Question

2. The prices, in dollars per unit, of the three commodities X, Y and Z are x, y and z,

Answer 1

Answer:

Price of X is $24.81

Price of Y is $3.66

Price of Z is $11.36

Step-by-step explanation:

for person A, we know that earns $40, then we can write the equation:

-4*z + 3*x + 3*y = $40

For person B, we know that earns $50, then:

1*z + 2*x - 3*y = $50

For person C, we know that earns $130, then:

6*z - 1*x + 4*y = $130

Then we have a system of equations:

-4*z + 3*x + 3*y = $40

1*z + 2*x - 3*y = $50

6*z - 1*x + 4*y = $130

To solve the system, we need to isolate one of the variables in one of the equations.

Let's isolate z in the second equation:

z = $50 - 2*x + 3*y

now we can replace this in the other two equations:

-4*z + 3*x + 3*y = $40

6*z - 1*x + 4*y = $130

So we get:

-4*($50 - 2*x + 3*y) + 3*x + 3*y = $40

6*($50 - 2*x + 3*y) - 1*x + 4*y = $130

Now we need to simplify both of these, so we get:

-$200 + 11x - 9y = $40

$350 - 13*x + 28*y = $130

Now again, we need to isolate one of the variables in one of the equations.

Let's isolate x in the first one:

-$200 + 11x - 9y = $40

11x - 9y = $40 + $200 = $240

11x = $240 + 9y

x = ($240 + 9y)/11

Now we can replace this in the other equation:

$350 - 13*x + 28*y = $130

$350 - 13*($240 + 9y)/11 + 28*y = $130

Now we can solve this for y.

- 13*($240 + 9y)/11 + 28*y  = $130 - $350 = -$220

-13*$240  - (13/11)*9y + 28y = - $220

y*(28 - (9*13/1) ) = -$220 + (13/11)*$240

y = ( (13/11)*$240 - $220)/(28 - (9*13/1) ) = $3.66

We know that:

x = ($240 + 9y)/11

Replacing the value of y, we get:

x = ($240 + 9*$3.66)/11  = $24.81

And the equation of z is:

z = $50 - 2*x + 3*y = $50 - 2* $24.81 + 3*$3.66 = $11.36

Then:

Price of X is $24.81

Price of Y is $3.66

Price of Z is $11.36