4}{15}
\displaystyle \therefore x+y=\frac{1}{u}=\frac{15}{4}\qquad\ldots\text{(5)}
\displaystyle \therefore x-y=\frac{1}{v}=\frac{3}{2}\qquad\ldots\text{(6)}
\displaystyle \text{Adding equations (5) and (6),}
\displaystyle 2x=\frac{15}{4}+\frac{3}{2}
\displaystyle \Rightarrow 2x=\frac{21}{4}
\displaystyle \Rightarrow x=\frac{21}{8}
\displaystyle \text{Substituting }x=\frac{21}{8}\text{ in equation (6),}
\displaystyle \frac{21}{8}-y=\frac{3}{2}
\displaystyle \Rightarrow y=\frac{21}{8}-\frac{12}{8}=\frac{9}{8}
\displaystyle \therefore \text{The solution is }x=\frac{21}{8},\ y=\frac{9}{8}.
\displaystyle \\

\displaystyle \text{(xvii) Let }\frac{1}{3x+y}=u\text{ and }\frac{1}{3x-y}=v.
\displaystyle \text{Then the given equations become}
\displaystyle u+v=\frac{3}{4}
\displaystyle \Rightarrow 4u+4v=3\qquad\ldots\text{(1)}
\displaystyle \frac{u}{2}-\frac{v}{2}=-\frac{1}{8}
\displaystyle \Rightarrow 4u-4v=-1\qquad\ldots\text{(2)}
\displaystyle \text{Adding equations (1) and (2),}
\displaystyle 4u+4v=3
\displaystyle \underline{(+)\hspace{0.5cm}4u-4v=-1}
\displaystyle 8u=2
\displaystyle \Rightarrow u=\frac{1}{4}
\displaystyle \text{Substituting }u=\frac{1}{4}\text{ in equation (1),}
\displaystyle 4\left(\frac{1}{4}\right)+4v=3
\displaystyle \Rightarrow 1+4v=3
\displaystyle \Rightarrow 4v=2
\displaystyle \Rightarrow v=\frac{1}{2}
\displaystyle \therefore 3x+y=\frac{1}{u}=4\qquad\ldots\text{(3)}
\displaystyle \therefore 3x-y=\frac{1}{v}=2\qquad\ldots\text{(4)}
\displaystyle \text{Adding equations (3) and (4),}
\displaystyle 6x=6
\displaystyle \Rightarrow x=1
\displaystyle \text{Substituting }x=1\text{ in equation (4),}
\displaystyle 3(1)-y=2
\displaystyle \Rightarrow y=1
\displaystyle \therefore \text{The solution is }x=1,\ y=1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve the following systems of equations using the cross multiplication method:}
\displaystyle \text{(i) }ax+by=a-b\text{ and }bx-ay=a+b
\displaystyle \text{(ii) }x+y=a+b\text{ and }ax-by=a^2-b^2
\displaystyle \text{(iii) }ax+by=1\text{ and }bx+ay=\frac{2ab}{a^2+b^2}
\displaystyle \text{Answer:}
\displaystyle \text{(i) The given system of equations may be written as}
\displaystyle ax+by-(a-b)=0
\displaystyle bx-ay-(a+b)=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{b[-(a+b)]-(-a)[-(a-b)]}
\displaystyle =\frac{-y}{a[-(a+b)]-b[-(a-b)]}
\displaystyle =\frac{1}{a(-a)-b(b)}
\displaystyle \Rightarrow \frac{x}{-ab-b^2-a^2+ab}
\displaystyle =\frac{-y}{-a^2-ab+ab-b^2}=\frac{1}{-a^2-b^2}
\displaystyle \Rightarrow \frac{x}{-(a^2+b^2)}
\displaystyle =\frac{-y}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}
\displaystyle \Rightarrow x=1\text{ and }y=-1
\displaystyle \therefore \text{The solution is }x=1,\ y=-1,
\displaystyle \text{provided }a^2+b^2\ne0.
\displaystyle \\

