the cost of three pencil and one eraser is #22, while the cost of four pencils and three eraser is #42, find the numbers

Question

the cost of three pencil and one eraser is #22, while the cost of four pencils and three eraser is #42, find the numbers

Answer 1

Well, it seems like those pencils and erasers are playing quite a money game! Let's break down their sneaky scheme.

Let's assume the cost of a pencil is "p" and the cost of an eraser is "e". According to the given information, we can set up two equations to represent the given scenarios:

Equation 1: 3p + 1e = 22
Equation 2: 4p + 3e = 42

Now, let's solve this puzzle using a humorous twist!

We shall proceed with the good ol' method of elimination.

First, let's multiply Equation 1 by 3 to eliminate the "e" term:

3(3p + 1e) = 3(22)
9p + 3e = 66 (Equation 3)

Now, let's multiply Equation 2 by 1 to eliminate the "p" term:

1(4p + 3e) = 1(42)
4p + 3e = 42 (Equation 4)

Great! Now we will subtract Equation 4 from Equation 3:

(9p + 3e) - (4p + 3e) = 66 - 42
9p + 3e - 4p - 3e = 24
5p = 24

Dividing both sides of the equation by 5:

p = 24/5

Well, this is a bit funny! We end up with a fraction for the cost of a pencil. So, we're going to need a decimal approximation.

p ≈ 4.8

Now, we can substitute the "cost of a pencil" value into either Equation 1 or 2. Let's use Equation 1:

3(4.8) + e = 22
14.4 + e = 22
e = 22 - 14.4
e ≈ 7.6

So, the cost of a pencil is approximately #4.8, and the cost of an eraser is approximately #7.6.

I hope this amusing arithmetic adventure made you smile!

Answer 2

To solve this problem, let's assign variables to represent the unknown quantities. Let's say:

- The cost of one pencil is p,
- The cost of one eraser is e.

According to the given information:

1) "The cost of three pencils and one eraser is #22":
3p + 1e = 22

2) "The cost of four pencils and three erasers is #42":
4p + 3e = 42

Now we have a system of equations. We can solve it using substitution or elimination. Let's use the elimination method:

To eliminate the variable "e", we can multiply equation 1 by 3, and equation 2 by 1:

Equation 1: 3(3p + 1e) = 3(22) => 9p + 3e = 66
Equation 2: 1(4p + 3e) = 1(42) => 4p + 3e = 42

Now, subtract equation 2 from equation 1 to eliminate the variable "e":

(9p + 3e) - (4p + 3e) = 66 - 42
9p - 4p + 3e - 3e = 24
5p = 24
p = 24/5
p ≈ 4.8

Substituting the value of p back into equation 1:
3(4.8) + 1e = 22
14.4 + 1e = 22
1e = 22 - 14.4
1e ≈ 7.6

Therefore, one pencil costs approximately #4.8 and one eraser costs approximately #7.6.

Answer 3

To solve this problem, we can use a system of equations. Let's represent the cost of a pencil as "p" and the cost of an eraser as "e."

According to the given information, we have two equations:

1) 3p + 1e = 22 --> Equation 1
2) 4p + 3e = 42 --> Equation 2

Now, we can solve these equations by either substitution or elimination. Let's use the elimination method:

First, we'll multiply Equation 1 by 4 and Equation 2 by 3 to make the coefficients of "p" the same:

4(3p + 1e) = 4(22) --> 12p + 4e = 88
3(4p + 3e) = 3(42) --> 12p + 9e = 126

Now, subtract the first equation from the second equation:

(12p + 9e) - (12p + 4e) = 126 - 88
12p - 12p + 9e - 4e = 38
5e = 38
e = 38 / 5
e = 7.6

Now, substitute the value of "e" back into one of the original equations. Let's use Equation 1:

3p + 1(7.6) = 22
3p + 7.6 = 22
3p = 22 - 7.6
3p = 14.4
p = 14.4 / 3
p = 4.8

Therefore, the cost of one pencil is #4.8 and the cost of one eraser is #7.6.

Answer 4

3p + e = 22

4p + 3e = 42

I would use substitution, from the first
e = 22-3p
plug into the 2nd

4p + 3(22-3p) = 42
4p + 66 - 9p = 42
-5p = -24
p = 4.80 , then e = 7.60