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.

- 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!!!!!!!!*

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestNot 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

\(7(3x-2y) = 2(2x+y) + 21\).(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 !!!

Log in to reply

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

thanks for this.

Log in to reply