Solve each pair of the equations given below, using substitution method:

Question 1: $\displaystyle x+y=12$ , $\displaystyle x-y=2$

$\displaystyle x+y=12$ … … … … …i)

$\displaystyle x-y=2$ … … … … …ii)

From ii)

$\displaystyle x=y+2$

Substituting in i)

$\displaystyle y +2+y=12$

$\displaystyle \Rightarrow 2y=10$

$\displaystyle \Rightarrow y=5$

Therefore $\displaystyle x=5+2=7$

Hence $\displaystyle x=7, y=5$

$\displaystyle \\$

Question 2: $\displaystyle 5x+3y=24$, $\displaystyle 3x-y=20$

$\displaystyle 5x+3y=24$ … … … … …i)

$\displaystyle 3x-y=20$ … … … … …ii)

From ii)

$\displaystyle y=3x-20$

Substituting in i)

$\displaystyle 5x+3(3x-20)= 24$

$\displaystyle \Rightarrow 14x-60=24$

$\displaystyle \Rightarrow x=6$

Therefore $\displaystyle y=3\times 6-20= -2$

Hence $\displaystyle x=6, y= -2$

$\displaystyle \\$

Question 3: $\displaystyle x-3y=2$ , $\displaystyle 2x+7y=30$

$\displaystyle x-3y=2$ … … … … …i)

$\displaystyle 2x+7y=30$ … … … … …ii)

From i)

$\displaystyle x=3y+2$

Substituting in ii)

$\displaystyle 2(3y+2)+ 7y=30$

$\displaystyle \Rightarrow 13y+4=30$

$\displaystyle \Rightarrow y=2$

Therefore $\displaystyle x=3\times 2+2=8$

Hence $\displaystyle x=8, y=2$

$\displaystyle \\$

Question 4: $\displaystyle x+4y=5$, $\displaystyle 4x+y=50$

$\displaystyle x+4y=5$ … … … … …i)

$\displaystyle 4x+y=50$ … … … … …ii)

From i)

$\displaystyle x=5-4y$

Substituting in ii)

$\displaystyle 4(5-4y)+ y=50$

$\displaystyle \Rightarrow 20-25y=50$

$\displaystyle \Rightarrow y= -2$

Therefore $\displaystyle x=5-4\times (-2)= 13$

Hence $\displaystyle x=13, y= -2$

$\displaystyle \\$

Question 5: $\displaystyle 2x+3y=6$, $\displaystyle 3x+5y=15$

$\displaystyle 2x+3y=6$ … … … … …i)

$\displaystyle 3x+5y=15$ … … … … …ii)

From i)

$\displaystyle x=3- \frac{3}{2} y$

Substituting in ii)

$\displaystyle 3(3- \frac{3}{2} y )+ 5y=15$

$\displaystyle \Rightarrow 9-4.5y+5y=15$

$\displaystyle \Rightarrow y=12$

Therefore $\displaystyle x=3- \frac{3}{2} \times 12=3-18=-15$

Hence $\displaystyle x= -15, y=12$

$\displaystyle \\$

Question 6: $\displaystyle 5x-7y=9$ , $\displaystyle 2x+5y=12$

$\displaystyle 5x-7y=9$ … … … … …i)

$\displaystyle 2x+5y=12$ … … … … …ii)

From i)

$\displaystyle x= \frac{7}{5} y- \frac{9}{5}$

Substituting in ii)

$\displaystyle 2( \frac{7}{5} y- \frac{9}{5} )+ 5y=12$

$\displaystyle \Rightarrow ( \frac{14}{5} +5 )y- \frac{18}{5} =12$

$\displaystyle \Rightarrow \frac{39}{5} y= \frac{78}{5}$

$\displaystyle \Rightarrow y=2$

$\displaystyle x= \frac{7}{5} \times 2- \frac{9}{5} = \frac{5}{5} =1$

Hence $\displaystyle x=1, y=2$

$\displaystyle \\$

Solve each pair of the equations given below, using elimination method:

Question 7: $\displaystyle x+2y=39$ , $\displaystyle 2x-3y=1$

$\displaystyle x+2y=39$ … … … … …i)

$\displaystyle 2x-3y=1$ … … … … …ii)

Multiply i) by 2 and then subtract ii) from i)

$\displaystyle 2x+4y=78$

$\displaystyle (-) \underline{2x-3y=1}$

$\displaystyle 7y=77$

$\displaystyle \Rightarrow y=11$

Substituting in i)

