Algebra 1 part 1 and 2
Algebra 1 part 1 and 2
Algebra 1 is a fundamental and relatively basic form of algebra taught to students who are presumed to have little or no formal knowledge of mathematics beyond arithmetic. The major different between algebra and arithmetic is the inclusion of variables.In arithmetic operations we use only numbers and arithmetic operations, while in algebra we use variables instead of numbers.
Algebra is a branch of mathematics. Algebra plays an important role in our day to day life. The algebra 1 part 1 executes the four basic operations such as addition, subtraction, multiplication and division. The most important terms of practice algebra course are variables, constant, coefficients, exponents, terms and expressions. In Algebra, besides numerals we use symbols and alphabets in place of unknown numbers to make a statement. Hence, algebra 1 part 1 may be regarded as an extension of Arithmetic.
Algebra 1 part 2 is the division of algebra regarding the study of the regulations of operations and relations, and the creations and concepts arising from them, including Inequalities, polynomials, system of equations and algebraic structures. (Source: From Wikipedia)
Basic term involving algebra 1 part 1:
Variables
Algebraic variables are the alphabetical characters which are used for assigning the value. While solving the algebraic equation value of the variable will be changed. Widely used variables are x, y, z
Constant
Algebraic constants are the value whose value never changes while solving the algebraic equation. In 3y+2, the value 2 is the constant.
Term
Terms of the algebraic expression is concatenated to form the algebraic expression by the arithmetic operations such as addition, subtraction, multiplication and division. In the following example 5n2 + 4n the terms 5n2, 4n are combined to form the algebraic expression 5n2 + 4n by the addition operation ( + )
Order of the operation in algebra 1 part 1:
Step 1: Solve the operations within the parentheses,
Step 2: Solve all exponential operation in the equation or expression,
Step 3: Solve all multiplication or division operations,
Step 4.: Finally, Solve all addition or subtraction operations.
Procedure for solving algebraic equation:
Step 1: If we want to solve the equation for x, we have to move the x to on side of the equal sign, which is similar to, 5x+5+25=125.
Step 2:Do the 'reverse operations.' This means we would want to add -25. So we will get 5x+5=100.
Step 3: Now we have to add -5 on both sides of the equation. So we will get 5x=95.
Step 4: Now we have to divide both sides by 5. That is 5x/5=95/5. Therefore, x=19.
Example problems for algebra 1 part 1
Example 1:
x+y=9
-2x+y=0
Substitution method:
x+y=9 ---------------------- equation 1
-2x+y=0---------------------- equation 2
if we add the equations 1 and 2, we will get
3x=9
3x/3=9/3 (both sides are divided by 3)
x=3
Substitue x=3 in the equation 1, so we will get
3+y=9
3-3+y=9-3 (-3 is added on both sides)
y=6
Elimination method:
x+y=9 ---------------------- equation 1
-2x+y=0---------------------- equation 2
take the equation 1
x+y=9
x+y-y=9-y (-y is added on the both sides)
x=9-y
substitute x=9-y in the equation 2, we will get
-2(9-y ) +y=0
-18+2y+y=0
-18+3y=0
-18+18+3y=0+18
3y=18
3y/3=18/3 (both sides are divided by 3)
Y=6
Substitute y=6 in the equation 1
X+6=9
X+6-6=9-6 ( Add -3 on the both sides)
X=3
Example 2:
Solve for the variable x, 2x+7=25
Solution
2x+7=25
2x+7-7=25 -7(Add -7 on both sides)
2x = 18
2x/2=18/2(both sides divided by 2)
x=9
Example 3:
Simplify the following expression 4*(44÷2)+x=0
Solution:
4*(44÷2)+x=0 (evaluate the expression inside the parenthesis)
4*(22)+x=0
88+x=0
88-88+x=0-88 (add -28 on both sides)
X=-88
Example problems in algebra 1 part 2
Algebra 1 part 2 problem 1:
The product of two consecutive positive odd integers is 99. Find those two numbers.
Solution:
Step 1:
In the given problem, product of two consecutive positive odd integers is 99.
