solving simultaneously...?

**suckatmaths** · February 24th 2009, 12:02 AM

"Solve Simultaneously."

"2x+3y=8
3x+2y=7"

"5x-4y=27
4x-5y=18"

I know I have to add/subtract somewhere...But not sure where...

Can someone help me, please?

**ADARSH** · February 24th 2009, 12:23 AM

Originally Posted by suckatmaths

"Solve Simultaneously."

"2x+3y=8
3x+2y=7"

"5x-4y=27
4x-5y=18"

I know I have to add/subtract somewhere...But not sure where...

Can someone help me, please?

-We are gonna get the value of x from first equation

-We will put that in second equation

-This gives an wequation in y

-Get value of y

-And thus the value of x

1st one:

2x+3y = 8

2x = 8 - 3y

Hence

$x= \frac{(8-3y)}{2}..................(1)$

$3x+2y =7<br />$

$3x = 7 -2y$

Put the value of x as in 1

$3 \frac{(8-3y)}{2} = 7 -2y$

Hence
$3(8-3y) = 2(7-2y)$

$24 - 9 y = 14 - 4y$

Hence

$-5y = - 10$

Thus
$y = \frac{10}{5}= ~2$

And
$x = \frac{8-3y}{2} = \frac{ 8 -\frac{3\times 10}{5}}{2} =1$

2nd One ;

$5x - 4y = 27$

$5x = 27 + 4y$

$x =\frac{27+4y}{5} .................1$

$4x - 5y = 18$

Put the value of x from 1 here

$\frac{4( 27 +4y)}{5} - 5y = 18$

$4(27+4y) = 5(5y+ 18)$

$108 + 16y = 25y +90$

Hence
$9y = 18$

Thus $y= 2$

$<br /> x = \frac{(27+4y)}{5} = 35/5=7 <br />$

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