\displaystyle \text{(ii) The given system of equations may be written as}
\displaystyle x+y-(a+b)=0
\displaystyle ax-by-(a^2-b^2)=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{1[-(a^2-b^2)]-(-b)[-(a+b)]}
\displaystyle =\frac{-y}{1[-(a^2-b^2)]-a[-(a+b)]}
\displaystyle =\frac{1}{1(-b)-1(a)}
\displaystyle \Rightarrow \frac{x}{-a^2+b^2-ab-b^2}
\displaystyle =\frac{-y}{-a^2+b^2+a^2+ab}=\frac{1}{-(a+b)}
\displaystyle \Rightarrow \frac{x}{-a(a+b)}
\displaystyle =\frac{-y}{b(a+b)}=\frac{1}{-(a+b)}
\displaystyle \Rightarrow x=a\text{ and }y=b
\displaystyle \therefore \text{The solution is }x=a,\ y=b,
\displaystyle \text{provided }a+b\ne0.
\displaystyle \text{If }a+b=0,\text{ the equations are dependent and have}
\displaystyle \text{infinitely many solutions satisfying }x+y=0.
\displaystyle \\

\displaystyle \text{(iii) The given system of equations may be written as}
\displaystyle ax+by-1=0
\displaystyle bx+ay-\frac{2ab}{a^2+b^2}=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{b\left(-\frac{2ab}{a^2+b^2}\right)-a(-1)}
\displaystyle =\frac{-y}{a\left(-\frac{2ab}{a^2+b^2}\right)-b(-1)}
\displaystyle =\frac{1}{a(a)-b(b)}
\displaystyle \Rightarrow \frac{x}{-\frac{2ab^2}{a^2+b^2}+a}
\displaystyle =\frac{-y}{-\frac{2a^2b}{a^2+b^2}+b}=\frac{1}{a^2-b^2}
\displaystyle \Rightarrow \frac{x}{\frac{-2ab^2+a^3+ab^2}{a^2+b^2}}
\displaystyle =\frac{-y}{\frac{-2a^2b+b^3+a^2b}{a^2+b^2}}
\displaystyle =\frac{1}{a^2-b^2}
\displaystyle \Rightarrow \frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}}
\displaystyle =\frac{-y}{\frac{b(b^2-a^2)}{a^2+b^2}}
\displaystyle =\frac{1}{a^2-b^2}
\displaystyle \Rightarrow x=\frac{a(a^2-b^2)}{a^2+b^2}
\displaystyle \times\frac{1}{a^2-b^2}=\frac{a}{a^2+b^2}
\displaystyle \Rightarrow y=-\frac{b(b^2-a^2)}{a^2+b^2}
\displaystyle \times\frac{1}{a^2-b^2}=\frac{b}{a^2+b^2}
\displaystyle \therefore \text{The solution is }x=\frac{a}{a^2+b^2},
\displaystyle y=\frac{b}{a^2+b^2},\text{ provided }a^2+b^2\ne0
\displaystyle \text{and }a^2\ne b^2.
\displaystyle \text{If }a=b\ne0,\text{ there are infinitely many solutions satisfying}
\displaystyle x+y=\frac{1}{a}.
\displaystyle \text{If }a=-b\ne0,\text{ there are infinitely many solutions satisfying}
\displaystyle x-y=\frac{1}{a}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve the following systems of equations using the cross multiplication method:}
\displaystyle \text{(iv) }\frac{a}{x}-\frac{b}{y}=0\text{ and }\frac{ab^2}{x}+\frac{a^2b}{y}=a^2+b^2,
\displaystyle \text{where }x,y\ne0
\displaystyle \text{(v) }3x+2y+25=0\text{ and }2x+y+10=0
\displaystyle \text{(vi) }ax+by=a^2\text{ and }bx+ay=b^2
\displaystyle \text{Answer:}
\displaystyle \text{(iv) Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{The given system of equations may be written as}
\displaystyle au-bv+0=0
\displaystyle ab^2u+a^2bv-(a^2+b^2)=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{u}{(-b)[-(a^2+b^2)]-(a^2b)(0)}
\displaystyle =\frac{-v}{a[-(a^2+b^2)]-(ab^2)(0)}
\displaystyle =\frac{1}{a(a^2b)-(ab^2)(-b)}
\displaystyle \Rightarrow \frac{u}{b(a^2+b^2)}
\displaystyle =\frac{-v}{-a(a^2+b^2)}
\displaystyle =\frac{1}{ab(a^2+b^2)}
\displaystyle \Rightarrow u=\frac{b(a^2+b^2)}{ab(a^2+b^2)}=\frac{1}{a}
\displaystyle \Rightarrow \frac{1}{x}=\frac{1}{a}
\displaystyle \Rightarrow x=a
\displaystyle \Rightarrow -v=\frac{-a(a^2+b^2)}{ab(a^2+b^2)}=-\frac{1}{b}
\displaystyle \Rightarrow v=\frac{1}{b}
\displaystyle \Rightarrow \frac{1}{y}=\frac{1}{b}
\displaystyle \Rightarrow y=b
\displaystyle \therefore \text{The solution is }x=a,\ y=b,
\displaystyle \text{provided }ab(a^2+b^2)\ne0.
\displaystyle \\

