Answer:
80.
[tex]4s+5c=46\\3s+c=22[/tex]
81. Each apple cost $0.50
82. One pound of chocolate = $3
Step-by-step explanation:
80.
Let sold of 1 senior citizen ticket be "s", and
cost of 1 child ticket be "c"
First Day, 4 senior citizen ticket and 5 child tickets sold, equaled $46, so the equation we can write is:
[tex]4s+5c=46[/tex]
Second Day, 3 senior citizen ticket and 1 child ticket sold, equaled $22, so the equation we can write is:
[tex]3s+c=22[/tex]
So, the system of equation (2 equations) written above can be solved simultaneously to find our answer.
[tex]4s+5c=46\\3s+c=22[/tex]
81.
Let price of 1 apple be "a" and price of 1 banana be "b"
Since 3 apples and 2 banana cost $2, so we can write:
3a + 2b = 2
Then,
4 apples and 4 bananas cost $3, so we can write:
4a + 4b = 3
We can solve for the variable b in equation 1. Shown below:
[tex]3a + 2b = 2\\2b=2-3a\\b=\frac{2-3a}{2}[/tex]
Now we put this value of b into equation 2 and solve for cost of 1 apple, a. Shown below:
[tex]4a+4b=3\\4a+4(\frac{2-3a}{2})=3\\4a+4-6a=3\\2a=1\\a=\frac{1}{2}\\a=0.5[/tex]
So each apples cost $0.5
82.
Let 1 pound of chocolate be cost "c"
Let 1 pound of peanut be cost "p"
4 lb of chocolate and 2 lb of peanut = $16, thus we can write:
4c + 2p = 16
Next,
2 lb of chocolate and 3 lb of peanut costed $12, so we can write:
2c + 3p = 12
We solve first equation for p:
[tex]4c + 2p = 16\\2p=16-4c\\p=\frac{16-4c}{2}\\p=8-2c[/tex]
Now, we put this expression for p into equation 2 and solve for c:
[tex]2c + 3p = 12\\2c + 3(8-2c) = 12\\2c+24-6c=12\\4c=12\\c=3[/tex]
So, 1 pound of chocolates cost $3
