Question

4 Pencils and 2 rulers cost $8. 2 pencils and 4 rulers cost $10. Find the cost of 3 rulers and 3 pencils

Answer 1

$9

Explanation:

pencil - p

ruler - r

4p + 2r = 8

2p + 4r = 10

4p + 2r = 8 {finding p in terms of r so we can substitiute it}

- 2r -2r

4p = 8 - 2r

÷ 2 ÷2

2p = 4 - r {now, we can replace "2p" in our other equation}

2p + 4r = 10 becomes...

[4 - r] + 4r = 10

4 - r + 4r = 10

4 + 3r = 10

- 4 -4

3r = 6

÷3 ÷3

r = 2

when we have 4 pencils and 2 rulers, we have:

4p + 2[2] = 8

4p + 4 = 8

-4 -4

4p = 4

÷4 ÷ 4

p = 1

{checking with our other equation:

p = 1 ; r = 2

2p + 4r = 10

2[1] + 4[2] = 10

2 + 8 = 10

10 = 10

TRUE STATEMENT;

confirms our variables}

so , we have found that the cost of a pencil is $1, and the cost of a ruler is $2,

p = 1 ; r = 2

so, let's solve our original problem:

3 rulers + 3 pencils = ?

3(2) + 3(1) = ?

6 + 3 = 9

so, the cost of this combination would be $9

hope this helps!! have a lovely day :)

Answer 2

4 pencil & 2 ruler = $8

2 pencil & 4 ruler = $10

let the no. of Pencil be X and no. of ruler be Y

4x + 2y = 8

2x + 4y= 10

solving the equation by elimination method,

multiplying eq 2 with 2

4x + 2y = 8

4x + 8y = 20

subtracting,

- 6 y = -12

y = 2

2x + 4(2)= 10

2x + 8 = 10

2x = 10 - 8

2x = 2

x = 1

Cost of 3 ruler + 3 pencil

= 3(1) + 3(2)

= 3 + 6

= 9

= $9