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