\displaystyle \text{(v) The given system of equations is}
\displaystyle 3x+2y+25=0
\displaystyle 2x+y+10=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{2(10)-1(25)}
\displaystyle =\frac{-y}{3(10)-2(25)}
\displaystyle =\frac{1}{3(1)-2(2)}
\displaystyle \Rightarrow \frac{x}{-5}=\frac{-y}{-20}=\frac{1}{-1}
\displaystyle \Rightarrow x=\frac{-5}{-1}=5
\displaystyle \Rightarrow y=-\left(\frac{-20}{-1}\right)=-20
\displaystyle \therefore \text{The solution is }x=5,\ y=-20.
\displaystyle \\

\displaystyle \text{(vi) The given system of equations may be written as}
\displaystyle ax+by-a^2=0
\displaystyle bx+ay-b^2=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{b(-b^2)-a(-a^2)}
\displaystyle =\frac{-y}{a(-b^2)-b(-a^2)}
\displaystyle =\frac{1}{a(a)-b(b)}
\displaystyle \Rightarrow \frac{x}{a^3-b^3}
\displaystyle =\frac{-y}{a^2b-ab^2}=\frac{1}{a^2-b^2}
\displaystyle \Rightarrow x=\frac{a^3-b^3}{a^2-b^2}
\displaystyle =\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}
\displaystyle =\frac{a^2+ab+b^2}{a+b}
\displaystyle \Rightarrow y=-\frac{a^2b-ab^2}{a^2-b^2}
\displaystyle =-\frac{ab(a-b)}{(a-b)(a+b)}
\displaystyle =-\frac{ab}{a+b}
\displaystyle \therefore \text{The solution is }x=\frac{a^2+ab+b^2}{a+b},
\displaystyle y=-\frac{ab}{a+b},\text{ provided }a^2\ne b^2.
\displaystyle \text{If }a=b\ne0,\text{ there are infinitely many solutions satisfying}
\displaystyle x+y=a.
\displaystyle \text{If }a=-b\ne0,\text{ the system has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve the following systems of equations using the cross multiplication method:}
\displaystyle \text{(vii) }(a+2b)x+(2a-b)y=2\text{ and }(a-2b)x+(2a+b)y=3
\displaystyle \text{(viii) }x\left(a-b+\frac{ab}{a-b}\right)=y\left(a+b-\frac{ab}{a+b}\right)
\displaystyle \text{and }x+y=2a^2
\displaystyle \text{(ix) }bx+cy=a+b\text{ and}
\displaystyle ax\left(\frac{1}{a-b}-\frac{1}{a+b}\right)+cy\left(\frac{1}{b-a}-\frac{1}{b+a}\right)=\frac{2a}{a+b}
\displaystyle \text{Answer:}
\displaystyle \text{(vii) The given system of equations may be written as}
\displaystyle (a+2b)x+(2a-b)y-2=0
\displaystyle (a-2b)x+(2a+b)y-3=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{(2a-b)(-3)-(2a+b)(-2)}
\displaystyle =\frac{-y}{(a+2b)(-3)-(a-2b)(-2)}
\displaystyle =\frac{1}{(a+2b)(2a+b)-(a-2b)(2a-b)}
\displaystyle \Rightarrow \frac{x}{-6a+3b+4a+2b}
\displaystyle =\frac{-y}{-3a-6b+2a-4b}
\displaystyle =\frac{1}{2a^2+5ab+2b^2-(2a^2-5ab+2b^2)}
\displaystyle \Rightarrow \frac{x}{-2a+5b}
\displaystyle =\frac{-y}{-a-10b}=\frac{1}{10ab}
\displaystyle \Rightarrow x=\frac{-2a+5b}{10ab}
\displaystyle \Rightarrow y=\frac{a+10b}{10ab}
\displaystyle \therefore \text{The solution is }x=\frac{-2a+5b}{10ab},
\displaystyle y=\frac{a+10b}{10ab},\text{ provided }ab\ne0.
\displaystyle \\

