Why Existence and Uniqueness Need Separate Checks
Solving two linear equations can produce an intersection, but that does not automatically give a feasible market equilibrium. The solution might imply a negative price or a negative quantity. It is useful, therefore, to answer two questions separately: do the schedules intersect at exactly one point, and is that point economically feasible?
This distinction also clarifies what slope information tells us. A downward-sloping demand schedule and an upward-sloping supply schedule typically guarantee a unique algebraic intersection. Their intercepts determine where that intersection lies, including whether it falls in the nonnegative price-and-quantity region. The earlier material on equilibrium calculations provides the basic procedure; here the focus is on the conditions under which that procedure gives one feasible answer.
The intercept \(a\) is the quantity demanded at a price of zero; \(c\) is the quantity supplied at a price of zero. The slope of demand with respect to price is \(-b\), while the slope of supply is \(d\). In an economically conventional application, the schedules are generally considered over the relevant nonnegative-price region; a linear equation can still be solved outside that region, but such a solution may not describe a feasible market outcome.
Existence and Uniqueness for the Standard Linear Model
Equilibrium requires the quantity buyers demand to equal the quantity sellers supply at the same price. Setting the schedules equal gives one equation in price:
If \(b+d>0\), this equation has exactly one solution for price. In the standard case \(b>0\) and \(d\geq 0\), that condition always holds. Once the price is found, substituting it into either schedule determines the associated quantity.
The theorem gives two distinct tests. Because \(b+d>0\), the price is nonnegative exactly when \(a-c\geq 0\), or \(a\geq c\). The quantity test is \(ad+bc\geq 0\). A positive denominator makes these numerator tests straightforward, but neither should be skipped if intercept assumptions are not specified.
Why the Formulas Hold
Subtracting \(c\) from both sides of the equilibrium equation and collecting the price terms yields \((b+d)P=a-c\). Dividing by the nonzero coefficient \(b+d\) gives the unique price \(P^*=(a-c)/(b+d)\). For the quantity, substitute that price into demand:
The cancellation in the numerator shows why both intercepts affect equilibrium quantity. The same quantity results if the price is substituted into supply, since the price was obtained by setting the two schedules equal. If \(b+d=0\), division is not valid: the schedules have the same slope. They are then parallel and distinct, so there is no intersection, or they coincide, so there are infinitely many algebraic intersections. That case does not have a unique equilibrium.
If both intercepts are nonnegative, the feasibility test can be simplified. With \(a\geq 0\), \(c\geq 0\), \(b>0\), and \(d\geq 0\), the quantity numerator \(ad+bc\) is nonnegative. Thus \(a\geq c\) is enough for a feasible equilibrium. It is still important to check the quantity when the intercepts are not both known to be nonnegative. A unique intersection can otherwise have a negative quantity even when its price is positive.
Worked Calculations
Worked Example: A Unique Feasible Intersection
A hypothetical market has weekly demand \(Q_D(P)=96-4P\) and supply \(Q_S(P)=12+3P\). Price is measured in dollars per unit, and quantities are units per week. Find the equilibrium and determine whether it is feasible.
Method. The schedules have \(a=96\), \(b=4\), \(c=12\), and \(d=3\). The slopes satisfy \(b+d=4+3=7>0\), so there is exactly one algebraic intersection.
The equilibrium price is \(P^*=(a-c)/(b+d)=(96-12)/(4+3)=84/7=12\) dollars per unit.
Substitute into demand: \(Q_D(12)=96-4(12)=96-48=48\) units per week. Substitution into supply gives \(Q_S(12)=12+3(12)=12+36=48\) units per week.
Both schedules give the same quantity, and price and quantity are nonnegative. The unique intersection is a feasible equilibrium: \(P^*=12\) and \(Q^*=48\). The check avoids a common mistake: finding the price alone and assuming that the schedules agree without verifying the quantity.
Worked Example: A Unique Intersection at a Negative Price
A hypothetical market has \(Q_D(P)=30-2P\) and \(Q_S(P)=42+P\), with price in dollars and quantity in units per week. Find the algebraic intersection and decide whether it is a feasible equilibrium.
Here \(a=30\), \(b=2\), \(c=42\), and \(d=1\). Since \(b+d=2+1=3>0\), the intersection is unique. Its price is \(P^*=(30-42)/3=-12/3=-4\) dollars.
