Initial Publication Date: April 12, 2016

Simultaneous Equations: Sample Problems

Problem 1:

You are given an economic model for a market that explains price in terms of quantity supplied and quantity demanded. You want to know what the quantity and price will be if the market reaches equilibrium.

Use the substitution method to find the equilibrium price and quantity, given the following supply and demand functions.

Supply: $P = 40 - 4Q$
Demand: $3P = 29 + 2Q$

Show Problem 1 Solution

Problem 1 Solution

Read both equations and identify the unknown variables.

The unknown variables in both equations are Q (quantity) and P (price).
Pick an equation and isolate one of its variables.

Let's choose the equation P = 40 - 4Q and solve it for Q.
$P \color{red}{+ 4Q} = 40 - 4Q \color{red}{+ 4Q}$

$\require{cancel}P + 4Q = 40 \cancel{- 4Q + 4Q}$

$P + 4Q = 40$

$P + 4Q \color{red}{ - P} = 40 \color{red}{ - P}$

$\cancel{P} + 4Q - \cancel{P} = 40 - P$

$4Q = 40 - P$

$4Q\color{red}{/4} = (40 - P)\color{red}{/4}$

$\cancel{4}Q/\cancel{4} = (40 - P)/4$

$Q = (40 - P)/4$

$Q = 40/4 - P/4$

$Q = 10 - P/4$
Substitute this new equation for the variable in the other equation and simplify.

Substitute Q = 10 - P/4 into the equation 3P = 29 + 2Q and simplify.
$3P = 29 + 2(10 - P/4)$

$3P = 29 + 20 -2P/4$

$3P = 49 - P/2$

$3P \color{red}{+ P/2} = 49 - P/2 \color{red}{+ P/2}$

$3P + P/2= 49 - \cancel{P/2 + P/2}$

$3P + P/2 = 49$

$6P/2 + P/2 = 49$

$7P/2 = 49$

$\color{red}{(2/7) \times} (7P/2) = \color{red}{(2/7) \times} 49$

$\cancel{(2/7)} \times (\cancel{7}P/\cancel{2}) = (2/7) \times 49$

$P = 14$
Take the equation you derived in Step #2, plug in the value you found in Step #3, and simplify.

Plug P = 14 into Q = 10 - P/4 and simplify.
$Q = 10 - 14/4$

$Q = 10 - 3.5$

$Q = 6.5$
Plug the values you found for both variables back into each equation and verify that the equations both hold true.

Plug in Q = 6.5 and P = 14 into P = 40 - 4Q and simplify.
$14 = 40 - 4(6.5)$

$14 = 40 - 26$

$14 = 14$

Plug in Q = 6.5 and P = 14 into 3P = 29 + 2Q and simplify.
$3(14) = 29 + 2(6.5)$

$42 = 29 + 13$

$42 = 42$

Problem 2:

You are given an economic model for a market that explains price in terms of quantity supplied and quantity demanded. You want to know what the quantity and price will be if the market reaches equilibrium.

In this case, use the elimination method --- notice that it will be easy to use because the coefficients and sign for Q are the same.

Supply: $Q = 10 + P$ Demand: $Q = 100 - 2P$

Show Problem 2 Solution

Problem 2 Solution

Read both equations and identify the unknown variables.

The unknown variables in both equations are Q (quantity) and P (price).
Choose the variable you want to eliminate first and compare the signs and magnitudes of the coefficients on this variable in both equations.

The coefficients on Q are the same magnitude and the same sign, so we subtract the equations and then solve for the remaining variable.
$Q = 10 + P$ $-(Q = 100 - 2P)$

$0 = -90 + 3P$

$0 \color{red}{+ 90} = -90 + 3P \color{red}{+ 90}$

$0 + 90 = \cancel{-90} + 3P + \cancel{90}$

$90 = 3P$

$90\color{red}{/3} = 3P\color{red}{/3}$

$90/3 = \cancel{3}P/\cancel{3}$

$P = 30$
Plug the value you got from step #3 into one of your equations and solve for the other variable.