\displaystyle \text{(viii) First, we simplify the first equation:}
\displaystyle x\left(a-b+\frac{ab}{a-b}\right)
\displaystyle =y\left(a+b-\frac{ab}{a+b}\right)
\displaystyle \Rightarrow x\left(\frac{a^2-ab+b^2}{a-b}\right)
\displaystyle -y\left(\frac{a^2+ab+b^2}{a+b}\right)=0
\displaystyle \Rightarrow x\left(\frac{a^3+b^3}{a^2-b^2}\right)
\displaystyle -y\left(\frac{a^3-b^3}{a^2-b^2}\right)=0
\displaystyle \text{The given system of equations may be written as}
\displaystyle \frac{a^3+b^3}{a^2-b^2}x
\displaystyle -\frac{a^3-b^3}{a^2-b^2}y+0=0
\displaystyle x+y-2a^2=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{\left(-\frac{a^3-b^3}{a^2-b^2}\right)(-2a^2)-0(1)}
\displaystyle =\frac{-y}{\left(\frac{a^3+b^3}{a^2-b^2}\right)(-2a^2)-0(1)}
\displaystyle =\frac{1}{\frac{a^3+b^3}{a^2-b^2}(1)-\left(-\frac{a^3-b^3}{a^2-b^2}\right)(1)}
\displaystyle \Rightarrow \frac{x}{\frac{2a^2(a^3-b^3)}{a^2-b^2}}
\displaystyle =\frac{-y}{-\frac{2a^2(a^3+b^3)}{a^2-b^2}}
\displaystyle =\frac{1}{\frac{2a^3}{a^2-b^2}}
\displaystyle \Rightarrow \frac{x}{a^3-b^3}
\displaystyle =\frac{y}{a^3+b^3}=\frac{1}{a}
\displaystyle \Rightarrow x=\frac{a^3-b^3}{a}
\displaystyle \Rightarrow y=\frac{a^3+b^3}{a}
\displaystyle \therefore \text{The solution is }x=\frac{a^3-b^3}{a},
\displaystyle y=\frac{a^3+b^3}{a},\text{ provided }a\ne0\text{ and }a^2\ne b^2.
\displaystyle \\

