Can you substitute this 3x-4y=-1 and x+y=5
x=(4y-1)/3 (4y-1)/3+y=5 (y-1)/3+y=5/4 2y-1=15/4 2y=4 y=2 x=7/3
one has a slope of 3/4 the other a slope of -1; to they do cross at a point .... thats always a good start
3x -4y = -1 x + y = 5 (-3) 3x -4y = - 1 -3x -3y = -15 -------------- -7y = -16 y = 16/7 ... if i did it right
3x -4y = -1 x + y = 5 (-4) 3x -4y = -1 4x +4y = 20 ----------- 7x = 19 x = 19/7 perhaps ?
and by (-4) i mean (+4) :)
then where did I go wrong?
my first guess? typing in the reply box ;)
I need it in algebraic method..
that is in "algebraic" method.
Linear?
3x-4y=-1 x = (4y -1)/3.... I did not use "substitution"; but i used algebraic methods
I need substitution :(
(4y-1)/3+y=5 ; *3 4y -1 +3y = 15 ; combine like terms 7y - 1 = 15 ; +1 7y = 16 ; /7 y = 16/7
Llort: (y-1)/3+y=5/4 this is fictitious :)
if you "need" substitution; then substitute ... the answers are the same regardless
ohhh, thanks, I forgot PEMDAS... or BEDMAS, whatever you call it.
id have picked the simpler equation to determine the substitutes ... but its all good :)
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