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help!!!!!!

eh guys, i'm here again today. This time around it's simultaneous equation.......

give the solution if you know you can solve this two simultaneous equation.

  1. find \(x\) and \(y\) in the equation:

\( \frac{3x-2y}{2} = \frac{2x+y}{7} + \frac{3}{2} \)

\( 7 - \frac{2x-y}{6} = x + \frac{y}{4} \)..

ii. find \(x\) and \(y\) in the equation:

\( \frac{3x+2}{4} - \frac{x+2y}{2} = \frac{x-3}{12} \)

\( \frac{2y+1}{5} + \frac{x-3y}{4} = \frac{3x+1}{10} \)

so solve if you can!!!!!!!!

Note by Samuel Ayinde
2 years, 9 months ago

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Not very difficult ... In first case, multiplying 1st eqn by 14, and 2nd equation by 12, you get two simultaneous equations, solve by eliminating any 1 variable !

In 2nd case, multiply 1st eqn by 12, 2nd by 20 and again you get everything as integer co-efficients of x and y.

I will solve both here as I have time :P

  1. \(7(3x-2y) = 2(2x+y) + 21\).(multiplying both sides by 14)...i.e. \(17x-8y=21\)

    , and \(84 -2(2x-y) = 12x +3y\)...(multiplying both sides by 12), i.e. \(16x +y =84\), and has solution \((x,y) = \biggl(\dfrac{693}{145} , \dfrac{1092}{145}\biggr)\) ... (get this by putting value of \(y\) as \(84-16x\) in 1st equation.

In the second case,

\( 3(3x+2) - 6(x+2y) = x-3\) .. (multiplying both sides by 12) ... i.e. \(2x -12y = -9\)

and \(4(2y+1)+5(x-3y) = 2(3x+1)\) .... (multiplying both sides by 20)... i.e. \( x +7y=2\) or \(2x+14y=4\) ... subtract 1st from 2nd to get \(26y=13\), thus the solution \((x,y) = (\dfrac{1}{2} , \dfrac{-3}{2})\)

That's it !!! Aditya Raut · 2 years, 9 months ago

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@Aditya Raut That's what i did but i was stuck somewhere. i'll do it again.

thanks for this. Samuel Ayinde · 2 years, 9 months ago

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