# Determine the equilibrium quantity and equilibrium price and show the points of equilibrium from the following equations. 1. x^2=3y 2. 5x+3y-10=0

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You are allowed to ask only one question at a time. I have deleted your second set of equations and my response is for the first set.

You have asked for the equilibrium quantity and price but what the two variables x and y denote is not given.

I can only determine the value of the two variables using the simultaneous equations.

x^2 = 3y...(1)

5x + 3y - 10 = 0 ...(2)

Substitute 3y = x^2 in (2)

=> x^2 + 5x - 10 = 0

x1 = [ -5 + sqrt (25 + 40)]/2 = 1.5311

=> x1 = -5/2 + sqrt 65 / 2 = -6.5312

x2 = -5/2 - sqrt 65 / 2

y1 = 0.7814

y2 = 14.2185

From the values derived you can determine the price and quantity that you require, though I do not think it would make any sense.

**The points of equilibrium are ( 1.5311 , 0.7814) and (-6.5312, 14.2185). **

assuming x is the quantity and y is the price

1.)

x^2=3y

5x+3y-10=0

substitute x^2 in for 3y

5x+x^2-10=0

to solve this problem the quadratic equation must be used.

x = (-b+-sqrt(b^2 - 4ac))/(2a)

a = 1

b = 5

c = -10

x = -(5+-sqrt(25+40))/2

x = -(5+sqrt(65))/2 is negative and cannot be a price or quantity

x = (-5+sqrt(65))/2 is our solution

5x+3y-10=0

5(-5+sqrt(65))/2 -10 = -3y

(-35+5sqrt(65))/2 = -3y

(35-5sqrt(65))/6 = y

2.)

3x^2+y-10=0

x^2+2x-y+4=0

solve the bottom for y

x^2+2x-y+4=0

y = x^2+2x+4

substitute into the top

3x^2+y-10=3x^2+x^2+2x+4-10 = 0

0 = 4x^2+2x-6

factor

0 = (2x-2)(2x+3)

x = 0 at 1 and -3/2

assuming x is the quantity and y is the price

1.)

x^2=3y

5x+3y-10=0

substitute x^2 in for 3y

5x+x^2-10=0

to solve this problem the quadratic equation must be used.

x = (-b+-sqrt(b^2 - 4ac))/(2a)

a = 1

b = 5

c = -10

x = -(5+-sqrt(25+40))/2

x = -(5+sqrt(65))/2 is negative and cannot be a price or quantity

x = (-5+sqrt(65))/2 is our solution

5x+3y-10=0

5(-5+sqrt(65))/2 -10 = -3y

(-35+5sqrt(65))/2 = -3y

(35-5sqrt(65))/6 = y

2.)

3x^2+y-10=0

x^2+2x-y+4=0

solve the bottom for y

x^2+2x-y+4=0

y = x^2+2x+4

substitute into the top

3x^2+y-10=3x^2+x^2+2x+4-10 = 0

0 = 4x^2+2x-6

factor

0 = (2x-2)(2x+3)

x = 0 at 1 and 3/2