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.
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.
