中一數學 algebraic equations

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。