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Mathematics 23 Online
OpenStudy (anonymous):

How much 87-octane gas and 93-octane gas should Kelsey mix in order to make 12 gal of 91-octane gas for her Legacy?

OpenStudy (anonymous):

4 gal 87, 8 gal 93

OpenStudy (anonymous):

would you like a method?

OpenStudy (anonymous):

yes that would help greatly, thanks

OpenStudy (anonymous):

i will provide two then

OpenStudy (anonymous):

we have to introduce a variable, call it \(x\) = number of gallons of 93 octane gas and it contributes \(93\times x\) amount of octane then since the total is 12 gallons, the amount of 87 octane gas must be \(12-x\) and it contributes \(87(12-x)\) amount of octane the total amount of octane you want is \(12\times 91=1092\) so you set \[93x+87(12-x)=1092\] and solve for x

OpenStudy (anonymous):

second method (which is actually identical) is to put \(x\) = amount of 93 octane, \(y\) = amount of 87 octane and since the total is 12 you can write \[x+y=12\] then as above you want the total octane to be 1092 and so you can write \[93x+87y=1092\] and solve the system \[x+y=12\] \[93x+87y=1092\]

OpenStudy (anonymous):

you will notice that to solve the system the first step is to solve \[x+y=12\] for \(y\) to get \[y=12-x\]and then substitute it in to the second equation and get \[93x+87(12-x)=1902\] so the work is identical and the first method actually does not require two variables and so is much better

OpenStudy (anonymous):

Ok, that makes sense. Thank You!

OpenStudy (anonymous):

yw oh and let me point out one more thing, you may notice that i use \(x\) =number of gallons of higher octane. you could have used \(x\)= number of gallons of lower octane and solved a different equation, but if you pick \{x\) as the higher value at the beginning you will not end up with \(-x\) in your equation

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