Answer:
He used 6 bananas at .50 which cost $3.00
12 oranges at .75 which cost $9.00 and
3 papayas at $1.25 which cost $3.75
Step-by-step explanation:
We can set up a system of equations using the information given.
b for banana, r for orange, p for papaya
b + r + p = 21. Also given r = 2b So substitute to simplify for calculating two unknowns:
b + 2b + p = 21 The number of items.
50b + 75(2b) +125p = 1575 The cost.
<em>I </em><em>multiplied</em><em> </em><em>everything</em><em> </em><em>by </em><em>100 </em><em>to </em><em>eliminate</em><em> </em><em>the</em><em> </em><em>decimals</em><em>.</em>
To solve by substituion, rewrite the first equation to get a value for p in terms of b
p = 21 - 3b
Substitute this value in the second equation and solve for b.
50b + 150b + 125(21–3b) = 1575
200b +2625 –375b = 1575 Subtract 2625 from both sides and combine the like terms:
–175b = 1575 –2625
–175b = –1050. Divide both sides by –175
b = 6
Substitute in the original equation to find the amount of each fruit:
b + r + p = 21
b = 6 bananas
r = 2b. 2(6) = 12 oranges. <em>I </em><em>used </em><em><u>r </u></em><em>so </em><em>as </em><em>not</em><em> to</em><em> </em><em>mix </em><em>up </em><em>o </em><em>with </em><em>0</em><em>.</em><em> </em>
6 + 12 + p = 21
p = 21 –18
p = 3 papayas