$\displaystyle x=39-2\times 11=17$

Hence $\displaystyle x=17 , y=11$

$\displaystyle \\$

Question 8: $\displaystyle 3x-y=5$ , $\displaystyle 5x-2y=4$

$\displaystyle 3x-y=5$ … … … … …i)

$\displaystyle 5x-2y=4$ … … … … …ii)

Multiply i) by 2 and then subtract ii) from i)

$\displaystyle 6x-2y=10$

$\displaystyle (-) \underline{5x-2y=4}$

$\displaystyle x=6$

$\displaystyle \Rightarrow x=6$

Substituting in i)

$\displaystyle y=3 (6) - 5 =13$

Hence $\displaystyle x=6 , y=13$

$\displaystyle \\$

Question 9: $\displaystyle 14x-3y=54$, $\displaystyle 21x-8y=95$

$\displaystyle 14x-3y=54$ … … … … …i)

$\displaystyle 21x-8y=95$ … … … … …ii)

Multiply i) by 3 and ii) by 2 and subtract ii) from i)

$\displaystyle 42x-9y=162$

$\displaystyle (-) \underline{42x-16y=190}$

$\displaystyle 7y=-28$

$\displaystyle \Rightarrow y = -4$

Substituting in i)

$\displaystyle 14x=3(-4)+54$

$\displaystyle x= \frac{42}{14} =3$

Hence $\displaystyle x=3, y=-4$

$\displaystyle \\$

Question 10: $\displaystyle 7x-6y=37$ , $\displaystyle 5x+4y=43$

$\displaystyle 7x-6y=37$ … … … … …i)

$\displaystyle 5x+4y=43$ … … … … …ii)

Multiply i) by 5 and ii) by 7 and subtract ii) from i)

$\displaystyle 35x-30y=185$

$\displaystyle (-) \underline{35x+28y=301}$

$\displaystyle -58y=-116$

$\displaystyle \Rightarrow$ y=2

Substituting in i)

$\displaystyle x= \frac{6\times 2+37}{7} =7$

Hence $\displaystyle x=7, y=2$

$\displaystyle \\$

Question 11: $\displaystyle 10x+3y=36$ , $\displaystyle 15x-14y=17$

$\displaystyle 10x+3y=36$ … … … … …i)

$\displaystyle 15x-14y=17$ … … … … …ii)

Multiply i) by 3 and ii) by 2 and subtract ii) from i)

$\displaystyle 30x+9y=108$

$\displaystyle (-) \underline{30x-28y=34}$

$\displaystyle 37y=74$

$\displaystyle \Rightarrow y=2$

Substituting in i)

$\displaystyle x= \frac{(36-3\times 2)}{10} =3$

Hence $\displaystyle x=3, y=2$

$\displaystyle \\$

Question 12: $\displaystyle 4x+3=14$ , $\displaystyle 9x-5y=55$

$\displaystyle 4x+3=14$ … … … … …i)

$\displaystyle 9x-5y=55$ … … … … …ii)

Multiply i) by 9 and ii) by 4 and subtract ii) from i)

$\displaystyle 36x+27y=126$

$\displaystyle (-) \underline{36x-20y=220}$

$\displaystyle 47y= -94$

$\displaystyle \Rightarrow y=-2$

Substituting in i)

$\displaystyle x= \frac{((14-3(-2))}{4} =5$

Hence $\displaystyle x=5, y= -2$

$\displaystyle \\$

Question 13: $\displaystyle 4x-3y=11$ , $\displaystyle 2x-5y= -5$

$\displaystyle 4x-3y=11$ … … … … …i)

$\displaystyle 2x-5y= -5$ … … … … …ii)

Multiply ii) by 2 and subtract ii) from i)

$\displaystyle 4x- 3y=11$

$\displaystyle (-) \underline{ 4x-10y=-10}$

$\displaystyle 7y=21$

$\displaystyle y=3$

Substituting in i)

$\displaystyle x= \frac{(11+3(3))}{4} =5$

Hence $\displaystyle x=5, y=3$

$\displaystyle \\$

Solve the following simultaneous equations:

Question 14: $\displaystyle 11x-8y=46$ , $\displaystyle 2x+7y=-17$

$\displaystyle 11x-8y=46$ … … … … …i)

$\displaystyle 2x+7y=-17$ … … … … …ii)

Multiply i) by 2 and ii) by 11 and subtract ii) from i)

