Algebra ii Math

Introduction to  algebra ii math:

Algebra ii is one of the important categories in mathematics.

Algebra ii contains the following topics

Solving equations and inequalities

Graphs and functions

Polynomials and factoring

Fractional Expressions

Powers and roots

Complex numbers

Quadratic equation

In this article we shall discuss about the some example problems on algebra ii maths. I like to share this Fractional Coefficients with you all through my article.


Problems on solving equations in algebra ii maths:

X+2y=5

2x-y=10

Solution:

Step 1: To make anyone of the variable as common coefficients

From the above problem multiply the equation 2 by 2 we get,

X+2y=5

4x-2y=20

------------

5x=25

Step 2: Divide by 5 on both sides

5x/5 =25/5

X=5

Step 3: Substitute the value of x in first equation

X+2y=5

5+2y=5

Step 4: Subtract 5 on both sides,

5+2y-5=5-5

2y=0

Y=0

Problems on polynomials and factoring in algebra ii maths:

Factor: x^2-5x-150

Solution:

Step 1: The above equation is quadratic equation because it has the highest degree of 2

Multiply the coefficient of x^2 and constant term

1*-150 = -150

Step 2: Find the factor for the product term. Sum of the factor is equal to coefficients of x

-150 = -15*10

-15+10 = -5

Step 3: Now the equation is written as,

x^2-15x+10x-150

Take the common term outside,

x(x-15)+10(x-15)

(x+10) (x-15)

Understanding Inverse Function Definition is always challenging for me but thanks to all math help websites to help me out.

Problems on fractional expressions in algebra ii maths:


(x+5)/(y-10) + (2x+3)/(y-2)

Solution:

Here the denominators are different. To make the common denominator take the LCD of (y-10) and (y-2).       LCD= (y-10)(y-2)

(x+5)/(y-10) + (2x+3)/(y-2)

(x+5)/(y-10) * (y-2)/(y-2) + (2x+3)/(y-2) * (y-10)/(y-10)

(x+5)(y-2)/(y-10)(y-2) +(2x+3)(y-10)/(y-2)(y-10)

((x+5)(y-2)+ (2x+3)(y-10)) /(y-10)(y-2)

(xy-2x+5y-10)+(2xy-20x+3y-30) / (y-10)(y-2)

(3xy-22x+8y-40) / (y-10)(y-2)

Problems on quadratic equation in algebra ii maths:

Solve the quadratic equation by factoring method and find roots?

2x^2+x-10=0

Solution:

Step 1: Multiply the coefficient of x^2 and the constant term,

2*-10 =-20 (product term)

Step 2: Find the factors for the product term

-20 ---- > -5 *4 = -20 (factors -5 and 4)

-5+4 = 1 (1 is equal to the coefficient of x)

Step 3: Split the coefficient of x as factors

2x^2+x-10=0

2x^2-5x+4x-10=0

Step 4: Taking the common term x for the first two term and 2 for the next two terms

x(2x-5) +2(2x-5) =0

(x+2) (2x-5)=0.

Now set (x+2) =0; x=-2;

(2x-5)=0; x=5/2.

The roots are x=-5/2 and -2.