Hi,
I have attached what I am trying to do hopefully it is clear any ideas?
Thanks
$\displaystyle x+8y=4$
$\displaystyle x+10y=6$
Subtract the two equations.
$\displaystyle -2y=-2$
Divide both sides of the equation by -2.
$\displaystyle \frac{-2y}{-2}=\frac{-2}{-2}$
$\displaystyle y=1$
Substitute y = 1 into the first equation and solve for x.
Thanks that really helped but I got stuck again on this one
$\displaystyle 5x-3y=47$ 1
$\displaystyle 7x-3y=67$ 2
So we ignore the 3y as $\displaystyle (-3y) - (-3y)$ is 0.
On the right: $\displaystyle 47-67 = -20$
On the left: $\displaystyle 5x-7x = -2x$
Therfore:
$\displaystyle \frac{-2x}{-20} = 10x$
Now we substitute $\displaystyle x = 10$ into formulae 1, because it has the lowest number on right
$\displaystyle 5 x 10 = 47$
$\displaystyle 50 = 47$
$\displaystyle 47 - 50$
$\displaystyle = -3$
$\displaystyle y = -3$
Although the answer says $\displaystyle y = -1 $
Hope this is clear, and thank you.