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Basics of Algebra

1. The 4 Important Exponents Rules

Here is a brief introduction to the four important rules for exponents in algebra:

Product of Powers Rule: When multiplying two terms with the same base, we add the exponents.
Example with numbers: \(2^3 \cdot 2^4 = 2^{3+4} = 2^7\)
Example with variables: \(x^3 \cdot x^4 = x^{3+4} = x^7\)

Quotient of Powers Rule: When dividing two terms with the same base, we subtract the exponents.
Example with numbers: \(\dfrac{3^6}{3^2} = 3^{6-2} = 3^4\)
Example with variables: \(\dfrac{y^6}{y^2} = y^{6-2} = y^4\)

Power of a Power Rule: When you have a power raised to another power, we multiply the exponents.
Example with numbers: \((2^3)^4 = 2^{3 \cdot 4} = 2^{12}\)
Example with variables: \((z^3)^4 = z^{3 \cdot 4} = z^{12}\)
Zero Power Rule: Any nonzero number raised to the power of zero is 1.
Example with numbers: \(7^0 = 1\)
Example with variables: \(a^0 = 1\)

Remember, understanding and applying these rules will help you simplify and solve more complex algebraic expressions and equations. Be careful to ensure the bases are the same when applying these rules.

2. Negative Exponents

Negative exponents can be understood by using the idea of reciprocals and the properties of exponents.

When a number or variable is raised to a negative exponent, it indicates that the reciprocal of that number or variable should be taken to the positive exponent.

Let's take an example: \(\frac{1}{{4^{-3}}}\)

To simplify this expression, we can rewrite it as the reciprocal with a positive exponent:

\[\frac{1}{{4^{-3}}} = \frac{1}{{\frac{1}{{4^3}}}}\]

Now, let's simplify the expression within the parentheses:

\(\frac{1}{{4^3}}\)

Here, \(4^3\) means \(4 \times 4 \times 4\), which is equal to 64.

Therefore, we have:

\[\frac{1}{{\frac{1}{{4^3}}}} = \frac{1}{{\frac{1}{64}}} = 64\]

So, \(\frac{1}{{4^{-3}}} = 64\).

In this case, the negative exponent flips the fraction and changes the sign of the exponent. It allows us to convert a negative exponent into a positive exponent by taking the reciprocal of the base.

3. Simplifying Exponent Expressions

Solve This:
Evaluate this expression \(\dfrac{5y^3}{\left(y^{-2}\right)^3}\)

To solve this problem, follow the order of operations and apply the exponent rules.

First, simplify the expression within the parentheses. To raise a power to another power, we multiply the exponents:
\[\left(y^{-2}\right)^3 = y^{-2 \times 3} = y^{-6}\]

Now, rewrite the original expression. To divide by a negative exponent, rewrite it as a positive exponent:
\[\frac{5y^3}{y^{-6}} = \frac{5y^3}{\frac{1}{y^6}}\]

Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[5y^3 \times \frac{y^6}{1} = 5y^{3 + 6} = 5y^9\]

Therefore, the simplified expression is \(5y^9\).

4. Solving Equations with Exponents


Solve the equation: \[ 7^{4a+3} = 7^{a+9}\]

To solve the equation, we can equate the exponents:
\[4a+3 = a+9 \Rightarrow 3a + 3 = 9\]

Solving for \(a\):
\[3a = 6 \Rightarrow a = 2\]

Therefore, the solution to the equation is \(a = 2\).

Solve This:
If \(16^{20} = 2^x\), then find the value of \(x\).

We know that \(16 = 2^4\) because \(2 \times 2 \times 2 \times 2 = 16\).

Using the property of exponents, we can rewrite the equation as \((2^4)^{20} = 2^x\).

Applying the power rule, we multiply the exponents:

\[2^{4 \times 20} = 2^x\]

Simplifying the exponent on the left side, we have:

\[2^{80} = 2^x \Rightarrow 80 = x\]

Therefore, in the equation \(16^{20} = 2^x\), the value of \(x\) is 80.


Solve This:
If \(4a - b = 2\), find the value of \(\frac{{81^a}}{{3^b}}\)

We are given that \(4a - b = 2\).

Let's simplify the expression \(\frac{{81^a}}{{3^b}}\) using the properties of exponents.

We know that \(81 = 3^4\) because \(3 \times 3 \times 3 \times 3 = 81\).

Substituting this into the expression, and using the power of a power rule, we multiply the exponents we have:

\[\frac{{(3^4)^a}}{{3^b}} = 3^{4a - b}\]

Since we are given that \(4a - b = 2\), we can substitute it back into the expression:

\[3^{4a - b} = 3^2\]

Therefore, the value of \(\frac{{81^a}}{{3^b}}\) when \(4a - b = 2\) is 9.

5. Convert Radicals to Fractional Exponents


Radicals can be represented using fractional exponents, providing a more concise and flexible notation. Here are a few examples:

  1. Square Root: The square root of a number can be expressed as the number raised to the power of \(1/2\). For instance, \(\sqrt{5}\) is equivalent to \(5^{1/2}\).

