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