Solve the given simultaneous equations:

Question 1.

$x+y=12 ... ... ... ... ...i)$

$x-y=2 ... ... ... ... ...ii)$

From ii)

$x=y+2$

Substituting in i)

$y +2+y=12$

$2y=10$

$y=5$

Therefore    $x=5+2=7$

Hence    $x=7, y=5$

$\\$

Question 2.

$5x+3y=24 ... ... ... ... ...i)$

$3x-y=20 ... ... ... ... ...ii)$

From ii)

$y=3x-20$

Substituting in i)

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

$14x-60=24$

$x=6$

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

Hence  $x=6, y= -2$

$\\$

Question 3.

$x-3y=2 ... ... ... ... ...i)$

$2x+7y=30 ... ... ... ... ...ii)$

From i)

$x=3y+2$

Substituting in ii)

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

$13y+4=30$

$y=2$

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

Hence $x=8, y=2$

$\\$

Question 4.

$x+4y=5 ... ... ... ... ...i)$

$4x+y=50 ... ... ... ... ...ii)$

From i)

$x=5-4y$

Substituting in ii)

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

$20-25y=50$

or $y= -2$

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

Hence $x=13, y= -2$

$\\$

Question 5.

$2x+3y=6 ... ... ... ... ...i)$

$3x+5y=15 ... ... ... ... ...ii)$

From i)

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

Substituting in ii)

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

$9-4.5y+5y=15$

or $y=12$

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

Hence $x= -15, y=12$

$\\$

Question 6.

$5x-7y=9 ... ... ... ... ...i)$

$2x+5y=12 ... ... ... ... ...ii)$

From i)

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

Substituting in ii)

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

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

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

or $y=2$

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

Hence $x=1, y=2$

$\\$

Question 7.

$x+2y=39 ... ... ... ... ...i)$

$2x-3y=1 ... ... ... ... ...ii)$

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

$2x+4y=78$

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

$7y=77$

or $y=11$

Substituting in i)

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

Hence $x=17 , y=11$

$\\$

Question 8.

$14x-3y=54 ... ... ... ... ...i)$

$21x-8y=95 ... ... ... ... ...ii)$

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

$42x-9y=162$

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

$7y=-28$

Substituting in i)

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

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

Hence $x=3, y=-4$

$\\$

Question 9.

$7x-6y=37 ... ... ... ... ...i)$

$5x+4y=43 ... ... ... ... ...ii)$

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

$35x-30y=185$

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

$-58y=-116$

or y=2

Substituting in i)

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

Hence $x=7, y=2$

$\\$

Question 10.

$10x+3y=36 ... ... ... ... ...i)$

$15x-14y=17 ... ... ... ... ...ii)$

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

$30x+9y=108$

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

$37y=74$

or $y=2$

Substituting in i)

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

Hence $x=3, y=2$

$\\$

Question 11.

$4x+3=14 ... ... ... ... ...i)$

$9x-5y=55 ... ... ... ... ...ii)$

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

$36x+27y=126$

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

$47y= -94$

$y=-2$

Substituting in i)

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

Hence $x=5, y= -2$

$\\$

Question 12.

$4x-3y=11 ... ... ... ... ...i)$

$2x-5y= -5 ... ... ... ... ...ii)$

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 $x=5, y=3$

$\\$

Question 13.

$11x-8y=46 ... ... ... ... ...i)$

$2x+7y=-17 ... ... ... ... ...ii)$

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

$22x-16y=92$

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

$-93y=279$

or $y= -3$

Substituting back in i)

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

Hence $x=2, y= -3$

$\\$

Question 14.

$5x-3y=13 ... ... ... ... ...i)$

$3x-2y=5 ... ... ... ... ...ii)$

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 $x=11, y=14$

$\\$

Question 15.

$5a+4b=22 ... ... ... ... ...i)$

$4a-5b=23 ... ... ... ... ...ii)$

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

$20a+16b=88$

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

$-9b= -27$

or $b=3$

Substituting in i)

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

Hence  $a=2, b=3$

$\\$

Question 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)

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

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

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

or $y= -8$

Therefore

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

Hence $x=3, y= -8$

$\\$

Question 17.

$8y-5z=7 ... ... ... ... ...i)$

$3y=4(z-2) ... ... ... ... ...ii)$

Simplify ii)

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

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

$24y-15z=21$

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

$17z=85$

or $z=5$

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

Hence $z=5, y=4$

$\\$

Question 18.

$2(a-3)+3(b-5)= 0 ... ... ... ... ...i)$

$5(a-1)+4(b-4)= 0 ... ... ... ... ...ii)$

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)

$10a+15b=105$

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

$7b=63$

or $b=9$

Substituting in iii)

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

Hence $a= -3, b=9$

$\\$

Question 19.

$4(3x-y)= 9x+5 ... ... ... ... ...i)$

$3(2x+3y)=13 (x+y-5) ... ... ... ... ...ii)$

Simplifying i) and ii)

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

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

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

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

Add iii) and  iv)

$10x=70$

$x=7$

Substitute in iii)

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

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

Hence $x=7, y=4$

$\\$

Question 20.

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

$4x-3y=22 ... ... ... ... ...ii)$

Simplifying i) multiply by 6

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

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

$12x-9y=66$

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

$-y= -6$

or $y=6$

Substituting in ii)

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

Hence  $x=10, y=6$

$\\$

Question 21.

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

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

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)

$8x-6y=0$

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

$-15y= -60$

or $y=4$

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

Hence $x=3, \ y=4$

$\\$

Question 22.

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

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

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)

$6x-15y=54$

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

$5y= -10$

or $y= -2$

Substituting in iii)

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

Hence $x=4, y= -2$

$\\$

Question 23.

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

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

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 $a= -1, b=2$

$\\$

Question 24.

$23x+31y=7 ... ... ... ... ...i)$

$31x+23y=47 ... ... ... ... ...ii)$

Add i) and ii)

$54x+54y=54$

or $x+y=1 ... ... ... ... ...iii)$

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

$23x+31y=7$

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

$8y= -16$

or $y= -2$

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

Hence $x=3, y= -2$

$\\$

Question 25.

$97x-78y=59 ... ... ... ... ...i)$

$78x-97y=116 ... ... ... ... ...ii)$

Add i) & ii)

$175x-175y=175$

or $x-y=1 ... ... ... ... ...iii)$

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 $x= -1, y= -2$

$\\$

Question 26.

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

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

Add i) & ii)

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

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

Substituting in i)

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

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

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

$\\$

Question 27.

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

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

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

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

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

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

or $y=5$

Substituting

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

or $x=2$

Hence $x=2, y=5$

$\\$

Question 28.

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

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

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

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

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

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

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

Substituting

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

or $x=12$

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

$\\$

Question 29.

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

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

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

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

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

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

or $x =-2$

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

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

$\\$

Question 30.

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

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

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

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

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

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

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

$x =1$

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

$y=5$

Hence $x=1, y=5$

$\\$

Question 31.

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

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

Simplify i) and ii)

$9x+6=2y+3$

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

$8x+8=3y-2$

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

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

$27x-6y=-9$

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

$11x-11$

$x=1$

Substituting

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

$\\$

Question 32.

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

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

Simplify i) and ii)

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

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

$-11x-5y= -2$

or $11x+5y=2 ... ... ... ... ...iii)$

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

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

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

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

$33x+15y=6$

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

$-29y=118$

or $y= -4$

Substituting in iii)

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

Hence $x=2, y= -4$

Advertisements