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Mathematics 14 Online
OpenStudy (anonymous):

How would I solve this using the substitution method? 4p=3q 3p-q=15

OpenStudy (anonymous):

4p=3q so p=3q/4 plug this p value in 2nd equation to get p val

OpenStudy (anonymous):

if p=3q/4, dont i already have myp value? what about my q value?

OpenStudy (ghazi):

now substitute value of p in 4p=3q

OpenStudy (ghazi):

you'll get q

OpenStudy (anonymous):

4(3q/4)=3q 12q/4=3q 3q=3q now what?

OpenStudy (ghazi):

dude substitute the value p=3/4*q in the second equation

OpenStudy (anonymous):

I think if you rearrange the first equation so q=4p/3 your substitution will be easier. Not more correct, just easier to calculate.

OpenStudy (ghazi):

q=60,p= 45

OpenStudy (ghazi):

sorry these were with negative sign

OpenStudy (amistre64):

substitutions: in this case, define p as a function of q: p(q) p(q) = 3q/4 3p(q) - q = 15

OpenStudy (anonymous):

p=5.5555555555556, q=4p/3

OpenStudy (ghazi):

no no p= -45, q=-60

OpenStudy (anonymous):

3(-45) -(-60)=15 -135+60=15 -75=15 So its not p=-45, q=-60

OpenStudy (ghazi):

4*-45=3*-60 therefore it is the answer

OpenStudy (anonymous):

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?

OpenStudy (ghazi):

of course it should i am also wondering how is it possible

OpenStudy (anonymous):

\[4p=3q ~and~ 3p-q=15\]So\[q=\frac{4p}{3}~and~3p-q=3p-\frac{4p}{3}=15 \implies 5p=45 \implies p=9 \implies q=12\]\[\implies (p,q)=(9,12)\]

OpenStudy (anonymous):

Don't make the easy ones hard.

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