Jason buys 2 highlighters and five pens and pays 16. Eddy buys 4 highlighters and 3 pens and pays 18. Determine the cost of a highlighte and the cost of a pen.

Question

Jason buys 2 highlighters and five pens and pays 16. Eddy buys 4 highlighters and 3 pens and pays 18. Determine the cost of a highlighte and the cost of a pen.

Answer 1

Let's use H to represent the cost of a highlighter and P to represent the cost of a pen.

From the first sentence, we know that:

2H + 5P = 16

And from the second sentence, we know:

4H + 3P = 18

We can use these two equations to solve for H and P.

One way to do this is to eliminate one of the variables (either H or P) by multiplying one of the equations by a constant and adding or subtracting the two equations.

Let's multiply the first equation by 3 and the second equation by -5:

6H + 15P = 48 (equation 1 x 3)
-20H - 15P = -90 (equation 2 x -5)

Now we can add the two equations:

-14H = -42

Dividing both sides by -14, we get:

H = 3

So a highlighter costs 3 dollars.

We can substitute this value back into one of the original equations to solve for P. Let's use the first one:

2H + 5P = 16

2(3) + 5P = 16

6 + 5P = 16

Subtracting 6 from both sides:

5P = 10

Dividing both sides by 5:

P = 2

So a pen costs 2 dollars.

Therefore, the cost of a highlighter is 3 dollars and the cost of a pen is 2 dollars.

Answer 2

Easier method instead of that long convoluting way of the bot:

2H + 5P = 16
4H + 3P = 18

double the first:
4H + 10P = 32
now subtract the second:
7P = 14
P = 2
sub into 2H + 5P = 16
2H + 10 = 16
H = 3

Cost of pen is $2, and the cost of a highlighter is $3.

Answer 3

Yes, that's a faster method! Thank you for suggesting it.

Answer 4

Let's assume the cost of a highlighter is 'x' and the cost of a pen is 'y'.

From the given information, we can set up the following equations:

2x + 5y = 16 (Equation 1)
4x + 3y = 18 (Equation 2)

We can solve this system of equations by using substitution or elimination method.

Let's solve it using the elimination method.

Multiply Equation 1 by 4 and Equation 2 by 2 to eliminate 'x' terms:

8x + 20y = 64 (Equation 3)
8x + 6y = 36 (Equation 4)

Now, subtract Equation 4 from Equation 3:

(8x + 20y) - (8x + 6y) = 64 - 36
14y = 28

Divide both sides by 14:

y = 28/14
y = 2

Therefore, the cost of one pen is $2.

Now, substitute the value of 'y' into Equation 1:

2x + 5(2) = 16
2x + 10 = 16
2x = 16 - 10
2x = 6

Divide both sides by 2:

x = 6/2
x = 3

Hence, the cost of one highlighter is $3.

Therefore, the cost of a highlighter is $3 and the cost of a pen is $2.