Plug P = 30 into Q = 10 + P and simplify.
$Q = 10 + 30$

$Q = 40$
Plug the values you found for both variables back into each equation and verify that the equations both hold true.

Plug in Q = 40 and P = 30 into Q = 10 + P and simplify.
$40 = 10 + 30$

$40 = 40$

Plug in Q = 40 and P = 30 into Q = 100 - 2P and simplify.
$40 = 100 - 2(30)$

$40 = 100 - 60$

$40 = 40$

Problem 3:

You are given the following equations for the labor supply and labor demand at a firm. If the market moves to equilibrium, what will be the wages (w) and quantity of labor (Q)?

In this case, use the substitution method.

Labor Supply: $Q = -100 + 4w$ Labor Demand: $2Q = 250 - w$

Show Problem 3 Solution

Problem 3 Solution

Read both equations and identify the unknown variables.

The unknown variables in both equations are Q (quantity) and w (wage).
Pick an equation and isolate one of its variables.

Let's choose the equation Q = 100 + 4w and solve for w.
$Q \color{red}{+100} = -100 + 4w \color{red}{+100}$

$Q +100 = \cancel{-100} + 4w \cancel{+100}$

$Q + 100 = 4w$

$(Q + 100)\color{red}{/4} = 4w\color{red}{/4}$

$(Q + 100)\color{red}{/4} = \cancel{4}w\color{red}{/\cancel{4}}$

$w = (Q + 100)/4$

$w = Q/4 + 100/4$

$w = Q/4 + 25$
Substitute this new equation for the variable in the other equation and simplify.

Substitute w = Q/4 + 25 into 2Q = 250 - w and simplify.
$2Q = 250 - (Q/4 + 25)$

$2Q = 250 - Q/4 - 25$

$2Q = 225 - Q/4$

$2Q \color{red}{+ Q/4} = 225 - Q/4 \color{red}{+ Q/4}$

$2Q + Q/4 = 225 - \cancel{Q/4 + Q/4}$

$2Q + Q/4 = 225$

$8Q/4 + Q/4 = 225$

$9Q/4 = 225$

$\color{red}{4/9 \times } (9Q/4) = \color{red}{4/9 \times }(225)$

$Q = 100$
Take the equation you derived in Step #2, plug in the value you found in Step #3, and simplify.

Plug Q = 100 into w = Q/4 + 25 and simplify.
$w = 100/4 + 25$

$w = 25 + 25$

$w = 50$
Plug the values you found for both variables back into each equation and verify that the equations both hold true.

Plug in Q = 100 and w = 50 into Q = -100 + 4w and simplify.
$100 = -100 + 4(50)$

$100 = -100 + 200$

$100 = 100$

Plug in Q = 100 and w = 50 into 2Q = 250 - w and simplify.
$2(100) = 250 - 50$

$200 = 200$

Problem 4:

You are given an economic model for a market that explains the quantity of loans available (Q) based in the interest rate (r). You want to know what the quantity and interest rate will be if the market reaches equilibrium.

In this case, use the elimination method --- notice that the equations for Q have the same magnitude and sign.

Supply: $Q = 150$ Demand: $Q = 350 - 5r$

Show Problem 4 Solution

Read both equations and identify the unknown variables.

The unknown variables in both equations are Q (quantity) and r (interest rate).
Choose the variable you want to eliminate first and compare the signs and magnitudes of the coefficients on this variable in both equations.

The coefficients on Q are the same magnitude and the same sign, so we subtract the equations and then solve for the remaining variable.
$Q = 150$ $-(Q = 350 - 5r)$

$0 = -200 + 5r$

$0 \color{red}{+ 200} = -200 + 5r \color{red}{+ 200}$

$0 + 200 = -\cancel{200} + 5r + \cancel{200}$

$200 = 5r$

$200\color{red}{/5} = 5r\color{red}{/5}$

$200/5 = \cancel{5}r\cancel{/5}$

$r = 40$
Plug the value you got from step #2 into one of your equations and solve for the other variable.