We are requested to find the two consecutive positive odd integers.
Let the first integer be y.
Then the consecutive positive odd integer is y + 2.
Given that the product of y and y + 2 = 99
=> y(y + 2) = 99
=> y2 + 2y - 99 = 0
So, we arrived at a quadratic equation y2 + 2y - 99 = 0.
Step 2:
We can solve the quadratic equation by using the following method,
Factoring method:
y2 + 11y – 9y - 99 = 0
=> y(y + 11) - 9(y + 11) = 0
=> (y + 11) (y - 9) = 0
=> y + 11 = 0 or y - 9 = 0
=> y = -11 or y = 9
Step 3:
In a quadratic equation, we can take only positive factor. So y= -11 is negligible.(product of two consecutive positive odd integers)
Therefore y = 9.
If y = 9, then y + 2 = 9 + 2 = 11.
Step 4:
Hence, the two consecutive positive odd numbers are 11 and 9.
Algebra 1 part 2 problem 2:
Use quadratic formula to solve y2 – 2y - 35 = 0.
Solution:
Step 1:
Given equation is y2 - 2y - 35 = 0.
From the problem a = 1, b = -2, and c = -35.
The quadratic formula is
` y= (-b+-sqrt(b^2-4ac))/(2a)`
Step 2:
Substitute the values of a, b, and c in the above quadratic formula to find the value of y.
` y= (-(-2)+-sqrt((-2)^2-4(1)(-35)))/(2(1))`
` y= (2+-sqrt(4+140))/(2)`
` y= (-2+-12)/(2)`
` y= (2(-1+-6))/(2)`
`y= -1+6 or y= -1-6`
`y= 5 or y= -7`
Step 3:
Therefore the solution set is {5, -7}.
Answer: {5,-7}
Practice problems in algebra 1 part 2:
Find differentiate (discriminate) value of y2 – 12y + 36 = 0 and determine the number of real roots.
Solve the following quadratic equation using quadratic formula:
6y2 + y – 48 = 0
Answer:
1. Discriminate value =0, Equation has double root.
2. (2.7, -2.9)
Algebra 1 is a fundamental and relatively basic form of algebra taught to students who are presumed to have little or no formal knowledge of mathematics beyond arithmetic. The major different between algebra and arithmetic is the inclusion of variables.In arithmetic operations we use only numbers and arithmetic operations, while in algebra we use variables instead of numbers.
Algebra is a branch of mathematics. Algebra plays an important role in our day to day life. The algebra 1 part 1 executes the four basic operations such as addition, subtraction, multiplication and division. The most important terms of practice algebra course are variables, constant, coefficients, exponents, terms and expressions. In Algebra, besides numerals we use symbols and alphabets in place of unknown numbers to make a statement. Hence, algebra 1 part 1 may be regarded as an extension of Arithmetic.
Algebra 1 part 2 is the division of algebra regarding the study of the regulations of operations and relations, and the creations and concepts arising from them, including Inequalities, polynomials, system of equations and algebraic structures. (Source: From Wikipedia)
Basic term involving algebra 1 part 1:
Variables
Algebraic variables are the alphabetical characters which are used for assigning the value. While solving the algebraic equation value of the variable will be changed. Widely used variables are x, y, z
Constant
Algebraic constants are the value whose value never changes while solving the algebraic equation. In 3y+2, the value 2 is the constant.
Term
Terms of the algebraic expression is concatenated to form the algebraic expression by the arithmetic operations such as addition, subtraction, multiplication and division. In the following example 5n2 + 4n the terms 5n2, 4n are combined to form the algebraic expression 5n2 + 4n by the addition operation ( + )
Order of the operation in algebra 1 part 1:
Step 1: Solve the operations within the parentheses,
Step 2: Solve all exponential operation in the equation or expression,
Step 3: Solve all multiplication or division operations,
Step 4.: Finally, Solve all addition or subtraction operations.
Procedure for solving algebraic equation:
Step 1: If we want to solve the equation for x, we have to move the x to on side of the equal sign, which is similar to, 5x+5+25=125.
