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Question 1 of 5
Sue went shopping on two occasions. On the first occasion she bought 5 apples and 2 bananas for $2.80. On the second occasion she paid $5.10 for 3 apples and 5 bananas. What is the cost of each fruit?
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First, let variables to represent apples and bananas.
a = cost of each apple
b = cost of each banana
Next, write the systems of equations being represented in the problem.
5 a + 2 b
=
2.80
First occasion
3 a + 5 b
=
5.10
Second occasion
( 5 a + 2 b ) × 3
=
2.80 × 3
15 a + 6 b
=
8.40
( 3 a + 5 b ) × 5
=
5.10 × 5
15 a + 25 b
=
25.50
Subtract the transformed equations.
15 a + 6 b
=
8.40
15 a + 25 b
=
25.50
- 19 b
=
- 17.10
b
=
0.90
Divide both sides by - 19
Solve for a , the cost of each apple.
3 a + 5 b
=
5.10
3 a + 5 ( 0.90 )
=
5.10
Substitute b = 0.90
3 a + 4.5 - 4.5
=
5.10 - 4.5
Subtract 4.5 from both sides
3 a
=
0.60
Divide both sides by 3
a
=
0.20
Apple = $ 0.20 , Banana = $ 0.90
Question 2 of 5
A customer bought 4 drinks and 3 pizzas for $ 40.80 . Another costumer bought 2 drinks and 1 pizza for $ 15.00 . Find the price of the following:
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First, let variables represent drinks and pizzas.
d = cost of each drink
p = cost of each pizza
Next, write the both of the customer’s purchases as systems of equations.
To make the solution easier, convert the dollar value into cents first..
4 d + 3 p
=
4080
Equation 1
2 d + 1 p
=
1500
Equation 2
Next, multiply the values of equation 2 by 2 and label the product as equation 3 .
2 d + 1 p
=
1500
Equation 2
( 2 d + 1 p ) × 2
=
1500 × 2
Multiply the values of both sides by 2
4 d + 2 p
=
3000
Equation 3
Then, subtract equation 3 from equation 1 .
4 d - 3 p
=
4080
- ( 4 d + 2 p )
=
3000
p
=
1080
4 d - 4 d cancels out
Now, substitute the value of p into any of the two equations.
2 d + 1 p
=
1500
Equation 2
2 d + 1 ( 1080 )
=
1500
p = 1080
2 d + 1080 - 1080
=
1500 - 1080
Subtract 1080 from both sides
2 d ÷ 2
=
420 ÷ 2
Divide both sides by 2
d
=
210
Finally, convert the values back to dollars by dividing each them by 100
Dollar value of each drink
Dollar value of each pizza
Price of each drink = $ 2.10
Price of each pizza = $ 10.80
Question 3 of 5
2 oranges and 5 apples cost $ 5.75 and 4 oranges and 6 apples cost $ 8.10 . Find the price of each apple and orange
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First, let variables represent apples and oranges.
x = cost of each orange
y = cost of each apple
Next, write the systems of equations being represented in the problem.
2 x + 5 y
=
575
First occasion
4 x + 6 y
=
810
Second occasion
( 2 x + 5 y ) × 2
=
575 × 2
4 x + 10 y
=
1150
Subtract the second equation from the transformed equation.
4 x + 10 y
=
1150
4 x + 6 y
=
810
4 y
=
340
y
=
85
Divide both sides by 4
Solve for x , the cost of each orange.
2 x + 5 y
=
575
2 x + 5 ( 85 )
=
575
Substitute y = 85
2 x + 425 - 425
=
575 - 425
Subtract 425 from both sides
2 x
=
150
Divide both sides by 2
x
=
75
Orange = 75 cents , Apple = 85 cents
Question 4 of 5
At a theatre, there are 400 patrons. If there were 60 more women than men, how many men attended the show?
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First, let variables to represent men and women.
x = number of women
y = number of men
Next, write the systems of equations being represented in the problem.
x + y
=
400
There are a total of 400 patrons
x
=
y + 60
60 more women than men
Rewrite equation 2 such that x and y are on the same side.
x
=
y + 60
x - y
=
y + 60 - y
Subtract y from both sides
x - y
=
60
x + 4 y
=
400
x - y
=
60
2 x
=
460
x
=
230
Divide both sides by 2
Solve for y , the number of men.
x - y
=
60
230 - y
=
60
Substitute x = 230
230 - y - 230
=
60 - 230
- y
=
- 170
y
=
170
There were 170 men who attended the show.
Question 5 of 5
Find the value of x and y
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First, write the systems of equations being represented in the problem.
3 x - y
=
12
Length
2 x + 3 y
=
19
Width
( 3 x - y ) × 3
=
12 × 3
9 x - 3 y
=
36
Add the second equation to the transformed equation.
2 x + 3 y
=
19
9 x - 3 y
=
36
11 x
=
55
x
=
5
Divide both sides by 11
2 x + 3 y
=
19
2 ( 5 ) + 3 y
=
19
Substitute x = 5
10 + 3 y - 10
=
19 - 10
Subtract 10 from both sides
3 y
=
9
Divide both sides by 3
y
=
3