# solving simultaneously...?

• Feb 24th 2009, 12:02 AM
suckatmaths
solving simultaneously...?
"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...

• Feb 24th 2009, 12:23 AM
Quote:

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...

-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

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

$\displaystyle 3x+2y =7$

$\displaystyle 3x = 7 -2y$

Put the value of x as in 1

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

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

$\displaystyle 24 - 9 y = 14 - 4y$

Hence

$\displaystyle -5y = - 10$

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

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

2nd One ;

$\displaystyle 5x - 4y = 27$

$\displaystyle 5x = 27 + 4y$

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

$\displaystyle 4x - 5y = 18$

Put the value of x from 1 here

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

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

$\displaystyle 108 + 16y = 25y +90$

Hence
$\displaystyle 9y = 18$

Thus $\displaystyle y= 2$

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