Systems of Linear Equations in Two Variables Question 2?

"A customer buys 3 hot dogs and 2 bags of potato chips for $9"
3h + 2p = 9

"Another customer buys 5 hot dogs and 4 bags of potato chips for $16"
5h + 4p = 16

the coefficients of p are 2 and 4.
4 is a multiple of 2, so the p terms can be easily eliminated.
multiply the first equation by -2 and add the result to the second equation:
-6h - 4p = -18
 5h + 4p = 16
----------------
-h    = -2

h = 2
hot dogs cost $2 each

3(2) + 2p = 9
2p = 3
p = 1.5
Potato chips cost $1.50 a bag.

Let $h be the cost of a hot dog, and $p be the cost of a bag of potato chips.

3h + 2p = 9.00 ...... [1]
5h + 4p = 16.00 ...... [2]

[1] × 2 :
6h + 4p = 18.00 ...... [3]

[3] - [2] :
h = 2.00

Put h = 2.00 into [1] :
3(2.00) + 2p = 9.00
p = 1.50

Cost of a hot dog = $2.00
Cost of a bag of potato chips = $1.50

3x + 2y = 900
5x + 4y = 1600

-6x - 4y = - 1800
5x + 4y = 1600______ADD
-x = - 200
x = 200_____hot dog $2.00

600 + 2y = 900
2y = 300
y = 150____potato chips $1.50

3x + 2y = 9, 5x + 4y = 16
Solution:
x = 2, y = 3/2
A hot dog costs $2 and a bag of potato chips costs $1.50.

Let

x = price of hot dog

y = price of potato chips

Then

3x + 2y = 9.......................(1)

5x + 4y = 16.....................(2)

Multiply equation (1) by 2

6x + 4y = 18 ....................(3)

Subtract (2) from (3)

x = 2

Put this in (1)

3(2) + 2y = 9

2y = 3

y = 1.50

hot dog = $2.00

potato chips = $1.50