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# Solve the following simultaneous equations. (1) 3a + 5b = 26; a + 5b = 22 (2) x + 7y = 10; 3x – 2y = 7 (3) 2x – 3y = 9; 2x + y = 13 (4) 5m – 3n = 19; m – 6n = –7 (5) 5x + 2y = –3; x + 5y = 4 (6) $\frac{1}{3}x+y=\frac{10}{3};2x+\frac{1}{4}y=\frac{11}{4}$ (7) 99x + 101y = 499; 101x + 99y = 501 (8) 49x – 57y = 172; 57x – 49y = 252

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## (1) 3a + 5b = 26; .....(I) a + 5b = 22 .....(II) Subtracting (II) from (I) 2a = 4 $⇒$ a = 2 Putting the value of a = 2 in (II) 5b = 22 $-$ 2 = 20 $⇒$b = $\frac{20}{5}=4$ Thus, a = 2 and b = 4. (2) x + 7y = 10; .....(I) 3x – 2y = 7 .....(II) Multiplying (I) with 3 3x + 21y = 30; .....(III) 3x – 2y = 7 .....(IV) Subtracting (IV) from (III) we get 23y = 23 $⇒$y = 1 Putting the value of y in (IV) we get 3x – 2 = 7 $⇒$3x = 7 + 2 = 9 $⇒$3x = 9 $⇒$x = 3 Thus, (x, y) = (3, 1) (3) 2x – 3y = 9 .....(I) 2x + y = 13 .....(II) Subtracting (II) from (I) we get – 3y − y = 9 − 13 $⇒-4y=-4\phantom{\rule{0ex}{0ex}}⇒y=1$ Putting this value in (I) we get $2x-3\left(1\right)=9\phantom{\rule{0ex}{0ex}}⇒2x=9+3=12\phantom{\rule{0ex}{0ex}}⇒x=\frac{12}{2}=6$ Thus, (x, y) = (6, 1) (4) 5m – 3n = 19 .....(I) m – 6n = –7 .....(II) Multiplying (I) with 2 we get 10m – 6n = 38 .....(III) m – 6n = –7 .....(IV) Subtracting (IV) from (III) we get $10m-m-6n-\left(-6n\right)=38-\left(-7\right)\phantom{\rule{0ex}{0ex}}⇒9m=45\phantom{\rule{0ex}{0ex}}⇒m=\frac{45}{9}=5$ Putting the value of m = 5 in (II) we get $5-6n=-7\phantom{\rule{0ex}{0ex}}⇒-6n=-7-5\phantom{\rule{0ex}{0ex}}⇒-6n=-12\phantom{\rule{0ex}{0ex}}⇒n=\frac{-12}{-6}=2$ Thus, (m, n) = (5, 2). (5) 5x + 2y = –3 .....(I) x + 5y = 4 .....(II) Multiply (II) with 5 we get 5x + 25y = 20 .....(III) Subtracting (III) from (I) we get $5x-5x+2y-25y=-3-20\phantom{\rule{0ex}{0ex}}⇒-23y=-23\phantom{\rule{0ex}{0ex}}⇒y=\frac{-23}{-23}=1$ Putting the value of y = 1 in (II) we get $x+5\left(1\right)=4\phantom{\rule{0ex}{0ex}}⇒x+5=4\phantom{\rule{0ex}{0ex}}⇒x=4-5=-1$ Thus, (x, y) = (−1, 1) (6) $\frac{1}{3}x+y=\frac{10}{3}.....\left(\mathrm{I}\right)\phantom{\rule{0ex}{0ex}}2x+\frac{1}{4}y=\frac{11}{4}.....\left(\mathrm{II}\right)$ Multiply (I) with 3 and (II) with 4 $x+3y=10.....\left(\mathrm{III}\right)\phantom{\rule{0ex}{0ex}}8x+y=11.....\left(\mathrm{IV}\right)$ Multiply (IV) with 3 24x + 3y = 33 .....(V) Subtracting (V) from (III) $x-24x+3y-3y=10-33\phantom{\rule{0ex}{0ex}}⇒-23x=-23\phantom{\rule{0ex}{0ex}}⇒x=1$ Putting the value of x = 1 in (III) $1+3y=10\phantom{\rule{0ex}{0ex}}⇒3y=10-1=9\phantom{\rule{0ex}{0ex}}⇒y=\frac{9}{3}=3$ Thus, (x, y) = (1, 3) (7) 99x + 101y = 499 .....(I) 101x + 99y = 501 .....(II) Adding (I) and (II) $200x+200y=1000\phantom{\rule{0ex}{0ex}}⇒x+y=5.....\left(\mathrm{III}\right)$ Subtracting (II) from (I) $99x-101x+101y-99y=499-501\phantom{\rule{0ex}{0ex}}⇒-2x+2y=-2\phantom{\rule{0ex}{0ex}}⇒-x+y=-1.....\left(\mathrm{IV}\right)$ Adding (III) and (IV) $x+y=5\phantom{\rule{0ex}{0ex}}-x+y=-1\phantom{\rule{0ex}{0ex}}⇒2y=4\phantom{\rule{0ex}{0ex}}⇒y=2$ Putting the value of y = 2 in (III) we get $x+2=5\phantom{\rule{0ex}{0ex}}⇒x=5-2=3$ Thus, (x, y) = (3, 2) (8) 49x – 57y = 172 .....(I) 57x – 49y = 252 .....(II) Adding (I) and (II) $49x+57x-57y-49y=172+252\phantom{\rule{0ex}{0ex}}⇒106x-106y=424\phantom{\rule{0ex}{0ex}}⇒x-y=4.....\left(\mathrm{III}\right)$ Subtracting (II) from (I) we have $49x-57y-57y-\left(-49y\right)=252-172\phantom{\rule{0ex}{0ex}}⇒-8x-8y=-80\phantom{\rule{0ex}{0ex}}⇒-x-y=-10\phantom{\rule{0ex}{0ex}}⇒x+y=10.....\left(\mathrm{IV}\right)$ Adding (III) and (IV) $x-y=4\phantom{\rule{0ex}{0ex}}x+y=10\phantom{\rule{0ex}{0ex}}⇒2x=14\phantom{\rule{0ex}{0ex}}⇒x=7$ Putting the value of x = 7 in (IV) we get $7+y=10\phantom{\rule{0ex}{0ex}}⇒y=10-7\phantom{\rule{0ex}{0ex}}⇒y=3$ Thus, (x, y) = (7, 3).  Suggest Corrections  5      Similar questions
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