Step 2:Do the 'reverse operations.' This means we would want to add -25. So we will get 5x+5=100.
Step 3: Now we have to add -5 on both sides of the equation. So we will get 5x=95.
Step 4: Now we have to divide both sides by 5. That is 5x/5=95/5. Therefore, x=19.
Example problems for algebra 1 part 1
Example 1:
x+y=9
-2x+y=0
Substitution method:
x+y=9 ---------------------- equation 1
-2x+y=0---------------------- equation 2
if we add the equations 1 and 2, we will get
3x=9
3x/3=9/3 (both sides are divided by 3)
x=3
Substitue x=3 in the equation 1, so we will get
3+y=9
3-3+y=9-3 (-3 is added on both sides)
y=6
Elimination method:
x+y=9 ---------------------- equation 1
-2x+y=0---------------------- equation 2
take the equation 1
x+y=9
x+y-y=9-y (-y is added on the both sides)
x=9-y
substitute x=9-y in the equation 2, we will get
-2(9-y ) +y=0
-18+2y+y=0
-18+3y=0
-18+18+3y=0+18
3y=18
3y/3=18/3 (both sides are divided by 3)
Y=6
Substitute y=6 in the equation 1
X+6=9
X+6-6=9-6 ( Add -3 on the both sides)
X=3
Example 2:
Solve for the variable x, 2x+7=25
Solution
2x+7=25
2x+7-7=25 -7(Add -7 on both sides)
2x = 18
2x/2=18/2(both sides divided by 2)
x=9
Example 3:
Simplify the following expression 4*(44÷2)+x=0
Solution:
4*(44÷2)+x=0 (evaluate the expression inside the parenthesis)
4*(22)+x=0
88+x=0
88-88+x=0-88 (add -28 on both sides)
X=-88
Example problems in algebra 1 part 2
Algebra 1 part 2 problem 1:
The product of two consecutive positive odd integers is 99. Find those two numbers.
Solution:
Step 1:
In the given problem, product of two consecutive positive odd integers is 99.
We are requested to find the two consecutive positive odd integers.
Let the first integer be y.
Then the consecutive positive odd integer is y + 2.
Given that the product of y and y + 2 = 99
=> y(y + 2) = 99
=> y2 + 2y - 99 = 0
So, we arrived at a quadratic equation y2 + 2y - 99 = 0.
Step 2:
We can solve the quadratic equation by using the following method,
Factoring method:
y2 + 11y – 9y - 99 = 0
=> y(y + 11) - 9(y + 11) = 0
=> (y + 11) (y - 9) = 0
=> y + 11 = 0 or y - 9 = 0
=> y = -11 or y = 9
Step 3:
In a quadratic equation, we can take only positive factor. So y= -11 is negligible.(product of two consecutive positive odd integers)
Therefore y = 9.
If y = 9, then y + 2 = 9 + 2 = 11.
Step 4:
Hence, the two consecutive positive odd numbers are 11 and 9.
Algebra 1 part 2 problem 2:
Use quadratic formula to solve y2 – 2y - 35 = 0.
Solution:
Step 1:
Given equation is y2 - 2y - 35 = 0.
From the problem a = 1, b = -2, and c = -35.
The quadratic formula is
` y= (-b+-sqrt(b^2-4ac))/(2a)`
Step 2:
Substitute the values of a, b, and c in the above quadratic formula to find the value of y.
` y= (-(-2)+-sqrt((-2)^2-4(1)(-35)))/(2(1))`
` y= (2+-sqrt(4+140))/(2)`
` y= (-2+-12)/(2)`
` y= (2(-1+-6))/(2)`
`y= -1+6 or y= -1-6`
`y= 5 or y= -7`
Step 3:
Therefore the solution set is {5, -7}.
Answer: {5,-7}
Practice problems in algebra 1 part 2:
Find differentiate (discriminate) value of y2 – 12y + 36 = 0 and determine the number of real roots.
Solve the following quadratic equation using quadratic formula:
6y2 + y – 48 = 0
Answer:
1. Discriminate value =0, Equation has double root.
2. (2.7, -2.9)
