Question 1104227
Let {{{ y }}} = gallons of gas in tank
Let {{{ x }}} = miles driven
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The more miles that are driven, the 
less gallons are in the tank
The general form for a linear equation is:
{{{ y = m*x + b }}}
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(1) {{{ 12 = m*120 + b }}}
(2) {{{ 8.66 = m*220 + b }}}
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Subtract (2) from (1)
(1) {{{ 12 = m*120 + b }}}
(2) {{{ -8.66 = -m*220 - b }}}
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{{{ 3.34 = -100m }}}
{{{ m = -.0334 }}}
and
(1) {{{ 12 = ( -.0334 )*120 + b }}}
(1) {{{ 12 = -4.008  + b }}}
(1) {{{ b = 16.008 }}}
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(a) {{{ y = -.0334x + 16.008 }}}
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check:
{{{ y = -.0334*120 + 16.008 }}}
{{{ y = -4.008 + 16.008 }}}
{{{ y = 12 }}}
and
{{{ y = -.0334*220 + 16.008 }}}
{{{ y = -7.348 + 16.008 }}}
{{{ y = 8.66 }}}
OK
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(b)
When {{{ x = 0 }}} miles driven, the tank is full
{{{ y = -.0334*0 + 16.008 }}}
There are 16.008 gallons in tank when full
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(c)
When does {{{ y = 0 }}} ?
{{{ 0 = -.0334x + 16.008 }}}
{{{ .0334x = 16.008 }}}
{{{ x = 479.281 }}}
I can drive 479.281 mi before running out of gas
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(d)
[ gallons in tank ] = [ -gallons/mi ]*[ miles driven ] + [ gallons w/full tank ]
{{{ y = m*x + b }}}
{{{ m = -.0334 }}}
[ miles / gallon ] = {{{ 1/.0334 }}}
{{{ 1/.0334 = 29.94 }}}
The fuel efficiency is 29.94 mi/gal