$\displaystyle 22x-16y=92$

$\displaystyle (-) \underline{22x+77y=187}$

$\displaystyle -93y=279$

$\displaystyle \Rightarrow y= -3$

Substituting back in i)

$\displaystyle x= \frac{(8(-3)+46)}{11} =2$

Hence $\displaystyle x=2, y= -3$

$\displaystyle \\$

Question 15: $\displaystyle 5x-3y=13$ , $\displaystyle 3x-2y=5$

$\displaystyle 5x-3y=13$ … … … … …i)

$\displaystyle 3x-2y=5$ … … … … …ii)

Multiply i) by 3 and ii) by 5 and subtract ii) from i)

$\displaystyle 15x-9y=39$

$\displaystyle (-) \underline{15x-10y=25}$

$\displaystyle y=14$

Substituting back in i)

$\displaystyle x= \frac{(3(14)+13)}{5} =11$

Hence $\displaystyle x=11, y=14$

$\displaystyle \\$

Question 16: $\displaystyle 5a+4b=22$ , $\displaystyle 4a-5b=23$

$\displaystyle 5a+4b=22$ … … … … …i)

$\displaystyle 4a-5b=23$ … … … … …ii)

Multiply i) by 4 and ii) by 5 and subtract ii) from i)

$\displaystyle 20a+16b=88$

$\displaystyle (-) \underline{20a+25b=115}$

$\displaystyle -9b= -27$

$\displaystyle \Rightarrow b=3$

Substituting in i)

$\displaystyle a= \frac{(22-4(3))}{5} =2$

Hence $\displaystyle a=2, b=3$

$\displaystyle \\$

Question 17: $\displaystyle 8x+3y=0$ , $\displaystyle 3x+5(y+3)= -16$

$\displaystyle 8x+3y=0$ … … … … …i)

$\displaystyle 3x+5(y+3)= -16$ … … … … …ii)

Simplifying ii)

$\displaystyle 3x+5y= -31 ... ... ... ... ...iii)$

From i)

$\displaystyle x=- \frac{3}{8} y$

Substituting in iii)

$\displaystyle 3(- \frac{3}{8} y)+5y= -31$

$\displaystyle (5- \frac{9}{8} )y=-31$

$\displaystyle \frac{31}{8} y= -31$

$\displaystyle \Rightarrow y= -8$

Therefore

$\displaystyle x= - \frac{3}{8} \times (-8)= 3$

Hence $\displaystyle x=3, y= -8$

$\displaystyle \\$

Question 18: $\displaystyle 8y-5z=7$ , $\displaystyle 3y=4(z-2)$

$\displaystyle 8y-5z=7$ … … … … …i)

$\displaystyle 3y=4(z-2)$ … … … … …ii)

Simplify ii)

$\displaystyle 3x-4z= -8 ... ... ... ... ...iii)$

Multiply i) by 3 and iii) by 8 and subtract iii) from i)

$\displaystyle 24y-15z=21$

$\displaystyle (-) \underline{y-32z=-64}$

$\displaystyle 17z=85$

$\displaystyle \Rightarrow z=5$

Substituting in i)

$\displaystyle y= \frac{(5(5)+7)}{8} =4$

Hence $\displaystyle z=5, y=4$

$\displaystyle \\$

Question 19: $\displaystyle 2(a-3)+3(b-5)= 0$ , $\displaystyle 5(a-1)+4(b-4)= 0$

$\displaystyle 2(a-3)+3(b-5)= 0$ … … … … …i)

$\displaystyle 5(a-1)+4(b-4)= 0$ … … … … …ii)

Simplify i) and ii)

$\displaystyle 2a+3b=21 ... ... ... ... ...iii)$

$\displaystyle 5a+4b=21 ... ... ... ... ...iv)$

Multiplying iii) by 5 and iv) by 2 and subtract iv) from iii)

$\displaystyle 10a+15b=105$

$\displaystyle (-) \underline{10a+8b=42}$

$\displaystyle 7b=63$

$\displaystyle \Rightarrow b=9$

Substituting in iii)

$\displaystyle a= \frac{(21-3(9))}{2} = -3$

Hence $\displaystyle a= -3, b=9$

$\displaystyle \\$

Question 20: $\displaystyle 4(3x-y)= 9x+5$ , $\displaystyle 3(2x+3y)=13 (x+y-5)$

$\displaystyle 4(3x-y)= 9x+5$ … … … … …i)

