সমাধান : \(8 x+5 y-11=0\)
\(\quad\) \(3 x-4 y-10=0\)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{-50-44}=\frac{y}{-33+80}=\frac{1}{-32-15}\)
বা, \(\frac{x}{-94}=\frac{y}{+47}=\frac{1}{-47}\)
\(\therefore x=\frac{-94}{-47}=2, \quad y=\frac{47}{-47}=-1\)
\(\therefore\) নির্ণেয় সমাধান \(x= 2, \quad y=-1\)

সমাধান : \(3 x-4 y-1=0\)
\(4 x-3 y-6=0\)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{24-3}=\frac{y}{-4+18}=\frac{1}{-9+16}\)
বা, \(\frac{x}{21}=\frac{y}{14}=\frac{1}{7}\)
\(\therefore x=\frac{21}{7}=3, \quad y=\frac{14}{7}=2\)
নির্ণেয় সমাধান \(x=3, \quad y=2\)

সমাধান : \(5 x+3 y-11=0\)
\(2 x-7 y+12=0\)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{3 \times 12-(-7) \times(-11)}=\frac{y}{-11 \times 2-5 \times 12}=\frac{1}{5 \times(-7)-3 \times 2}\)
বা, \(\frac{x}{36-77}=\frac{y}{-22-60}=\frac{1}{-35-6}\)
\(\frac{x}{-41}=\frac{y}{-82}=\frac{1}{-41}\)
\(\therefore x=\frac{-41}{-41}=1, \quad y=\frac{-82}{-41}=2\)
\(\therefore\) নির্ণেয় সমাধান \(x=1, y=2\)

সমাধান : \(7 x-3 y-31=0\)
\(9 x-5 y-41=0\)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{123-155}=\frac{y}{-279+287}=\frac{1}{-35+27}\)
বা, \(\frac{x}{-32}=\frac{y}{8}=\frac{1}{-8}\)
\(\therefore x= \frac{-32}{-8}=4, \quad y=\frac{8}{-8} =-1\)
\(\therefore\) নির্ণেয় সমাধান \(x=4, \quad y=-1\)

\(\frac{x}{6}-\frac{y}{3}=\frac{x}{12}-\frac{2 y}{3}=4\)
বা, \(12\left(\frac{x}{6}-\frac{y}{3}\right)=12\left(\frac{x}{12}-\frac{2 y}{3}\right)=12 \times 4\)
বা, \(2 x-4 y=x-8 y=48\)
\(\therefore 2 x-4 y=48\)
বা, \(2 x-4 y-48=0\ldots(i)\)
এবং \(x-8 y=48\)
বা, \(x-8 y-48=0\ldots(ii)\)
(i) ও (ii) নং সমীকরণে বজ্রগুণন পদ্ধতি প্রয়োগ করে পাই,
\(\frac{x}{(-4) \times(-48)-(-8) \times(-48)}=\)
\(\frac{y}{1 \times(-48)-2 \times(-48)}=\frac{1}{2 \times(-8)-1 \times(-4)}\)
বা, \(\frac{x}{192-384}=\frac{y}{-48+96}=\frac{1}{-16+4}\)
বা, \(\frac{x}{-192}=\frac{y}{48}=\frac{1}{-12}\)
\(\therefore \frac{x}{-192}=\frac{1}{-12}\)
এবং \(\frac{y}{48}=\frac{1}{-12}\)
বা, \(x=\frac{-192}{-12}=16\)
বা, \(y=\frac{48}{-12}=-4\)
\(\therefore\) নির্ণেয় সমাধান \(x=16, \mathrm{y}=-4\)

সমাধান : \(\frac{x}{5}+\frac{y}{3}=0\)
বা, \(3 x+5 y=0\)
\(3 x+5 y-0=0\)........(i)
\(\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)
বা, \(15 x-20 y-9=0\) ........(ii)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{5 \times (-9)-(-20) \times 0}=\frac{y}{15 \times 0-3 \times (-9)}=\frac{1}{3 \times (-20)-15 \times 5 }\)
বা, \(\frac{x}{-45}=\frac{y}{27}=\frac{1}{-60-75}\)
\(\therefore x=\frac{-45}{-135}=\frac{1}{3}, y=\frac{27}{-135}=-\frac{1}{5}\)
নির্ণেয় সমাধান \(x=\frac{1}{3}, y=-\frac{1}{5}\)

\(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8\)
বা, \(4 x+8+7 y-7 x=56 x-224\)
বা, \(56 x+3 x-7 y-224-8=0\)
বা, \(59 x-7 y-232=0\)...........(i)
\(\frac{2 y-3 x}{3}+2 y=3 x+4\)
বা, \(2 y-3 x+6 y=9 x+12\)
বা, \(12 x-8 y+12=0\)
বা, \(3 x-2 y+3=0\)...........(ii)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{-21-464}=\frac{y}{-696-177}=\frac{1}{-118+21}\)
বা, \(\frac{x}{-485}=\frac{y}{-873}=\frac{1}{-97}\)
\(\therefore x=\frac{485}{97}=5 \quad y=\frac{873}{97}=9\)
নির্ণেয় সমাধান : \(x=5, \quad y=9\)

সমাধান : \(x+5 y=36\)
\(x+5 y-36=0\).........(i)
\(\frac{x+y}{x-y}=\frac{5}{3}\)
বা, \(3 x+3 y=5 x-5 y\)
বা, \(2 x-8 y=0\)
\(x-4 y-0=0\) .............(ii)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{0-144}=\frac{y}{-36-0}=\frac{1}{-4-5}\)
বা, \(\frac{x}{-144}=\frac{y}{-36}=\frac{1}{-9}\)
\(\therefore x= \frac{-144}{-9}=16, \quad y=\frac{-36}{-9}=4\)
\(\therefore\) নির্ণেয় সমাধান : \(x=16, \quad y=4\)

