Pre algebra is mathematics course studied in middle school consisting of basic algebra operations such as factorization, simplification, solving equations. Get Pre Algebra help online with TutorVista. Solve problems, work on basic concepts and get help with your homework too. TutorVista's intensive Pre Algebra tutoring prepares you for high school algebra. Study with highly qualified tutors with many years of experience in tutoring students who will work with you to ensure that you understand the subject and score well in it.

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Pre algebra is the branch of the mathematics used to prepare the student for the study of algebra. Topics covered in pre algebra are Factors and Primes, GCF and LCM, Fractions, Square Roots, Basic Equations.

__Our intensive Pre Algebra tutoring covers topics like:__

- Relations and functions
- Factorization
- Linear equations
- Quadratic equations
- Simple roots and powers
- Variables

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Curriculum of pre algebra is given below:

- Introducing Whole Numbers
- Naming Numbers
- Rounding Whole Numbers
- Operations on Whole Numbers
- Word Problems with Whole Numbers.

- Comparing Integer Values
- Arithmetic Operations on Integers
- Introducing Exponents

- Reducing Fractions
- Converting Between Improper Fractions and Mixed Numbers
- Multiplying Fractions and Mixed Numbers
- Dividing Fractions and Mixed Numbers
- Finding Common Denominators
- Arithmetic Operations on Fractions and Mixed Numbers
- Comparing Fractions

- Naming Decimals
- Rounding Decimals
- Operations on Decimels
- Converting Between Fractions and Decimals

- Powers and Exponents
- Square Roots and Cube Roots
- Scientific Notation

- Converting Between Percents, Decimals, and Fractions
- Solving Percent Problems

- Evaluating Algebraic Expressions
- Combining Like Terms
- Solving Equations

- Introducing Ratios and Proportions
- Solving Proportions
- Solving Problems Using Ratios and Proportions

- Adding and Subtracting Polynomials
- Multiplying and Dividing Polynomials

- Introducing Geometric Figures
- Finding Perimeter and Circumferences
- Finding Areas and Volumes
- Working with Right Triangle
- Measuring Lengths, Weight, Capacities and Time.

- Plotting Points and Lines on the Coordinate Plane
- Graphing Linear Equations
- Finding the Slopes of Lines
- Solving Systems of Linear Equations

**Coefficient**is a numerical part of a variable term.**Variable**represents one or more numbers.**Factor**takes place when two numbers are multiplied.**Expression**is mathematical sentence which contain more than one operation.**Quotient**is an answer to a division problem.**Power**is made up of an exponent.**Exponent**is a number or a variable that represents the number of times a base is used as a factor in a repeated multiplication.

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The value of x for the quadratic equation, $ax^2 + bx + c = 0$, $x \neq 0$ is

x = $\frac{-b \pm \sqrt{b^2} -4ac}{2a}$.

Let us solve x - 4 = 0

x - 4 = 0

$\Rightarrow$ x = 4.

Let us solve 4x - 12 = 0

4x - 12 = 0

$\Rightarrow$ 4x = 12 (by adding 12 both the sides)

$\Rightarrow$ x = $\frac{12}{4}$ = 3 (divide each side by 4)

x = 3.

Let us solve x$^2$ + x - 6 = 0

x$^2$ + x - 6 = 0

$\Rightarrow$ x$^2$ + 3x - 2x - 6 = 0 (by factoring)

$\Rightarrow$ x(x + 3) - 2(x + 3) = 0

$\Rightarrow$ (x - 2)(x + 3) = 0

$\Rightarrow$ either x - 2 = 0 or x + 3 = 0

$\Rightarrow$ x = 2, -3.

→ Read More Given below are some of the problems of pre algebra:

Let $P(x) = x^3 - 2x^2 + 4x + 5$

Step 1:

Put x - 3 = 0

$\Rightarrow$ x = 3

Step 2:

$P(3) = 3^3 - 2 \times 3^2 + 4 \times 3 + 5$

= 27 - 2 $\times$ 9 + 12 + 5

= 27 - 18 + 12 + 5

= 26

When P(x) is divided by (x - 3), then remainder is 26.

Step 1:

Put x - 3 = 0

$\Rightarrow$ x = 3

Step 2:

$P(3) = 3^3 - 2 \times 3^2 + 4 \times 3 + 5$

= 27 - 2 $\times$ 9 + 12 + 5

= 27 - 18 + 12 + 5

= 26

When P(x) is divided by (x - 3), then remainder is 26.

Let the fraction be $\frac{x}{y}$

**Step 1:**

Case 1:

When 2 is added to the numerator and 1 to the denominator it reduces to $\frac{1}{2}$

$\Rightarrow$ $\frac{x + 2}{y + 1} = \frac{1}{2}$

2x + 4 = y + 1

2x - y + 3 = 0 .........................(i)

Case 2:

Fraction becomes $\frac{3}{5}$, if we add 3 to the denominator.

$\frac{x}{y + 3} = \frac{3}{5}$

5x = 3y + 9

5x - 3y - 9 = 0 .......................(ii)

**Step 2: **

Solve equation (i) and (ii).

