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.

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!!!!!!!!

#Algebra #Sam

Easy Math Editor

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Comments

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

$7(3x-2y) = 2(2x+y) + 21$ .(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 - 6 years, 11 months ago

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

thanks for this.

samuel ayinde - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$