Category Archives: MAC1140

7.1 Systems of Linear Equations in Two Variables, 8.1 Systems of Linear Equations: Substitution and Elimination

Application of Systems of Linear Equations

Problem:

Benjamin & Associates, a real estate developer, recently built condominiums in McCall, Idaho.  The condos were either two-bedroom units or three-bedroom units.  If the total number of bedrooms in the entire complex is 498, how many two-bedroom units are there?  How many three-bedroom units are there?

Solution:

Assign variables to the values we are looking for in the equation.

Let x be the number of two-bedroom units.
Let y be the number of three-bedroom units.

Create equations using the information given in the problem.

Since there are 199 condos built in the complex, the number of two-bedroom units plus the three bedroom units should equal the total units of 199.

x+y=199

Since there are a total of 498 bedrooms in the complex, 2x represents number of bedrooms coming from two-bedroom units, and 3x represents number of bedrooms coming from three-bedroom units,  the number of bedrooms from two-bedroom units plus the number of bedrooms from three-bedroom units should equal to the total number of bedrooms of 498.

2x+3y=498

 

Solve the system of equations.

x+y=199
2x+3y=498

 

Solve one equation for one of the variables.  Choose to solve for x in the first equation since it doesn’t have a coefficient and fractions can be avoided that way.
x+y=199
x+y-y=199-y
x=199-y
Substitute the expression into the other equation.
2x+3y=498
2(199-y)+3y=498
Solve for the other variable. 

  • Use the distributive property to remove parenthesis.
  • Combine like terms
  • Isolate the variable on one side of the equation
2(199-y)+3y=498
398-2y+3y=498
398+y=498
398+y-398=498-398
y=100

y represents the number of three-bedroom units.  There are 100 three-bedroom units.

x represents the number of two-bedroom units.  There are 199-100=99 two-bedroom units.

1.6 Other Types of Equations, 5.1 Polynomial Functions and Models

Polynomial Equation (Solve by factoring with the grouping method)

Example:  Solve the polynomial equation

y^3+y^2=4y+4

Solution:  Solve the polynomial equation by factoring.

The original equation.
y^3+y^2=4y+4
Write the equation so that all the terms are on the same side.

  • Subtract 4y
  • Subtract 4
y^3+y^2=4y+4
y^3+y^2-4y=4y-4y+4
y^3+y^2-4y=4
y^3+y^2-4y-4=4-4
y^3+y^2-4y-4=0
Group two terms pairs of terms and factor the greatest common factor from each group.

  • The greatest common factor for the first group is y^2
  • The greatest common factor for the second group is -4
y^3+y^2-4y-4=0
(y^3+y^2)+(-4y-4)=0
(y^3+y^2)+(-4y-4)=0
y^2(y+1)+-4(y+1)=0
Factor the common binomial from each term.

  • The common binomial is (y+1)
  • The other factor is formed using the coefficients of the parenthesis
y^2(y+1)+-4(y+1)=0
(y+1)(y^2-4)=0
Continue to factor completely by factoring the difference of squares in the second parenthesis
(y+1)(y^2-4)=0
(y+1)(y+2)(y-2)=0
Apply the zero product property by setting each factor equal to zero.
y+1=0 or y+2=0 or y-2=0
Solve each remaining equation.
y+1=0 or y+2=0 or y-2=0
y+1-1=0-1 or y+2-2=0-2 or y-2+2=0+2
y=-1 or y=-2 or y=2

The solutions to the polynomial equation y^3+y^2=4y+4 are y=-1 or y=-2 or y=2.