Multiply equation (i) by 3 and subtract from (ii).

5x - 3y - 9 = 0

-(6x - 3y + 9) = 0

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

-x + 0 - 18 = 0

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

$\Rightarrow$ x = - 18

Plug the value of x in equation (i)

$\Rightarrow$ 2 $\times$ (-18) - y + 3 = 0

-36 - y + 3 = 0

y = - 33

The required fraction is $\frac{18}{33}$.

Case 1:

When 2 is added to the numerator and 1 to the denominator it reduces to $\frac{1}{2}$

$\Rightarrow$ $\frac{x + 2}{y + 1} = \frac{1}{2}$

2x + 4 = y + 1

2x - y + 3 = 0 .........................(i)

Case 2:

Fraction becomes $\frac{3}{5}$, if we add 3 to the denominator.

$\frac{x}{y + 3} = \frac{3}{5}$

5x = 3y + 9

5x - 3y - 9 = 0 .......................(ii)

Solve equation (i) and (ii).

Multiply equation (i) by 3 and subtract from (ii).

5x - 3y - 9 = 0

-(6x - 3y + 9) = 0

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

-x + 0 - 18 = 0

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

$\Rightarrow$ x = - 18

Plug the value of x in equation (i)

$\Rightarrow$ 2 $\times$ (-18) - y + 3 = 0

-36 - y + 3 = 0

y = - 33

The required fraction is $\frac{18}{33}$.

Let the number be 'x'.

Step 1:

The problem states that:

3(square of a number) - 10 = 10 $\times$ number + 15

$\Rightarrow$ 3x$^2$ - 10 = 10x + 15

$\Rightarrow$ 3x$^2$ - 10x = 15 + 10

$\Rightarrow$ 3x$^2$^{ }- 10x = 25

3x$^2$^{ }- 10x - 25 = 0

Which is quadratic equation

Step 2:

Solve for x, 3x$^2$ - 10x - 25 = 0

3x$^2$^{ }- 10x - 25 = 0

$\Rightarrow$ 3x$^2$ - 15x + 5x - 25 = 0

$\Rightarrow$ 3x(x - 5) + 5(x - 5) = 0

$\Rightarrow$ (3x + 5)(x - 5) = 0

Step 3:

Either 3x + 5 = 0 or x - 5 = 0

$\Rightarrow$ 3x = - 5 or x = 5

$\Rightarrow$ x = $\frac{-5}{3}$ or x = 5

Hence, the number is $\frac{-5}{3}$ or 5.

Step 1:

The problem states that:

3(square of a number) - 10 = 10 $\times$ number + 15

$\Rightarrow$ 3x$^2$ - 10 = 10x + 15

$\Rightarrow$ 3x$^2$ - 10x = 15 + 10

$\Rightarrow$ 3x$^2$

3x$^2$

Which is quadratic equation

Step 2:

Solve for x, 3x$^2$ - 10x - 25 = 0

3x$^2$

$\Rightarrow$ 3x$^2$ - 15x + 5x - 25 = 0

$\Rightarrow$ 3x(x - 5) + 5(x - 5) = 0

$\Rightarrow$ (3x + 5)(x - 5) = 0

Step 3:

Either 3x + 5 = 0 or x - 5 = 0

$\Rightarrow$ 3x = - 5 or x = 5

$\Rightarrow$ x = $\frac{-5}{3}$ or x = 5

Hence, the number is $\frac{-5}{3}$ or 5.

Let the three consecutive odd integer be x, x + 2 and x + 4.

That is,

First integer = x

Second integer = x + 2

Third integer = x + 4

Step 1:

The problem states that:

Sum of the first and third integers = Second integer + 7

$\Rightarrow$ x + (x + 4) = (x + 2) + 7

$\Rightarrow$ x + x + 4 = x + 2 + 7

Step 2:

Combine the like terms:

$\Rightarrow$ 2x + 4 = x + 9

Subtract x from both side

$\Rightarrow$ 2x - x + 4 = x + 9 - x

$\Rightarrow$ x + 4 = 9

Subtract 4 from both sides

$\Rightarrow$ x + 4 - 4 = 9 - 4

$\Rightarrow$ x = 5, first integer.

Step 3:

Largest or third integer = x + 4 = 5 + 4 = 9

Hence, the largest integer is 9

That is,

First integer = x

Second integer = x + 2

Third integer = x + 4

Step 1:

The problem states that:

Sum of the first and third integers = Second integer + 7

$\Rightarrow$ x + (x + 4) = (x + 2) + 7

$\Rightarrow$ x + x + 4 = x + 2 + 7

Step 2:

Combine the like terms:

$\Rightarrow$ 2x + 4 = x + 9

Subtract x from both side

$\Rightarrow$ 2x - x + 4 = x + 9 - x

$\Rightarrow$ x + 4 = 9

Subtract 4 from both sides

$\Rightarrow$ x + 4 - 4 = 9 - 4

$\Rightarrow$ x = 5, first integer.

Step 3:

Largest or third integer = x + 4 = 5 + 4 = 9

Hence, the largest integer is 9

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