  2. Cube Root: The cube root of a number can be represented as the number raised to the power of \(1/3\). For example, \(\sqrt[3]{5}\) is equal to \(5^{1/3}\).

  3. Higher Roots: The same concept applies to higher roots. To find the seventh root of \(10^3\), we write it as \(\sqrt[7]{10^3}\), which simplifies to \(10^{3/7}\).

\[ \sqrt{5} = 5^{1/2} \] \[ \sqrt[3]{5} = 5^{1/3} \] \[ \sqrt[7]{10^3} = 10^{3/7} \]

Converting radicals to fractional exponents allows for easier calculations and manipulation of expressions involving roots.

Solve This:
If \(\sqrt[7]{10^5} \cdot \sqrt[7]{10^9} = 10^x\), find \(x\).

To solve this problem, we'll simplify each radical expression:

\[\sqrt[7]{10^5} = (10^5)^{\frac{1}{7}} = 10^{\frac{5}{7}}\]
\[\sqrt[7]{10^9} = (10^9)^{\frac{1}{7}} = 10^{\frac{9}{7}}\]

Now, we can multiply the simplified radicals:

\[\sqrt[7]{10^5} \cdot \sqrt[7]{10^9} = 10^{\frac{5}{7}} \cdot 10^{\frac{9}{7}}\]

To multiply exponential expressions with the same base, we add the exponents:

\[10^{\frac{5}{7}} \cdot 10^{\frac{9}{7}} = 10^{\frac{5}{7} + \frac{9}{7}}\]

Combining the fractions:

\[10^{\frac{5+9}{7}} = 10^{\frac{14}{7}}\]

Simplifying the exponent:

\[10^{\frac{14}{7}} = 10^2\]

Therefore, the expression \(\sqrt[7]{10^5} \cdot \sqrt[7]{10^9}\) simplifies to \(10^2\).

Since this expression is equal to \(10^x\), we can equate the exponents:

\[x = 2\]

Hence, the value of \(x\) in the equation \(10^x\) is \(2\).

Solve This:
If \(2\sqrt{x+3} = \sqrt{8x}\), find \(x\).

To solve this equation, we will isolate the square root terms and then square both sides:

\[2\sqrt{x+3} = \sqrt{8x}\]
\[(2\sqrt{x+3})^2 = (\sqrt{8x})^2\]
\[4(x+3) = 8x\]

Expanding and simplifying the equation:

\[4x + 12 = 8x\]

Bringing like terms together:

\[4x - 8x = -12\]

Combining like terms:

\[-4x = -12\]

Dividing both sides by \(-4\) to solve for \(x\):

\[x = 3\]

Hence, the solution to the equation \(2\sqrt{x+3} = \sqrt{8x}\) is \(x = 3\).

1. FOIL Method and Distributive Property


The FOIL method is used to multiply two binomial expressions, while the distributive property allows us to multiply a constant with a binomial. Let's demonstrate with an example:

\[ (2x+1)(3x-2) \]

Applying the FOIL method:

\[ \begin{align*} \text{First:} & \quad 2x \cdot 3x = 6x^2 \\ \text{Outer:} & \quad 2x \cdot (-2) = -4x \\ \text{Inner:} & \quad 1 \cdot 3x = 3x \\ \text{Last:} & \quad 1 \cdot (-2) = -2 \\ \end{align*} \]

Combining these results:

\[ 6x^2 - 4x + 3x - 2 \]

Simplifying further:

\[ 6x^2 - x - 2 \]

Now, let's consider the expression multiplied by a constant:

\[ 3(2x+1)(3x-2) \]

Using the distributive property:

\[ \begin{align*} & \quad 3 \cdot 2x + 3 \cdot 1 \cdot 3x - 3 \cdot 2 \\ = & \quad 6x + 9x^2 - 6 \\ \end{align*} \]

Therefore, \( 3(2x+1)(3x-2) \) simplifies to \( 6x + 9x^2 - 6 \), which is 3 times the result of the original expression.

2. Simplifying an Expression


Let's simplify the expression:

\[ (3x^2-1)(2x+3) - (4x-2)(2x+1) \]

To simplify this expression, we will apply the distributive property and perform the necessary multiplications and subtractions.

Using the distributive property:

\[ \begin{align*} & \quad (3x^2)(2x) + (3x^2)(3) - (1)(2x) - (1)(3) - (4x)(2x) - (4x)(1) + (2)(2x) + (2)(1) \\ = & \quad 6x^3 + 9x^2 - 2x - 3 - 8x^2 - 4x + 4x + 2 \\ \end{align*} \]

Combining like terms:

\[ 6x^3 + (9x^2 - 8x^2) + (-2x + 4x) + (-3 + 2) \]

Simplifying further:

\[ 6x^3 + x^2 + 2x - 1 \]

Therefore, the simplified form of \((3x^2-1)(2x+3) - (4x-2)(2x+1)\) is \(6x^3 + x^2 + 2x - 1\).