$\displaystyle 3(2x+3y)=13 (x+y-5)$ … … … … …ii)

Simplifying i) and ii)

$\displaystyle 12x-4y=9x+5$

$\displaystyle 3x-4y=5 ... ... ... ... ...iii)$

$\displaystyle 6x+9y=13x+13y-65$

$\displaystyle 7x+4y=65 ... ... ... ... ...iv)$

$\displaystyle 10x=70$

$\displaystyle \Rightarrow x=7$

Substitute in iii)

$\displaystyle 4y=3x-5=21-5$

$\displaystyle \Rightarrow y= \frac{16}{4} =4$

Hence $\displaystyle x=7, y=4$

$\displaystyle \\$

Question 21: $\displaystyle \frac{x}{2} - \frac{y}{3} =3$ , $\displaystyle 4x-3y=22$

$\displaystyle \frac{x}{2} - \frac{y}{3} =3$ … … … … …i)

$\displaystyle 4x-3y=22$ … … … … …ii)

Simplifying i) multiply by 6

$\displaystyle 3x-2y=18 ... ... ... ... ...iii)$

Multiply ii) by 3 and iii) by 4 and subtract iii) from ii)

$\displaystyle 12x-9y=66$

$\displaystyle (-) \underline{12x-8y=72}$

$\displaystyle -y= -6$

$\displaystyle \Rightarrow y=6$

Substituting in ii)

$\displaystyle x= \frac{(3\times 6+22)}{4} =10$

Hence $\displaystyle x=10, y=6$

$\displaystyle \\$

Question 22: $\displaystyle \frac{x}{3} - \frac{y}{4} =0$ , $\displaystyle \frac{2x}{3} + \frac{3y}{4} =5$

$\displaystyle \frac{x}{3} - \frac{y}{4} =0$ … … … … …i)

$\displaystyle \frac{2x}{3} + \frac{3y}{4} =5$ … … … … …ii)

Simplify i) and ii) by multiplying i) by 12 and ii) also by 12

$\displaystyle 4x-3y=0 ... ... ... ... ...iii)$

$\displaystyle 8x+9y=60 ... ... ... ... ...iv)$

Multiply iii) by 2 and subtract iv) from iii)

$\displaystyle 8x-6y=0$

$\displaystyle (-) \underline{8x+9y=60}$

$\displaystyle -15y= -60$

$\displaystyle \Rightarrow y=4$

$\displaystyle x=6\times \frac{4}{8} =3$

Hence $\displaystyle x=3, \ y=4$

$\displaystyle \\$

Question 23: $\displaystyle \frac{x}{3} - \frac{5y}{6} =3$ , $\displaystyle \frac{3x}{4} - \frac{5y}{2} =8$

$\displaystyle \frac{x}{3} - \frac{5y}{6} =3$ … … … … …i)

$\displaystyle \frac{3x}{4} - \frac{5y}{2} =8$ … … … … …ii)

Simplify i) and ii) by multiplying i) by 6 and ii) by 4

$\displaystyle 2x-5y=18 ... ... ... ... ...iii)$

$\displaystyle 3x-10=32 ... ... ... ... ...iv)$

Now multiply iii) by 3 and iv) by 2 and subtract iv) from iii)

$\displaystyle 6x-15y=54$

$\displaystyle (-) \underline{6x-20y=64}$

$\displaystyle 5y= -10$

$\displaystyle \Rightarrow y= -2$

Substituting in iii)

$\displaystyle x= \frac{(5(-2)+18)}{2} =4$

Hence $\displaystyle x=4, y= -2$

$\displaystyle \\$

Question 24: $\displaystyle \frac{(3a+5)}{4} = \frac{(2b-1)}{6}$ , $\displaystyle \frac{4a}{3} + \frac{b}{6} = -1$

$\displaystyle \frac{(3a+5)}{4} = \frac{(2b-1)}{6}$ … … … … …i)

$\displaystyle \frac{4a}{3} + \frac{b}{6} = -1$ … … … … …ii)

Simplify i) and ii), multiply i) by 12, and ii) by 6

$\displaystyle 3(3a+5)= 2(2b-1)$

$\displaystyle 9a+15=4b-2 ... ... ... ... ...iii)$

$\displaystyle 9a-4b= -17$

$\displaystyle 8a+b= -6 ... ... ... ... ...iv)$

Multiply iv) by 4 and add iii) and iv)