\displaystyle \text{(ix) First, we simplify the second equation:}
\displaystyle ax\left(\frac{1}{a-b}-\frac{1}{a+b}\right)
\displaystyle +cy\left(\frac{1}{b-a}-\frac{1}{b+a}\right)=\frac{2a}{a+b}
\displaystyle \Rightarrow ax\left(\frac{2b}{a^2-b^2}\right)
\displaystyle +cy\left(\frac{2a}{b^2-a^2}\right)=\frac{2a}{a+b}
\displaystyle \Rightarrow \frac{2abx}{a^2-b^2}
\displaystyle -\frac{2acy}{a^2-b^2}=\frac{2a}{a+b}
\displaystyle \Rightarrow \frac{bx-cy}{a^2-b^2}=\frac{1}{a+b}
\displaystyle \Rightarrow bx-cy=a-b
\displaystyle \text{The given system of equations may be written as}
\displaystyle bx-cy-(a-b)=0
\displaystyle bx+cy-(a+b)=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{(-c)[-(a+b)]-c[-(a-b)]}
\displaystyle =\frac{-y}{b[-(a+b)]-b[-(a-b)]}
\displaystyle =\frac{1}{b(c)-b(-c)}
\displaystyle \Rightarrow \frac{x}{c(a+b)+c(a-b)}
\displaystyle =\frac{-y}{-b(a+b)+b(a-b)}=\frac{1}{2bc}
\displaystyle \Rightarrow \frac{x}{2ac}=\frac{-y}{-2b^2}=\frac{1}{2bc}
\displaystyle \Rightarrow x=\frac{2ac}{2bc}=\frac{a}{b}
\displaystyle \Rightarrow y=\frac{2b^2}{2bc}=\frac{b}{c}
\displaystyle \therefore \text{The solution is }x=\frac{a}{b},\ y=\frac{b}{c},
\displaystyle \text{provided }abc\ne0\text{ and }a^2\ne b^2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve the following systems of equations using the cross multiplication method:}
\displaystyle \text{(x) }(a-b)x+(a+b)y=2a^2-2b^2\text{ and}
\displaystyle (a+b)(x+y)=4ab
\displaystyle \text{(xi) }\frac{a^2}{x}-\frac{b^2}{y}=0\text{ and}
\displaystyle \frac{a^2b}{x}+\frac{ab^2}{y}=a+b,\text{ where }x,y\ne0
\displaystyle \text{Answer:}
\displaystyle \text{(x) The given system of equations may be written as}
\displaystyle (a-b)x+(a+b)y-2(a^2-b^2)=0
\displaystyle (a+b)x+(a+b)y-4ab=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{x}{(a+b)(-4ab)-(a+b)[-2(a^2-b^2)]}
\displaystyle =\frac{-y}{(a-b)(-4ab)-(a+b)[-2(a^2-b^2)]}
\displaystyle =\frac{1}{(a-b)(a+b)-(a+b)(a+b)}
\displaystyle \Rightarrow \frac{x}{-4ab(a+b)+2(a+b)(a^2-b^2)}
\displaystyle =\frac{-y}{-4ab(a-b)+2(a+b)(a^2-b^2)}
\displaystyle =\frac{1}{-2b(a+b)}
\displaystyle \Rightarrow \frac{x}{2(a+b)(a^2-2ab-b^2)}
\displaystyle =\frac{-y}{2(a-b)(a^2+b^2)}
\displaystyle =\frac{1}{-2b(a+b)}
\displaystyle \Rightarrow x=\frac{2(a+b)(a^2-2ab-b^2)}{-2b(a+b)}
\displaystyle =\frac{2ab+b^2-a^2}{b}
\displaystyle \Rightarrow y=-\frac{2(a-b)(a^2+b^2)}{-2b(a+b)}
\displaystyle =\frac{(a-b)(a^2+b^2)}{b(a+b)}
\displaystyle \therefore \text{The solution is }x=\frac{2ab+b^2-a^2}{b},
\displaystyle y=\frac{(a-b)(a^2+b^2)}{b(a+b)},
\displaystyle \text{provided }b(a+b)\ne0.
\displaystyle \\

\displaystyle \text{(xi) Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{The given system of equations may be written as}
\displaystyle a^2u-b^2v+0=0
\displaystyle a^2bu+ab^2v-(a+b)=0
\displaystyle \text{By cross multiplication,}
\displaystyle \frac{u}{(-b^2)[-(a+b)]-(ab^2)(0)}
\displaystyle =\frac{-v}{a^2[-(a+b)]-(a^2b)(0)}
\displaystyle =\frac{1}{a^2(ab^2)-(-b^2)(a^2b)}
\displaystyle \Rightarrow \frac{u}{b^2(a+b)}
\displaystyle =\frac{-v}{-a^2(a+b)}
\displaystyle =\frac{1}{a^3b^2+a^2b^3}
\displaystyle \Rightarrow \frac{u}{b^2(a+b)}
\displaystyle =\frac{-v}{-a^2(a+b)}
\displaystyle =\frac{1}{a^2b^2(a+b)}
\displaystyle \Rightarrow u=\frac{b^2(a+b)}{a^2b^2(a+b)}=\frac{1}{a^2}
\displaystyle \Rightarrow \frac{1}{x}=\frac{1}{a^2}
\displaystyle \Rightarrow x=a^2
\displaystyle \Rightarrow -v=\frac{-a^2(a+b)}{a^2b^2(a+b)}=-\frac{1}{b^2}
\displaystyle \Rightarrow v=\frac{1}{b^2}
\displaystyle \Rightarrow \frac{1}{y}=\frac{1}{b^2}
\displaystyle \Rightarrow y=b^2
\displaystyle \therefore \text{The solution is }x=a^2,\ y=b^2,
\displaystyle \text{provided }ab(a+b)\ne0.
\displaystyle \\

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