সমাধান : \(13 x-12 y+15=0\)
\(8 x-7 y+0=0\)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{0+105}=\frac{y}{120-0}=\frac{1}{-91+96}\)
বা, \(\frac{x}{105}=\frac{y}{120}=\frac{1}{5}\)
\(\therefore x=\frac{105}{5}=21, \quad y=\frac{120}{5}=24\)
\(\therefore\) নির্ণেয় সমাধান : \(x=21, \quad y=24\)

সমাধান : \(x+y-2 b=0\)
\(x-y-2 a=0\)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{-2 a-2 b}=\frac{y}{-2 b+2 a}=\frac{1}{-1-1}\)
বা, \(x=\frac{-2(b+a)}{-2}\)
\(\therefore x=a+b\)
বা, \(y=\frac{-2(b-a)}{-2}\)
\(\therefore y=b-a\)
\(\therefore\) নির্ণেয় সমাধান \(x=a+b, y=b-a\)

সমাধান : \(x-y-2 a=0\)
\(a x+b y-\left(a^{2}+b^{2}\right)=0\)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{a^{2}+b^{2}+2 a b}=\frac{y}{-2 a^{2}+a^{2}+b^{2}}=\frac{1}{b+a}\)
বা, \(\frac{x}{(a+b)^{2}}=\frac{y}{b^{2}-a^{2}}=\frac{1}{b+a}\)
\(\therefore x= \frac{(a+b)^{2}}{a+b}=a+b, \quad y=\frac {(b-a)(b+a)}{b-a}=b-a\)
\(\therefore\) নির্ণেয় সমাধান \(x=a+b, y=b-a\)

সমাধান : \(\frac{x}{a}+\frac{y}{b}=2\)
বা, \(b x+a y-2 a b=0\)..........(i)
\(a x-b y-\left(a^{2}-b^{2}\right)=0\) ..........(ii)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{-a\left(a^{2}-b^{2}\right)-2 a b^{2}}=\frac{y}{-2 a^{2} b+b\left(a^{2}-b^{2}\right)}=\frac{1}{-b^{2}-a^{2}}\)
বা, \(\frac{x}{-a\left(a^{2}-b^{2}+2 b^{2}\right)}=\frac{y}{b\left(-2 a^{2}+a^{2}-b^{2}\right)}=\frac{1}{-\left(b^{2}+a^{2}\right)}\)
বা, \(\frac{x}{-a\left(a^{2}+b^{2}\right)}=\frac{y}{-b\left(a^{2}+b^{2}\right)}=\frac{1}{-\left(b^{2}+a^{2}\right)}\)
বা, \(\frac{x}{-a\left(a^{2}+b^{2}\right)}=\frac{y}{-b\left(a^{2}+b^{2}\right)}=\frac{1}{-\left(b^{2}+a^{2}\right)}\)
\(\therefore x=\frac{-a\left(a^{2}+b^{2}\right)}{-\left(b^{2}+a^{2}\right)}= a, \quad y=\frac{-b\left(a^{2}+b^{2}\right)}{-\left(b^{2}+a^{2}\right)}= b\)
\(\therefore\) নির্ণেয় সমাধান \(x=a, \quad y=b\)

সমাধান : \(a x+b y=1\),
\(a x+b y-1=0\)..........(i)
\(b x+a y=\frac{2 a b}{a^{2}+b^{2}}\)
বা, \(b\left(a^{2}+b^{2}\right) x+a\left(a^{2}+b^{2}\right) y=2 a b\)
বা, \(b\left(a^{2}+b^{2}\right) x+a\left(a^{2}+b^{2}\right) y-2 a b=0\)...........(ii)
বজ্রগুণন পদ্ধতিতে পাই,
\(\frac{x}{b \times(-2 a b)-a\left(a^{2}+b^{2}\right)(-1)} =\frac{y}{b\left(a^{2}+b^{2}\right)(-1)-a(-2 a b)} =\frac{1}{a \times a\left(a^{2}+b^{2}\right)-b \times b\left(a^{2}+b^{2}\right)}\)
বা, \(\begin{aligned} \frac{x}{-2 a b^{2}+a^{3}+a b^{2}}= & \frac{y}{-a^{2} b-b^{3}+2 a^{2} b} \\ & =\frac{1}{\left(a^{2}+b^{2}\right)\left(a^{2}-b^{2}\right)}\end{aligned}\)
বা, \(\frac{x}{a\left(a^{2}-b^{2}\right)}=\frac{y}{b\left(a^{2}-b^{2}\right)}=\frac{1}{\left(a^{2}+b^{2}\right)\left(a^{2}-b^{2}\right)}\)
বা, \(\frac{x}{a}=\frac{y}{b}=\frac{1}{a^{2}+b^{2}}\)
\(\therefore \frac{x}{a}=\frac{1}{a^{2}+b^{2}}\)
বা, \(x=\frac{a}{a^{2}+b^{2}}\)
এবং, \(\frac{y}{b}=\frac{1}{a^{2}+b^{2}}\)
বা, \(y=\frac{b}{a^{2}+b^{2}}\)
নির্ণেয় সমাধান \(x=\frac{a}{a^{2}+b^{2}}, \quad y=\frac{b}{a^{2}+b^{2}}\)