Substituting \(P=-4\) into demand gives \(Q_D(-4)=30-2(-4)=30+8=38\). Substituting into supply gives \(Q_S(-4)=42+(-4)=38\). The schedules agree, so \((-4,38)\) is indeed the unique algebraic intersection.
It is not a feasible market equilibrium because its price is negative, even though the quantity is positive. The intercept condition predicts this result: \(a-c=30-42=-12<0\), so \(P^*<0\). A common error is to call any equal-quantity solution an equilibrium without checking that its price is nonnegative.
Worked Example: A Positive Price but Negative Quantity
Consider the mathematical schedules \(Q_D(P)=-2-P\) and \(Q_S(P)=-8+P\), with price in dollars and quantity in units. Determine whether their unique algebraic intersection is feasible.
The coefficients are \(a=-2\), \(b=1\), \(c=-8\), and \(d=1\). The denominator is \(b+d=1+1=2>0\), so there is exactly one intersection. Its price is \(P^*=(-2-(-8))/2=6/2=3\) dollars.
The demand quantity at that price is \(Q_D(3)=-2-3=-5\) units. Supply gives \(Q_S(3)=-8+3=-5\) units, confirming the intersection. The price is nonnegative, but the quantity is negative, so the point is infeasible.
This example uses intercepts that do not describe conventional nonnegative quantities at a zero price. It illustrates why a positive price alone is not a sufficient test when intercept restrictions are absent: \(ad+bc=(-2)(1)+(-8)(1)=-10<0\), so \(Q^*=-10/2=-5\).
Parallel Schedules and Common Mistakes
The standard slope conditions rule out parallel schedules because demand has slope \(-b<0\) and supply has slope \(d\geq0\). If schedules outside those standard conditions are considered, compare their slopes before applying the equilibrium formulas. For instance, \(Q_D=18-2P\) and \(Q_S=6-2P\) have the same slope but different intercepts, so they never intersect. If both schedules instead were \(18-2P\), they would coincide, and every point on the shared line would be an algebraic solution. Neither case has exactly one intersection.
For the standard schedules, \(b>0\) and \(d\geq0\), so \(b+d>0\) and the algebraic intersection is unique.
Use \(P^*=(a-c)/(b+d)\), retaining the sign rather than assuming the result is positive.
Use \(Q^*=(ad+bc)/(b+d)\), then substitute the price into both schedules to confirm equal quantities.
Accept the intersection as a feasible equilibrium only if both \(P^*\geq0\) and \(Q^*\geq0\).
A frequent exam trap is treating “unique” and “feasible” as synonyms. For example, a positive denominator establishes uniqueness, not nonnegative price or quantity. Another trap is checking only the price condition \(a\geq c\). That condition is sufficient for feasibility when the intercepts are both nonnegative, but in the general case the quantity numerator \(ad+bc\) also needs to be checked. Keep the algebraic existence test, the uniqueness test, and the feasibility test distinct.
Check Your Understanding
For each problem, distinguish the algebraic intersection from a feasible equilibrium. Quantities are measured in units unless stated otherwise.
- A market has \(Q_D=70-5P\) and \(Q_S=10+P\). Find the unique intersection and state whether it is feasible. (Answer: \(P^*=10\), \(Q^*=20\); unique and feasible.)
- Let \(Q_D=24-3P\) and \(Q_S=30+P\). Find both coordinates of the intersection and identify the feasibility issue. (Answer: \(P^*=-1.5\), \(Q^*=28.5\); unique algebraic intersection, but infeasible because price is negative.)
- For \(Q_D=5-2P\) and \(Q_S=-7+P\), calculate the equilibrium and check each schedule at the resulting price. (Answer: \(P^*=4\), \(Q^*=-3\); both schedules give \(-3\), so the unique intersection is infeasible because quantity is negative.)
- Two schedules are \(Q_D=16-3P\) and \(Q_S=4+3P\). Decide whether the intersection is unique and feasible, showing the price and quantity. (Answer: \(P^*=2\), \(Q^*=10\); unique and feasible.)
- Compare \(Q_D=20-2P\) with (a) \(Q_S=8-2P\) and (b) \(Q_S=20-2P\). State the number of algebraic intersections in each case. (Answer: (a) none, because the schedules are parallel and distinct; (b) infinitely many, because the schedules coincide.)