Algebra word problem?

Alright then...

total vehicles = 18

total number of wheels is 10 more than 3 times the number of vehicles = 3 X 18 + 10 = 64 wheels

let x be the total number of bikes & y be the total number of cars.

so x + y = 18 ........... (1) (given)

Again one equation will be formed.. in which we distribute the number of wheels to bike & car & also the total number of wheels.

So 2x + 4y = 64

x + 2y = 32 ................ (2)

Solving it by elimination method of linear equation in two variables, we get x = 4 (for motorcycles) & y = 14 ( cars)

Number of wheels = (3*18) + 10 = 64

4*c + 2*m = 64 [1]

c + m =18 [2]

From [2], c = 18 - m

Substitute into [1]

4*(18 - m) + 2*m = 64

72 - 2*m = 64

2*m = 8

m = 4

from [2], c = 14

ie 4 motorcycles, 14 cars <<<<<

the number of wheels = 3 x 18 + 10 = 54 + 10 = 64

let cars = x

let motorcycles = y

x + y = 18 (vehicles) or, y = 18 - x

4x + 2y = 64 simplified to 2x + y = 32

subst y = 18 - x into 2x + y = 32

2x + 18 - x = 32

x = 32 - 18

x = 14

subst. x = 14 into x + y = 18

14 + y = 18

y = 4

number of motorcycles = 4

Is that right??

Let the no of cars be x

AND the no of motorcycles be y

x + y = 18 (A)

4x + 2y = 18*3 + 10

4x + 2y = 64 (B)

MULTIPLY (A) by 2

2x + 2y = 36 (C)

(B) - (C)

2x = 28

x = 14

ANSWER cars = 14 , motorcycles = 4

C + M = 18 whence C = 18 -- M

4C + 2M = 10 + 3(18) Or 4(18 -- M) + 2M = 10 + 3(18) whence M = 4

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no of cars is 14

and

no of motorcycles is 4