# 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
4 years 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 !!!

- 4 years ago

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

thanks for this.

- 4 years ago