$\displaystyle 9a-4b=-17$

$\displaystyle (+) \underline{32a+4b=-24}$

$\displaystyle 41a = -41$

$\displaystyle a= -1$

Substituting in iv)

$\displaystyle b= -8(-1)-6=2$

Hence $\displaystyle a= -1, b=2$

$\displaystyle \\$

Question 25: $\displaystyle 23x+31y=7$ , $\displaystyle 31x+23y=47$

$\displaystyle 23x+31y=7$ … … … … …i)

$\displaystyle 31x+23y=47$ … … … … …ii)

$\displaystyle 54x+54y=54$

$\displaystyle \Rightarrow x+y=1 ... ... ... ... ...iii)$

Now multiply iii) by 23 and subtract iii) from i)

$\displaystyle 23x+31y=7$

$\displaystyle (-) \underline{23x+23y=23}$

$\displaystyle 8y= -16$

$\displaystyle \Rightarrow y= -2$

$\displaystyle x=1-y=1-(-2)=3$

Hence $\displaystyle x=3, y= -2$

$\displaystyle \\$

Question 26: $\displaystyle 97x-78y=59$ , $\displaystyle 78x-97y=116$

$\displaystyle 97x-78y=59$ … … … … …i)

$\displaystyle 78x-97y=116$ … … … … …ii)

$\displaystyle 175x-175y=175$

$\displaystyle \Rightarrow x-y=1 ... ... ... ... ...iii)$

Now multiply iii) by 97 and subtract iii) from i)

$\displaystyle 97x-78y=59$

$\displaystyle (-) \underline{97x-97y=97}$

$\displaystyle 19y= -38$

$\displaystyle y= -2$

Substitute $\displaystyle x=y+1= -2+1=-1$

Hence $\displaystyle x= -1, y= -2$

$\displaystyle \\$

Question 27: $\displaystyle \frac{1}{x} + \frac{1}{y} =7$ , $\displaystyle \frac{1}{x} - \frac{1}{y} =1$

$\displaystyle \frac{1}{x} + \frac{1}{y} =7$ … … … … …i)

$\displaystyle \frac{1}{x} - \frac{1}{y} =1$ … … … … …ii)

$\displaystyle \frac{2}{x} =8$

$\displaystyle x= \frac{1}{4}$

Substituting in i)

$\displaystyle \frac{1}{y} =7- \frac{1}{\frac{1}{4}} =7-4=3$

$\displaystyle \Rightarrow y= \frac{1}{3}$

Hence, $\displaystyle x= \frac{1}{4} \ and\ y= \frac{1}{3}$

$\displaystyle \\$

Question 28: $\displaystyle \frac{2}{x} + \frac{10}{y} =3$ , $\displaystyle \frac{8}{x} - \frac{15}{y} =1$

$\displaystyle \frac{2}{x} + \frac{10}{y} =3$ … … … … …i)

$\displaystyle \frac{8}{x} - \frac{15}{y} =1$ … … … … …ii)

Multiply i) by 4 and subtract ii) from i)

$\displaystyle \frac{8}{x} + \frac{40}{y} =12$

$\displaystyle (-) \underline{ \frac{8}{x}- \frac{15}{y}=1}$

$\displaystyle \frac{55}{y} =11$

$\displaystyle \Rightarrow y=5$

Substituting

$\displaystyle \frac{8}{x} =1+ \frac{15}{y} =4$

$\displaystyle \Rightarrow x=2$

Hence $\displaystyle x=2, y=5$

$\displaystyle \\$

Question 29: $\displaystyle 2x+ \frac{3}{y} =20$ , $\displaystyle 4x- \frac{9}{y} =10$

$\displaystyle 2x+ \frac{3}{y} =20$ … … … … …i)

$\displaystyle 4x- \frac{9}{y} =10$ … … … … …ii)

Multiply i) by 2 & then subtract ii) from i)

$\displaystyle 4x+ \frac{6}{y} =40$

$\displaystyle (-) \underline{4x- \frac{9}{y}=10}$

$\displaystyle \frac{15}{y} =30$

$\displaystyle \Rightarrow y= \frac{1}{2}$

Substituting

$\displaystyle 2x=20- \frac{3}{\frac{1}{2}} = 20-6=24$

$\displaystyle \Rightarrow x=12$

Hence $\displaystyle x=12, \ and\ y= \frac{1}{2}$

$\displaystyle \\$

Question 30: $\displaystyle \frac{6}{x} -4y=9$ , $\displaystyle \frac{4}{x} - y=1$

