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

Question 1:  $x+y=12$$x-y=2$

$x+y=12$    … … … … …i)

$x-y=2$    … … … … …ii)

From ii)

$x=y+2$

Substituting in i)

$y +2+y=12$

$\Rightarrow$ $2y=10$

$\Rightarrow$ $y=5$

Therefore $x=5+2=7$

Hence $x=7, y=5$

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

$5x+3y=24$    … … … … …i)

$3x-y=20$    … … … … …ii)

From ii)

$y=3x-20$

Substituting in i)

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

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

$\Rightarrow$ $x=6$

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

Hence $x=6, y= -2$

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

$x-3y=2$    … … … … …i)

$2x+7y=30$    … … … … …ii)

From i)

$x=3y+2$

Substituting in ii)

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

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

$\Rightarrow$ $y=2$

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

Hence $x=8, y=2$

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Question 4: $x+4y=5$$4x+y=50$

$x+4y=5$    … … … … …i)

$4x+y=50$    … … … … …ii)

From i)

$x=5-4y$

Substituting in ii)

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

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

$\Rightarrow$ $y= -2$

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

Hence $x=13, y= -2$

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

$2x+3y=6$    … … … … …i)

$3x+5y=15$    … … … … …ii)

From i)

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

Substituting in ii)

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

$\Rightarrow$  $9-4.5y+5y=15$

$\Rightarrow$ $y=12$

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

Hence $x= -15, y=12$

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

$5x-7y=9$    … … … … …i)

$2x+5y=12$    … … … … …ii)

From i)

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

Substituting in ii)

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

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

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

$\Rightarrow$ $y=2$

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

Hence $x=1, y=2$

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

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

$x+2y=39$    … … … … …i)

$2x-3y=1$    … … … … …ii)

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

$2x+4y=78$

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

$7y=77$

$\Rightarrow$ $y=11$

Substituting in i)

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

Hence $x=17 , y=11$

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

$3x-y=5$    … … … … …i)

$5x-2y=4$    … … … … …ii)

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

$6x-2y=10$

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

$x=6$

$\Rightarrow$ $x=6$

Substituting in i)

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

Hence $x=6 , y=13$

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Question 9:  $14x-3y=54$$21x-8y=95$

$14x-3y=54$    … … … … …i)

$21x-8y=95$    … … … … …ii)

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

$42x-9y=162$

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

$7y=-28$

$\Rightarrow$ $y = -4$

Substituting in i)

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

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

Hence $x=3, y=-4$

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

$7x-6y=37$    … … … … …i)

$5x+4y=43$    … … … … …ii)

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

$35x-30y=185$

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

$-58y=-116$

$\Rightarrow$ y=2

Substituting in i)

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

Hence $x=7, y=2$

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

$10x+3y=36$    … … … … …i)

$15x-14y=17$    … … … … …ii)

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

$30x+9y=108$

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

$37y=74$

$\Rightarrow$ $y=2$

Substituting in i)

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

Hence $x=3, y=2$

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

$4x+3=14$    … … … … …i)

$9x-5y=55$    … … … … …ii)

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

$36x+27y=126$

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

$47y= -94$

$\Rightarrow$ $y=-2$

Substituting in i)

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

Hence $x=5, y= -2$

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

$4x-3y=11$    … … … … …i)

$2x-5y= -5$    … … … … …ii)

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

$4x- 3y=11$

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

$7y=21$

$y=3$

Substituting in i)

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

Hence $x=5, y=3$

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Solve the following simultaneous equations:

Question 14:  $11x-8y=46$$2x+7y=-17$

$11x-8y=46$    … … … … …i)

$2x+7y=-17$    … … … … …ii)

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

$22x-16y=92$

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

$-93y=279$

$\Rightarrow$ $y= -3$

Substituting back in i)

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

Hence $x=2, y= -3$

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Question 15:  $5x-3y=13$ , $3x-2y=5$

$5x-3y=13$    … … … … …i)

$3x-2y=5$    … … … … …ii)

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

$15x-9y=39$

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

$y=14$

Substituting back in i)

