http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf go to link and to question 17 help me solve it.
http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf go to link and to question 17 help me solve it.
$\displaystyle 3x+6y=30$
$\displaystyle x+6y=20 $
See how there are 2 6y's? You can subtract the two equations from each other to get rid of the 6ys/
$\displaystyle 2x=10 $
$\displaystyle x=5 $ Now you have x=5 just substitute it back into any equation to find y.
What I did was $\displaystyle 3x-x = 2x $ and $\displaystyle 30-20=10$ The 6y cancelled each other.
3x + 6y = 30... (1)
x + 6y = 20.....(2)
Multiply (2) by - on both sides
=> -(x+6y)=-20
=> -x - 6y = -20....(2)
Then, on solving (1) & (2)-
3x+6y = 30
-x - 6y = -20
+6y and -6y gets cancelled...
so.. we get 3x= 30..(1)
-x= -20..(2)
(1) + (2)
3x-x = 30-20
=> 2x=10
=> x=5
Subsituting x in any of the two equations.-
3x+6y = 30
=> 3 x 5 + 6y = 30
=> 15+6y = 30
=> 6y= 30-15
=> y= 15/6
=> y = 2.5