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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.

### 2 個解答

• 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

Second one:

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

6X+4X=330

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

• 1.

K=Kelvin's original money=Cheri's original money.

L=Lunch Box price.

K=6*L+12

K=4*L+58

6L+12=4L+58

2L=46

L=23 (a)

K=6*23+12=\$150 (b)

2.

C=Ice-cream price

D=Softdrink price

D=1/3*C -> C=3D

\$110=2C+4D

110=6D+4D

110=10D

D=11

C=\$33 (a)

(b) 3*33+5*11=99+55=\$154

請努力學習，今次諳住良心害你一次。