How do I solve this simultaneous equation

6 pencils and 4 rubbers cost $58, while 5 pencils and 2 rubbers cost $35.

How much do ten pencils and 4 rubbers cost?

How much do 4 pencils cost?

x = pencils
y = rubbers.

6x + 4y = 58 eq(1)
5x + 2y = 35 eq(2)

Solve by elimination.

Multiply Equation 2 by -2, and then add to equation 1 to eliminate the y's.

6x + 4y = 58
-10x - 4y = -70
-4x = -12
x = 3

Back substitute:
6x + 4y = 58
6(3) + 4y = 58
4y = 40
y = 10

pencil cost: $3
rubber cost: $10

10(3) + 4(10) = 30 + 40 = $70

4(3) = $12

x = pencils
y = rubbers

cross-multiply

2 (6x + 4y = 58) ------- equation 1
4 (5x + 2y = 35) ---------- equation 2

12x + 8y = 116
20x + 8y = 140

change signs

12x + 8y = 116
-20x - 8y = -140

-8x = -24

x = -24/-8
x = 3

substitute x=3 in equation 1

6x + 4y = 58
(6*3) + 4y = 58
18 + 4y = 58
4y = 58 - 24
y = 40/4
y=10

x = 3
y = 10

10x + 4y = ?
(10*3) + (4*10) = 70 <-------------first answer

4x = 4*3 = 12 <--------------- second answer

one pencil costs $3 and one rubber costs $10

1) How much do 10 pencils and 4 rubbers cost?
ans:- you have to multiply 10 into the cost of one pencil (which is 3) and multiply 4 into the cost of one rubber (which is 10) and add the two values, which is equal to 70

2) How much do 4 pencils cost?
If one pencil costs $3, you have to multiply 4 into 3, which is equal to 12