How would I solve this using the substitution method? 4p=3q 3p-q=15
4p=3q so p=3q/4 plug this p value in 2nd equation to get p val
if p=3q/4, dont i already have myp value? what about my q value?
now substitute value of p in 4p=3q
you'll get q
4(3q/4)=3q 12q/4=3q 3q=3q now what?
dude substitute the value p=3/4*q in the second equation
I think if you rearrange the first equation so q=4p/3 your substitution will be easier. Not more correct, just easier to calculate.
q=60,p= 45
sorry these were with negative sign
substitutions: in this case, define p as a function of q: p(q) p(q) = 3q/4 3p(q) - q = 15
p=5.5555555555556, q=4p/3
no no p= -45, q=-60
3(-45) -(-60)=15 -135+60=15 -75=15 So its not p=-45, q=-60
4*-45=3*-60 therefore it is the answer
but not if it doesn't agree with BOTH equations. it only agrees with one of them. doesnt it have to be right on both equations?
of course it should i am also wondering how is it possible
\[4p=3q ~and~ 3p-q=15\]So\[q=\frac{4p}{3}~and~3p-q=3p-\frac{4p}{3}=15 \implies 5p=45 \implies p=9 \implies q=12\]\[\implies (p,q)=(9,12)\]
Don't make the easy ones hard.
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