Question

Solve the following systems of linear equations using substitution  12p + 3q = 15  6q +...

Solve the following systems of linear equations using substitution  12p + 3q = 15  6q + p = 30 Remember that p = price, and q = quantity, the relationship of both variables is linear. Find the value of both variables.

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Answer #1

Given equation:

12P + 3Q = 15 -------------> EQUATION 1

P + 6Q = 30 -------------------->EQUATION 2

EQUATION 1*2 => 24 P + 6Q = 30

EQUATION 2*1 => P + 6Q = 30

24P+6Q =30

-(P+6Q=30)

SO,

24P+6Q =30

-P - 6Q = -30

Reducing the equation,

23P= 0

P = 0

Substitute P = 0 in equation 1

12 P + 3Q = 15

12(0) + 3Q = 15

3Q = 15

Q = 15/ 3

Q=5

P = Price = 0

Q = Quantity = 5

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