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中一數學 algebraic equations
1. Kelvin and Cherie have the same amount of money. They go to a fast food
restaurant to buy lunch boxes. Kelvin buys six lunch boxes and has $12 left. Cheri
buys four lunch boxes and has $58 left.
a) Find the price of each lunch box.
b) How much does kelvin have originally?
2)Sam pays $110 for 2 boxes of ice cream and 4 bottles of soft drink in a supermarket. If the price of each bottle of soft drink is one-third that of each box of ice cream,find the prices of
a) a box of ice cream,
b) 3 boxes of ice ice cream and 5 bottles of soft drink.
- PeterLv 71 十年前最愛解答
First one: This is no need to use algebra to solve.
Cherie brought 2 less lunch boxes than Kelvin and has $46 ($58-$12) more. Thus each lunch box costs $23 ($46/2). So Kelvin and Cherie each has $150 ($23x6+$12) originally.
Of course if you insist to solve it with algebra, you can do this:
Let Kelvin and Cherie each has X dollars.
Let each lunch box costs Y dollars
equation 1: X=6Y+12 (Kelvin's case)
equation 2: X=4Y+58 (Cherie's case)
Then 6Y+12=4Y+58, and Y=23, X=150
Let each box of ice cream costs X dollars.
Let each bottle of soft drink costs Y dollars.
equation 1: 2X+4Y=110 (Sam's purchase in supermarket)
equation 2: Y=1/3(X) (price of each bottle of soft drink is one-third that of each box of ice cream)
Then sub equation 2 into equation 1: 2X+4/3(X)=110
X=33 (price of a box of ice cream)
Y=11 (price of a bottle of soft drink)
Price of 3 boxes of ice cream and 5 bottles of soft drink = 3x33+5x11 = 154
- ?Lv 71 十年前
K=Kelvin's original money=Cheri's original money.
L=Lunch Box price.
D=1/3*C -> C=3D