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

Hence $x=11, y=14$

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Question 16:  $5a+4b=22$$4a-5b=23$

$5a+4b=22$    … … … … …i)

$4a-5b=23$    … … … … …ii)

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

$20a+16b=88$

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

$-9b= -27$

$\Rightarrow$ $b=3$

Substituting in i)

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

Hence $a=2, b=3$

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

$8x+3y=0$    … … … … …i)

$3x+5(y+3)= -16$    … … … … …ii)

Simplifying ii)

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

From i)

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

Substituting in iii)

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

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

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

$\Rightarrow$ $y= -8$

Therefore

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

Hence $x=3, y= -8$

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

$8y-5z=7$    … … … … …i)

$3y=4(z-2)$    … … … … …ii)

Simplify ii)

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

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

$24y-15z=21$

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

$17z=85$

$\Rightarrow$ $z=5$

Substituting in i)

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

Hence $z=5, y=4$

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

$2(a-3)+3(b-5)= 0$    … … … … …i)

$5(a-1)+4(b-4)= 0$    … … … … …ii)

Simplify i) and ii)

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

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

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

$10a+15b=105$

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

$7b=63$

$\Rightarrow$ $b=9$

Substituting in iii)

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

Hence $a= -3, b=9$

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

$4(3x-y)= 9x+5$    … … … … …i)

$3(2x+3y)=13 (x+y-5)$    … … … … …ii)

Simplifying i) and ii)

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

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

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

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

$10x=70$

$\Rightarrow$ $x=7$

Substitute in iii)

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

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

Hence $x=7, y=4$

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

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

$4x-3y=22$    … … … … …ii)

Simplifying i) multiply by 6

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

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

$12x-9y=66$

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

$-y= -6$

$\Rightarrow$ $y=6$

Substituting in ii)

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

Hence $x=10, y=6$

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

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

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

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

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

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

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

$8x-6y=0$

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

$-15y= -60$

$\Rightarrow$ $y=4$

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

Hence $x=3, \ y=4$

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

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

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

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

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

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

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

$6x-15y=54$

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

$5y= -10$

$\Rightarrow$ $y= -2$

Substituting in iii)

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

Hence $x=4, y= -2$

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

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

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

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

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

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

$9a-4b= -17$

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

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

$9a-4b=-17$

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

$41a = -41$

$a= -1$

Substituting in iv)

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

Hence $a= -1, b=2$

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

$23x+31y=7$    … … … … …i)

$31x+23y=47$    … … … … …ii)

$54x+54y=54$

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

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

$23x+31y=7$

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

$8y= -16$

$\Rightarrow$ $y= -2$

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

Hence $x=3, y= -2$

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

$97x-78y=59$    … … … … …i)

$78x-97y=116$    … … … … …ii)

$175x-175y=175$

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

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

$97x-78y=59$

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

$19y= -38$

$y= -2$

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

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

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

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

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

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

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

Substituting in i)

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

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

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

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

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

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

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

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

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

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

$\Rightarrow$ $y=5$

Substituting

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

$\Rightarrow$ $x=2$

Hence $x=2, y=5$

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

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

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

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

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

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

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

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

Substituting

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

$\Rightarrow$ $x=12$

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

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

$\frac{6}{x}$ $-4y=9$    … … … … …i)

$\frac{4}{x}$ $- y=1$    … … … … …ii)

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

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

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

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

$\Rightarrow$ $x =-2$

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

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

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

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

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

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

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

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

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

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

$\Rightarrow$ $x =1$

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

$\Rightarrow$ $y=5$

Hence $x=1, y=5$

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

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

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

Simplify i) and ii)

$9x+6=2y+3$

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

$8x+8=3y-2$

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

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

$27x-6y=-9$

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

$11x-11$

$\Rightarrow$ $x=1$

Substituting

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

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

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

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

Simplify i) and ii)

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

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

$-11x-5y= -2$

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

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

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

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

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

$33x+15y=6$

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

$-29y=118$

$\Rightarrow$ $y= -4$

Substituting in iii)

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

Hence $x=2, y= -4$