$\displaystyle \frac{6}{x} -4y=9$ … … … … …i)

$\displaystyle \frac{4}{x} - y=1$ … … … … …ii)

Multiply ii) by 4 & then subtract ii) from i)

$\displaystyle \frac{6}{x} - 4y=9$

$\displaystyle (-) \underline{\frac{16}{x}- 4y=4}$

$\displaystyle - \frac{10}{x} =5$

$\displaystyle \Rightarrow x =-2$

$\displaystyle y= \frac{4}{x} - 1= -2-1= -3$

Hence $\displaystyle x =-2, y= -3$

$\displaystyle \\$

Question 31: $\displaystyle \frac{3}{2x} - \frac{5}{3y} = \frac{7}{6}$ , $\displaystyle \frac{4}{5x} + \frac{1}{y} =1$

$\displaystyle \frac{3}{2x} - \frac{5}{3y} = \frac{7}{6}$ … … … … …i)

$\displaystyle \frac{4}{5x} + \frac{1}{y} =1$ … … … … …ii)

Multiply ii) by $\displaystyle \frac{5}{3}$ and add i) & ii)

$\displaystyle \frac{3}{2x} - \frac{5}{3y} = \frac{7}{6}$

$\displaystyle (+) \underline{ \frac{20}{15x}+ \frac{5}{3y}= \frac{5}{3} }$

$\displaystyle ( \frac{3}{2} + \frac{4}{3} )( \frac{1}{x} )=( \frac{7}{6} + \frac{5}{3} ) )$

$\displaystyle \frac{17}{6} ( \frac{1}{x} )= \frac{51}{18}$

$\displaystyle \Rightarrow x =1$

Substituting $\displaystyle \frac{1}{y} =1-( \frac{4}{5} )=1/5$

$\displaystyle \Rightarrow y=5$

Hence $\displaystyle x=1, y=5$

$\displaystyle \\$

Question 32: $\displaystyle \frac{3x+2}{2y+3} =\frac{1}{8}$ , $\displaystyle \frac{x+1}{3y-2} =\frac{1}{8}$

$\displaystyle \frac{3x+2}{2y+3} =\frac{1}{8}$ … … … … …i)

$\displaystyle \frac{x+1}{3y-2} =\frac{1}{8}$ … … … … …ii)

Simplify i) and ii)

$\displaystyle 9x+6=2y+3$

$\displaystyle \Rightarrow 9x-2y=-3 ... ... ... ... ...iii)$

$\displaystyle 8x+8=3y-2$

$\displaystyle \Rightarrow 8x-3y= -10 ... ... ... ... ...iv)$

Multiply iii) by 3 and iv) by 2 and subtract iv) from iii)

$\displaystyle 27x-6y=-9$

$\displaystyle (-) \underline{16x-6y= -20}$

$\displaystyle 11x-11$

$\displaystyle \Rightarrow x=1$

Substituting

$\displaystyle y= \frac{(9x+3)}{2} = \frac{12}{2} =6$

$\displaystyle \\$

Question 33: $\displaystyle \frac{(2x+1)}{5} - \frac{(3x-y)}{2} =y$ , $\displaystyle \frac{(3x+2)}{2} + \frac{(2-y)}{3} =x-y$

$\displaystyle \frac{(2x+1)}{5} - \frac{(3x-y)}{2} =y$ … … … … …i)

$\displaystyle \frac{(3x+2)}{2} + \frac{(2-y)}{3} =x-y$ … … … … …ii)

Simplify i) and ii)

$\displaystyle 2(2x+1)-5(3x-y)=10y$

$\displaystyle 4x+2-15x+5y=10y$

$\displaystyle -11x-5y= -2$

$\displaystyle \Rightarrow 11x+5y=2 ... ... ... ... ...iii)$

$\displaystyle 3(3x+2)+2(2-y)=6(x-y)$

$\displaystyle 9x+6+4-2y=6x-6y$

$\displaystyle 3x+4y=-10 ... ... ... ... ...iv)$

Multiply ii) by 3 and iv) by 11 and then subtract iv) from ii)

$\displaystyle 33x+15y=6$

$\displaystyle (-) \underline{33x+4y= -110}$

$\displaystyle -29y=118$

$\displaystyle \Rightarrow y= -4$

Substituting in iii)

$\displaystyle x= \frac{2-5(-4)}{11} = \frac{22}{11} =2$

Hence $\displaystyle x=2, y= -4$