Plug r = 40 into Q = 350 -5r and simplify.
$Q = 350 - 5(40)$

$Q = 350 - 200$

$Q = 150$
Plug the values you found for both variables back into the equation and verify that the equation holds true.

Plug in Q = 150 and r = 40 into Q = 350 - 5r and simplify.
$150 = 350 - 5(40)$

$150 = 350 - 200$

$150 = 150$

Problem 5:

A economist wants to know how a consumer's utility will be maximized by consuming two goods called X and Y. The maximizing choice of the two goods is described by the following two equations.

$Y/X = 1/2$ $5X + 10Y = 100$

Use the substitution method to solve for X and Y.

Show Problem 5 Solution

Problem 5 Solution

Read both equations and identify the unknown variables.

The unknown variables in both equations are goods X and Y.
Pick an equation and isolate one of its variables.

Let's choose the equation Y/X = 1/2 and solve it for Y.
$(Y/X)\color{red}{ \times X} = (1/2)\color{red}{ \times X}$

$(Y/\cancel{X) \times X} = (1/2) \times X$

$Y = X/2$
Substitute this new equation for the variable in the other equation and simplify.

Substitute Y = X/2 into the equation 5X + 10Y = 100 and simplify.
$5X + 10(X/2) = 100$

$5X + 5X = 100$

$10X = 100$

$10X\color{red}{/10} = 100\color{red}{/10}$

$\cancel{10}X/\cancel{10} = 100/10$

$X = 10$
Take the equation you derived in Step #2, plug in the value you found in Step #3, and simplify.

Plug X= 10 into Y = X/2 and simplify.
$Y = 10/2$

$Y = 5$
Plug the values you found for both variables back into each equation and verify that the equations both hold true.

Plug in X = 10 and Y = 5 into Y/X = 1/2 and simplify.
$5/10 = 1/2$

$1/2 = 1/2$

Plug in X = 10 and Y = 5 into 5X + 10Y = 100 and simplify.
$5(10) + 10(5) = 100$

$50 + 50 = 100$

$100 = 100$

Problem 6:

A firm's cost minimizing choice of labor (L) and capital (K) is described by the following two equations.

$2K/L = 5$ $2{,}500 = L^2K$

Use the substitution method to solve for L and K that minimizes cost.

Show Problem 6 Solution

Problem 6 Solution

Read both equations and identify the unknown variables.

The unknown variables in both equations are goods L and K.
Pick an equation and isolate one of its variables.

Let's choose the equation 2K/L and solve it for K.
$(2K/L)\color{red}{ \times L} = 5\color{red}{ \times L}$

$(2K/\cancel{L) \times L} = 5L$

$2K = 5L$

$2K\color{red}{/2} = 5L\color{red}{/2}$

$\cancel{2}K\color{red}{/\cancel{2}} = 5L\color{red}{/2}$

$K = 2.5L$
Substitute this new equation for the variable in the other equation and simplify.

Substitute K = 2.5L into the equation 2,500 = L²K and simplify.
$2{,}500 = L^2(2.5L)$

$2{,}500 = 2.5L^3$

$2{,}500\color{red}{/2.5} = (2.5L^3)\color{red}{/2.5}$

$1{,}000 = L^3$

$1{,}000^{1/3} = (L^3)^{1/3}$

$1{,}000^{1/3} = L^{3 \times 1/3}$

$L = 1{,}000^{1/3}$

$L = 10$
Take the equation you derived in Step #2, plug in the value you found in Step #3, and simplify.

Plug L = 10 into K = 2.5L and simplify.
$K = 2.5(10)$

$K = 25$
Plug the values you found for both variables back into each equation and verify that the equations both hold true.

Plug in L = 10 and K = 25 into 2K/L = 5 and simplify.
$2(25)/10 = 5$

$50/10 = 5$

$5 = 5$

Plug in L = 10 and K = 25 into 2,500 = L²K and simplify.
$2{,}500 = (10^2)(25)$

$2{,}500 = 100 \times 25$

$2{,}500 = 2{,}500$

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