Question 270008
Which combination of {{{100}}} animals cost $100?
Let {{{s}}} = number of sheep he bought
Let {{{c}}} = number of cows he bought
Let {{{h}}} = number of horses he bought
{{{50s + 100c + 1000h = 10000}}} (in cents)
{{{5s + 10c + 100h = 1000}}}
(1) {{{s + 2c + 20h = 200}}}
He buys {{{100}}} animals, so
(2) {{{s + c + h = 100}}}
This is 3 unknowns and only 2 equations, so it's
not directly solvable, but a little work will give
some answers
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If I subtract (2) from (1):
(1) {{{s + 2c + 20h = 200}}}
(2) {{{-s - c - h = -100}}}
{{{c + 19h = 100}}}
According to this, he could buy {{{5}}} horses
and {{{5}}} cows
{{{5 + 19*5 = 100}}}
Then, using (2),
(2) {{{s + c + h = 100}}}
{{{s + 5 + 5 = 100}}}
{{{s = 90}}}
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he could buy {{{5}}} horses, {{{5}}} cows, and {{{90}}} sheep
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check:
(1) {{{s + 2c + 20h = 200}}}
{{{90 + 2*5 + 20*5 = 200}}}
{{{90 + 10 + 100 = 200}}}
{{{200 = 200}}}
OK
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Go back to {{{c + 19h = 100}}}
Suppose he bought {{{4}}} horses?
{{{c + 19*4 = 100}}}
{{{c = 100 - 76}}}
{{{c = 24}}}
and
(2) {{{s + c + h = 100}}}
{{{s + 24 + 4 = 100}}}
{{{s = 100 - 28}}}
{{{s = 72}}}
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he could buy {{{4}}} horses, {{{24}}} cows, and {{{72}}} sheep
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check:  
(1) {{{s + 2c + 20h = 200}}}
{{{72 + 2*24 + 20*4 = 200}}}
{{{72 + 48 + 80 = 200}}}
{{{200 = 200}}}
OK
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Suppose he bought {{{3}}} horses?
{{{c + 19*3 = 100}}}
{{{c = 100 - 57}}}
{{{c = 43}}}
and
(2) {{{s + c + h = 100}}}
{{{s + 43 + 3 = 100}}}
{{{s = 100 - 46}}}
{{{s = 54}}}
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he could buy {{{3}}} horses, {{{43}}} cows, and {{{54}}} sheep
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check:
(1) {{{s + 2c + 20h = 200}}}
{{{54 + 2*43 + 20*3 = 200}}}
{{{54 + 86 + 60 = 200}}}
{{{200 = 200}}}
OK