Question 360584
Let {{{e}}} = the amount he charges per event
Let {{{p}}} = the amount he charges per picture
given:
For Stacy's wedding:
(1) {{{60000 = e + 200p}}} (in cents)
For Shelly's wedding:
(2) {{{71250 = e + 275p}}} (in cents)
Subtract equation (1) from (2)
{{{71250 = e + 275p}}}
{{{-60000 = -e - 200p}}}
{{{11250 = 75p}}}
{{{p = 150}}}
Keven charges {{{150}}} cents, or $1.50 per picture
check answer:
(1) {{{60000 = e + 200p}}}
{{{60000 = e + 200*150}}}
{{{60000 = e + 30000}}}
{{{e = 60000 - 30000}}}
{{{e = 30000}}}
and
(2) {{{71250 = e + 275p}}}
{{{71250 = 30000 + 275p}}}
{{{275p = 41250}}}
{{{p = 150}}}
OK