Category Archives: MAC1105

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.

1.4 Quadratic Equations

Solving a Quadratic Equation using the Quadratic Formula: Example 1 of 1

Example:  Solve the quadratic equation with the quadratic formula.

2x^2+7=4x

Solution:

The original equation.
2x^2+7=4x
Write the equation so that all of the terms are on the same side.
2x^2+7=4x
2x^2+7-4x=4x-4x
2x^2-4x+7=0
Identify a, b and c.
a=2, b=-4, c=7
2x^2-4x+7=0
The quadratic formula.
x={-b pm sqrt{b^2-4ac}}/{2a}
Substitute the values into the quadratic formula.
x={-b pm sqrt{b^2-4ac}}/{2a}
x={-(-4) pm sqrt{(-4)^2-4(2)(7)}}/{2(2)}
Simplify using order of operations by applying powers, multiplying and then subtracting.
x={-(-4) pm sqrt{(-4)^2-4(2)(7)}}/{2(2)}
x={4 pm sqrt{16-56}}/{4}
x={4 pm sqrt{-40}}/{4}
Simplify the radical by looking for perfect square factors of 40.
x={4 pm sqrt{-40}}/{4}
x={4 pm sqrt{-1(4)(10)}}/{4}
x={4 pm 2i sqrt{10}}/{4}
Simplify by canceling the common factor of 2 out of the terms in the numerator and denominator.
x={4 pm 2i sqrt{10}}/{4}
x={2(2) pm 2i sqrt{10}}/{2(2)}
x={2 pm i sqrt{10}}/{2}

The solutions to the quadratic equation are x={2 + i sqrt{10}}/{2} and x={2 - i sqrt{10}}/{2}.

1.4 Quadratic Equations

Solving Quadratic Equations by Factoring: Trinomial a=1

Example: Solve the quadratic equation by factoring.

x^2+x=20

Solution:

The original equation
x^2+x=20
Write the equation with all the terms on one side of the equation and zero on the other side of the equation. x^2+x=20
x^2+x-20=20-20
x^2+x-20=0
Factor the expression on one side. x^2+x-20=0
(x+5)(x-4)=0
Use the zero product property and set each factor equal to zero. (x+5)(x-4)=0
x+5=0 or x-4=0
Solve each equation. x+5=0 or x-4=0
x+5-5=0-5 or x-4+4=0+4
x=-5 or x=4

Check: x=4

x^2+x=20
(4)^2+4=20
16+4=20
20=20

Since the value of 4 makes the equation true, 4 is a solution to the equation.

Check: x=-5

x^2+x=20
(-5)^2+(-5)=20
25-5=20
20=20

Since the value of -5 makes the equation true, -5 is a solution to the equation.

1.4 Quadratic Equations

Solving a Quadratic Equation: The Square Root Method Example 1 of 1

Example: Solve the quadratic equation with completing the square.

x^2-9=0

The original quadratic equation.
x^2-9=0
Rewrite the quadratic equation so that the square and everything that the square applies to are on one side of the equation.  This is called isolating the square. x^2-9=0
x^2-9+9=0+9
x^2=9
Cancel out the square by square rooting both sides. x^2=9
sqrt{x^2}=sqrt{9}
x=pm 3
The remaining equations are already solved.  The solutions to the equation are 3 and -3. x=-3 or x=3