Algebra 1 Practice

Introduction for extra algebra 1 practice:

For extra practice we need to work out more problems in algebra 1. Only through this way we get extra practice in Algebra 1. Algebra 1 is one of the subdivisions of algebra which is mainly used to find the unknown variables with the known variables. Algebra 1 consists of algebraic expressions, conditions and polynomials. An algebraic expression represents a scale, a number gets added or subtracted or multiplied or divided on both the sides of the scale. The numbers are defined as constants. Algebra 1 has the part in complex numbers, matrices, vectors, real numbers etc


Example Problems for extra algebra 1 practice:

Example 1:

Evaluate the equation     7(-3x - 2) - (-x - 6) = -4(5x + 5) + 12

Solution:

Given the equation

7(-3x - 2) - (-x - 6) = -4(5x + 5) + 12

Multiply factors.
-21x - 14 + x + 6 = -20x - 20 +12

Grouping the terms.

-20x - 8 = -20x - 8

Add 20x + 8 to both sides, the above equation becomes

0 = 0

All real values are solution to this equation.

Example 2:

Simplify the expression    3(a -4) + 6b - 3(a -b -2) + 8

Solution:

Given the algebraic expression

3(a -4) + 6b - 3(a -b -2) + 8

Multiply factors.

= 3a - 12 + 6b -3a + 3b + 6 + 8

Grouping the above terms.

= 9b + 2

Example 3:

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Solution:

Given the expression

|x - 4| - 5|-4|

If x <; 4 then x - 4 < 4

and if

x - 4 < 4 the |x - 4| = -(x - 4).

Substitute |x - 4| by -(x - 4) and |-4| by 4

|x - 4| - 5|-4|

= -(x - 4) -5(4)

= -x -16

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Extra algebra 1 practice problems:


1) Evaluate the equation     6(-8x - 3) - (-5x - 5) = -8(2x + 4) + 9

Answer: x = 10/59

2) Simplify the expression    5(a -8) + 11b - 5(a -b +6) + 4

Answer: 16b + 64

Free Algebra Practice

Introduction to free algebra practice:

Algebra comes from the Arabic word al-gebra. It represents as the “reunion of the broken words”. In mathematics, Letters like a, b, c, x are been used to represent the numbers, values, complex numbers, integers and these unite with the general operations to form the algebraic expressions. If two algebraic expressions are equated we get the algebraic equations. In free algebra practice we shall discuss about the examples and practice problems.

Solved examples - free algebra practice:

Example problem: 1

Solve the two consecutive numbers have a sum of 81. find the  numbers?

Solution:

Let x = The first Consecutive Number

Since the numbers are consecutive, meaning one number comes right after the other, the second number must be one more than the first. So, x + 1 equals the second number.

Let x + 1 is a 2nd consecutive

The problem says that the sum of the two numbers is 81. This can be shown in the equation like the following:

x + (x + 1) = 81

The equation which you just wrote can be solved as follows:

Initial Equation

x + (x + 1) = 81

After combine like terms

2x + 1 = 81

After subtracting 1 from each side

2x = 80

After dividing each side by 2

x = 40

Answer : x = 40


Solving linear equations - free algebra practice:

Solve the following linear equations: x + 2y + 3z = 14, 3x + y + 2z = 11, 2x + 3y + z = 11.

Solution:

x + 2y + 3z = 14 (1)

3x + y + 2z = 11 (2)

2x + 3y + z = 11 (3)

Consider the equations (1) and (3)

(1) ? x + 2y + 3z = 14

(3) × 3 ? 6x + 9y + 3z = 33 (subtracting)

–5x – 7y = –19

5x + 7y = 19 (4)

Consider the equations (2) and (3)

(2) ? 3x + y + 2z = 11

(3) × 2 ? 4x + 6y + 2z = 22 (subtracting)

–x – 5y = –11

x + 5y = 11 (5)

Consider the equations (4) and (5)

(4) ? 5x + 7y = 19

(5) × 5 ? 5x + 25y = 55 (subtracting)

–18y = –36

? y = 2

Substitute y = 2 in (5) we get

x + 5(2) = 11

x + 10 = 11

x + 10 - 10 = 11 - 10

? x = 1

Substitute x = 1, y = 2 in (3) we get

2(1) + 3(2) + z = 11

2 + 6 + z = 11

8 - 8 + z = 11 - 8

? z = 3

The solution is x = 1, y = 2, z = 3.

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Problems for practice - free algebra practice:

Problem 1:

The quadratic polynomial is divided by (x–1), (x+1) and (x–2) leaves the remainders 2, 4, 4 respectively, find the quadratic polynomial.

solution:The required quadratic polynomial is q(x) = x2 – x + 2.

Problem: 2

Solve the value of a and b if ax3 + bx2 + 7x + 9 and x3 + ax2 – 2x + b – 4 when divided by x + 2 leave remainders –13 and –16 respectively.

Solution: a –1, b = –4