Section 2 - Valdosta State University · 2019-09-06 · Solve the system by substitution. = − = +...

Section 2.3

Systems of Linear Equation in Two

Variables

Copyright © 2017, 2013, 2010 Pearson Education, Inc.

2Copyright © 2017, 2013, 2010 Pearson Education, Inc.

Objectives

Solve systems of linear equations graphicallySolve systems of linear equations algebraically with the substitution methodSolve systems of linear equations algebraically by eliminationModel systems of equations to solve problemsDetermine if a system of linear equations is inconsistent or dependent


System of Linear Equations

A system of linear equations is a collection of linear equations containing the same set of variables.

A solution to a system of equations in two variables is an ordered pair that satisfies both equations of the system.


Break-Even

A company is said to break even when the total revenue equals the total cost − that is, when R(x) = C(x).

Because profit P(x) = R(x) – C(x), we can also say that company breaks even if the profit for the product is zero.


Example: Break-Even

Suppose a company has its total revenue for a product given by R = 5585x and its total cost given by C = 61,740 + 440x, where x is the number of thousands of tons of the product that are produced and sold per year. The company is said to break even when the total revenue equals the total cost—that is, when R = C. Find the number of thousands of tons of the product that gives break-even and how much the revenue and cost are at that level of production.SolutionGraph the revenue as y1 and the cost function as y2.Use the intersection-method on the graphing calculator to find the break-even point.


Example (cont)

The company will break-even on this product if 12 thousand tons of the product are sold, when both the cost and revenue equal $67,020.


Example: Solving a System of Linear Equations

Solve the system.

SolutionTo solve this system with a graphing utility, we first solve both equations for y:

3 4 212 5 9

x yx y− =

+ = −

3 4 214 21 3

21 3 3 214 4

x yy x

x xy

− =− = −

− −= =

−

2 5 95 9 2

9 25

x yy x

xy

+ = −= − −− −

=


Example: Cont

Graphing these equations with a window that contains the point of intersection Figure (a)) and finding the point of intersection (Figure (b)) gives x = 3, y = –3, so the solution is (3, –3) .

(a) (b)


Solution by Substitution

One equation is solved for a variable and that variable is replaced by the equivalent expression in the other equation.


Market Equilibrium

Demand: The quantity of a product that is demanded by consumers.

Supply: The quantity that is supplied.

Surplus: If the price results in more units being supplied than demanded.

Shortfall: If the price results in fewer units being supplied than demanded.

Market equilibrium: Supply quantity equals demand quantity.


Example: Market EquilibriumSuppose that the daily demand for a product is given by p = 200 − 2q, where q is the number of units demanded and p isthe price per unit in dollars, and that the daily supply is given by p = 60 + 5q, where q is the number of units supplied and p is the price in dollars. Market equilibrium occurs when the supply quantity equals the demand quantity (and when the prices are equal)—that is, when q and p both satisfy the system.

a. If the price is $140, how many units are supplied and how many are demanded?b. Does this price give a surplus or a shortfall of the product?c. What price gives market equilibrium?

= − = +

200 260 5

p qp q


Example (cont)

a. If the price is $140, how many units are supplied and how many are demanded?

b. Does this price give a surplus or a shortfall of the product?Solutiona. 140 = 200 – 2q

q = 30and the number of units supplied satisfies 140 = 60 + 5q, or q = 16.

b. At this price $140, the quantity supplied is less than the quantity demanded, so a shortfall occurs.

= − = +

200 260 5

p qp q


Example (cont)

c. What price gives market equilibrium.SolutionSolve the system by substitution.

= − = +

200 260 5

p qp q

+ = −==

60 5 200 27 140

20

q qqq

= − = + =200 2(20) 60 5(20) 160p

The market equilibrium occurs when the number of units is 20 and the equilibrium price is $160 per unit.


Solution of Systems of Equations by Substitution


Example: Solution by Substitution

Solve the system by substitution.

SolutionSolve the second equation for y.

Substitute this expression for y in the first equation and solve.

+ = − + =

3 5 16 2 14

x yx y

+ == − += − +

6 2 142 6 14

3 7

x yy xy x

+ − + = −3 5( 3 7) 1x x


Example (cont)

Solve the system by substitution.

Solution

Substitute x = 3 into y = –3x + 7:y = –3(3) + 7 = –2

The solution is (3, –2).

+ = − + =

3 5 16 2 14

x yx y

+ − + = −3 5( 3 7) 1x x− + = −3 15 35 1x x

− = −12 36x= 3x


Solving a System of Two Equations in Two Variables by EliminationWe rewrite one or both of the equations in an equivalent form that allows us to eliminate one of the variables by adding or subtracting the equations.


Example: Solution by Elimination

Use the elimination method to solve the system.

SolutionMultiply the first equation by –2.

Add the equations.

+ = − + =

3 5 16 2 14

x yx y

− − =+ =

6 10 26 2 14x yx y

− − =+ =− =

6 10 26 2 14

8 16

x yx y

y= −2y


Example (cont)

Use the elimination method to solve the system.

SolutionSubstituting y = –2 into the first equation gives:

+ = − + =

3 5 16 2 14

x yx y

+ = −+ − = −

− = −==

3 5 13 5( 2) 1

3 10 13 9

3

x yx

xxx The solution is (3, –2).


