Mr. Harvey wants to mix screws costing 0.25 a piece with bolts costing 0.40 each. the mixture will cost 3.10$. The number of 0.25 pieces is 2 more than the number of 0.40 pieces. How many of each type should be included in the package?​



Answer :

Answer:

GIVEN THAT:

Number of screws costing $0.25 as x

Number of bolts costing $0.40 as y

From the given information:

The total cost of the mixture is $3.10

The number of $0.25 pieces is 2 more than the number of $0.40 pieces

The cost equation is: $0.25x + $0.40y = $3.10

The quantity equation is: x = y + 2

By substituting x = y + 2 into the cost equation:

$0.25(y + 2) + $0.40y = $3.10

Simplifying: $0.25y + $0.50 + $0.40y = $3.10

Combining like terms: $0.65y + $0.50 = $3.10

Solving for y: $0.65y = $2.60

y = 4 (number of bolts)

Substitute y = 4 back into x = y + 2:

x = 4 + 2

x = 6 (number of screws)

Therefore, Mr. Harvey should include 6 screws and 4 bolts in the package.

hope it's helpful

Answer:

6 screws

4 bolts

Step-by-step explanation:

To solve this problem, we can set up a system of equations based on the given information.

Let s be the number of screws costing $0.25 each.

Let b be the number of bolts costing $0.40 each.

Given that a mixture of screws and bolts costs $3.10, this can be expressed as :

[tex]0.25s + 0.40b = 3.10[/tex]

Given that the number of screws is 2 more than the number of bolts, then:

[tex]s = b + 2[/tex]

Therefore, the system of equations is:

[tex]\begin{cases}0.25s + 0.40b = 3.10\\s = b + 2\end{cases}[/tex]

To solve this system of equations, substitute the second equation into the first equation and solve for b:

[tex]0.25(b + 2) + 0.40b = 3.10\\\\0.25b + 0.50 + 0.40b = 3.10\\\\0.65b + 0.50 = 3.10\\\\0.65b = 3.10 - 0.50\\\\0.65b = 2.60\\\\b = 2.60 \dov 0.65\\\\b=4[/tex]

Therefore, 4 bolts are included in the package.

To find the number of screws, substitute b = 4 into the second equation:

[tex]s = 4 + 2\\\\s = 6[/tex]

Therefore, 6 screws are included in the package.