Asked by Tam
Sophia buys an eqaul number of oranges and pears for a party. The oranges are bought at a price of 7 for $2 and the pears are bought at a price of 5 for $3. She pays $33 more for the pears than the oranges. How much does Sophia pay in all? How many oranges and pears does she buy altogether?
Answers
Answered by
Henry
X = # of oranges.
X = # of pears.
($2/7) * X = $Y
Multiply both sides by 7:
2x = 7y
Eq1: 2x - 7y = 0
($3/5) * X = y + 33
Multiply both sides by 5:
3x = 5y + 165
Eq2: 3x - 5y = 165
Multiply Eq1 by 3 and Eq2 by -2 and add:
+6x - 21y = 0
-6x + 10y = -330
Sum: -11y = -330
Y = $30 = Cost of oranges
Y+33 = 30 + 33 = $63 = Cost of pears
Total cost = 30+63 = $93.
In Eq1, replace y with 30:
2x - 7*30 = 0
2x = 210
X = 105 of each fruit.
Tot. fruit = 2*105 = 210.
X = # of pears.
($2/7) * X = $Y
Multiply both sides by 7:
2x = 7y
Eq1: 2x - 7y = 0
($3/5) * X = y + 33
Multiply both sides by 5:
3x = 5y + 165
Eq2: 3x - 5y = 165
Multiply Eq1 by 3 and Eq2 by -2 and add:
+6x - 21y = 0
-6x + 10y = -330
Sum: -11y = -330
Y = $30 = Cost of oranges
Y+33 = 30 + 33 = $63 = Cost of pears
Total cost = 30+63 = $93.
In Eq1, replace y with 30:
2x - 7*30 = 0
2x = 210
X = 105 of each fruit.
Tot. fruit = 2*105 = 210.
Answered by
Anonymous
It’s so Confusing
Answered by
Anonymous
yup
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