Sagot :
Answer:
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
Answer:
$9
Step-by-step 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 :)