Modeling Systems of Equations

Solution of real problems sometimes requires us to create two or more equations whose simultaneous solution is the solution to the problem.


Example: Investments

An investor has $200,000 to invest, part at 8% and the remainder at 3%. If her investment goal is to have an annual income of $13,600, how much should she put in each investment?SolutionLet x be the amount invested at 8%.Let y be the amount invested at 3%.x + y = 200,000The income from each investment can be written as 0.08x + 0.03y = 13,600.


Example: (cont)

An investor has $200,000 to invest, part at 8% and the remainder at 3%. If her investment goal is to have an annual income of $13,600, how much should she put in each investment?

SolutionSystem of equations

Multiply the first equation by –0.08 and add the two equations.

+ = + =

200,0000.08 0.03 13,600

x yx y

− − = −+ =

0.08 0.08 16,0000.08 0.03 13,600

x yx y

− = −0.05 2400y= 48,000y


Example: (cont)

An investor has $200,000 to invest, part at 8% and the remainder at 3%. If her investment goal is to have an annual income of $13,600, how much should she put in each investment?

SolutionSubstitute y = 48,000 into the first equation and solve for x.

$152,000 should be invested at 8% and $48,000 should be invested at 3%.

+ =+ =

=

200,00048,000 200,000

152,000

x yx

x


Example: Medication

A nurse has two solutions that contain different concentrations of a certain medication. One is a 12% concentration, and the other is an 8% concentration. How many cubic centimeters (cc) of each should she mix together to obtain 20 cc of a 9% solution?

SolutionWe begin by denoting the amount of the first solution by x and the amount of the second solution by y. The total amount of solution is the sum of x and y, so

x + y = 20


Example: (cont)

The total medication in the combined solution is 9% of 20 cc, or 0.09(20) = 1.8 cc, and the mixture is obtained by adding 0.12x and 0.08y, so

0.12x = 0.08y = 1.8

We can use substitution to solve the system 20 0.12 0.08 1.8

x yx y+ =

+ =


Example: (cont)

Substituting 20 – x for y in 0.12x + 0.08y = 1.8 gives 0.12x + 0.08(20 – x) = 1.8, and solving this equation gives

Thus, combining 5 cc of the first solution with 20 – 5 = 15 cc of the second solution gives 20 cc of the 9% solution.

( )0.12 0.08 20 1.80.12 1.6 0.08 1.8

0.04 0.25

x xx x

xx

+ − =

+ − ===


Dependent and Inconsistent Systems

Unique solution No solution Graphs are intersecting lines. Graphs are parallel lines;

the system is inconsistent.


Dependent and Inconsistent Systems

Infinite solutions Graphs are the same line; the system is dependent.


Example: Systems with NonuniqueSolutions

Use the elimination method to solve each of the following systems, if possible. Verify the solution graphically.a. b.

Solutiona. Multiply the first equation by –2.

− = − =

2 5 74 10 14

x yx y

− = − = −

4 2 102 3

x yx y

− + = −− =

4 10 144 10 14

x yx y

=0 0There are infinitely many solutions.


Example: (cont)

Use the elimination method to solve each of the following systems, if possible. Verify the solution graphically.a. b.

Solutionb. Multiply the second equation by –2.

− = − =

2 5 74 10 14

x yx y

− = − = −

4 2 102 3

x yx y

− =− + =

4 2 104 2 6

x yx y

=0 16The system is inconsistent.


Example: InvestmentMembers of an investment club have set a goal of earning 15% on the money they invest in stocks. They are considering buying two stocks, for which the cost per share and the projected growth per share (both in dollars) are summarized in the table.

Utility TechnologyCost/share $30 $45Growth/share $4.50 $6.75


Example: (Cont)a. If they have $180,000 to invest, how many shares of each stock should they buy to meet their goal?

b. If they buy 1800 shares of the utility stock, how many shares of the technology stock should they buy to meet their goal?


Example: (Cont)Solutiona. The money available to invest in stocks is $180,000, so if x is the number of utility shares and y is the number of technology shares purchased, we have

30x + 45y = 180,000A 15% return on their investment would be 0.15(180,000) = 27,000 dollars, so we have

4.50x + 6.75y = 27,000To find x and y, we solve the system

30 45 180,0004.50 6.75 27,000

x yx y+ =

+ =


Example: (Cont)Multiplying 4.5 times both sides of the first equation and –30 times both sides of the second equation gives

Adding the equations gives 0 = 0, so the system is dependent, with many solutions.

135 202.5 810,000135 202.5 810,000

x yx y+ =

− − = −


Example: (Cont)The number of shares of each stock that can be purchased satisfies both of the two original equations. In particular, it satisfies 30x + 45y = 180,000, so

with x between 0 and 6000 shares and y between 0 and 4000 shares (because neither x nor y can be negative).

b. Substituting 1800 for x in the equation gives y = 2800, so if they buy 1800 shares of the utility stock, they should buy 2800 shares of the technology stock to meet their goal.

180,000 30 12,000 2, or 45 3

x xy − −=

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Section 2 - Valdosta State University · 2019-09-06 · Solve the system by substitution. = − = +...

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