Question 693720
I can call 1990, the year he started work, {{{ x = 0 }}}
1991 will be {{{ x = 1 }}}
The slope will be ( change in y ) / ( change in x ), so
{{{ m = 2100 / 1 }}} ( the 1 stands for 1 year )
{{{ m = 2100 }}}
given:
{{{ y(8) = 1.5*y(0) }}}
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So far I have
{{{ y = 2100x + b }}}
Now I can say
{{{ y(0) = 2100*0 + b }}}
{{{ y(0) = b }}}
and also
{{{ y(8) = 2100*8 + b }}}
{{{ 1.5*y(0) = 2100*8 + b }}}
{{{ 1.5*b = 16800 + b }}}
{{{ 1.5b - b = 16800 }}}
{{{ .5b = 16800 }}}
{{{ b = 16800 / .5 }}}
{{{ b = 33600 }}}
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Now I have the final equation
{{{ y = 2100x + 33600 }}}
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His salary in 1998 was
{{{ y = 2100*8 + 33600 }}}
{{{ y = 16800 + 33600 }}}
{{{ y = 50400 }}}
$50,400 
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In 2005 he will earn
{{{ y = 2100*15 + 33600 }}}
{{{ y = 31500 + 33600 }}}
{{{ y = 65100 }}}
$65,100
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Check:
Is the slope between {{{ x = 0 }}} and {{{ x = 15 }}}
equal to {{{ 2100 }}} ?
{{{ ( 65100 - 33600 ) / ( 15 - 0 ) }}}
{{{ 31500 / 15 = 2100 }}}
OK
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Here's a plot:
{{{ graph( 400, 400, -2, 16, -10000, 100000, 2100x + 33600 ) }}}