3. Difference of Squares


The difference of squares concept states that for any two numbers or algebraic expressions, the square of the first term minus the square of the second term can be factored as the product of their sum and difference:

\[a^2 - b^2 = (a + b)(a - b)\]

Let's apply this concept to the example expression: \(16x^2 - 9y^2\).

We can observe that \(16x^2\) is the square of \(4x\) and \(9y^2\) is the square of \(3y\). Therefore, we can rewrite the expression as:

\[16x^2 - 9y^2 = (4x)^2 - (3y)^2\]

Now, applying the difference of squares formula, we can factor it as:

\[(4x)^2 - (3y)^2 = (4x + 3y)(4x - 3y)\]

Thus, the expression \(16x^2 - 9y^2\) can be factored as \((4x + 3y)(4x - 3y)\) using the difference of squares concept.

Factor This:
\[81m^4 - 16n^6\]

To factor this expression, we can apply the difference of squares formula:

\[(9m^2)^2 - (4n^3)^2\]

Applying the difference of squares formula, we have:

\[(9m^2 + 4n^3)(9m^2 - 4n^3)\]

Therefore, the factored form of \(81m^4 - 16n^6\) is \((9m^2 + 4n^3)(9m^2 - 4n^3)\).

4. Fractions Addition and Multiplication


Addition Rule: To add fractions with different denominators, find a common denominator, add the numerators, and keep the common denominator.

\[ \frac{1}{3} + \frac{2}{5} = \frac{5}{15} + \frac{6}{15} = \frac{11}{15} \]

Multiplication Rule: To multiply fractions, multiply the numerators together and the denominators together.

\[ \frac{1}{3} \times \frac{2}{5} = \frac{1 \times 2}{3 \times 5} = \frac{2}{15} \]

These rules also apply to fractions with variables.

Rules for variables:

\[\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad}{bd} + \dfrac{cb}{bd} = \dfrac{ad + cb}{bd}\] \[\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}\]
\[ \text{Simplify: } \frac{4}{{x-1}} + \frac{1}{{x+1}} \]

To simplify the expression, we need to find a common denominator and combine the fractions. The common denominator in this case is \((x-1)(x+1)\).

\[ \frac{4}{{x-1}} + \frac{1}{{x+1}} = \frac{{4(x+1)}}{{(x-1)(x+1)}} + \frac{{1(x-1)}}{{(x-1)(x+1)}} \]

Now, we can combine the fractions by adding the numerators:

\[ \frac{{4(x+1) + 1(x-1)}}{{(x-1)(x+1)}} = \frac{{4x + 4 + x - 1}}{{(x-1)(x+1)}} \]

Simplifying the numerator:

\[ \frac{{5x + 3}}{{(x-1)(x+1)}} \]

Therefore, the simplified expression is: \( \dfrac{{5x + 3}}{{(x-1)(x+1)}} \)

Dividing by Fractions


Dividing one fraction by another is the same as multiplying by the reciprocal.
\[ \frac{4}{3} \div \frac{1}{2} = \frac{4}{3} \times \frac{2}{1} = \frac{8}{3} \]

This concept applies to fractions with variables as well.
\[ \frac{{x+1}}{{x^2+3}} \div \frac{{x+3}}{{x}} = \frac{{x+1}}{{x^2+3}} \times \frac{{x}}{{x+3}} \] \[ \dfrac{ \dfrac{{x+1}}{{x^2+3}}} { \dfrac{{x+3}}{{x}}} = \dfrac{{x+1}}{{x^2+3}} \times \dfrac{{x}}{{x+3}} \]

Solving equations with fractions


To solve the equation \( \dfrac{3}{4}x = \dfrac{2}{3} \), we multiply both sides by the reciprocal of \( \dfrac{3}{4} \), which is \( \dfrac{4}{3} \). This gives us:

\[ \dfrac{3}{4}x = \dfrac{2}{3}\] \[ \Rightarrow x = \dfrac{2}{3} \times \dfrac{4}{3} = \dfrac{8}{9} \]

For the equation \( \dfrac{3}{4}x + \dfrac{5}{3} = \dfrac{10}{3} \), we can simplify it by subtracting \( \dfrac{5}{3} \) from both sides and then multiplying both sides by the reciprocal of \( \dfrac{3}{4} \), which is \( \dfrac{4}{3} \). This leads to:

\[ \dfrac{3}{4}x + \dfrac{5}{3} = \dfrac{10}{3} \] \[\Rightarrow \dfrac{3}{4} x = \dfrac{10}{3} - \dfrac{5}{3} = \dfrac{5}{3} \] \[\Rightarrow x = \dfrac{5}{3} \times \dfrac{4}{3} = \dfrac{20}{9} \]

Therefore, the solutions to the equations are \( x = \dfrac{8}{9} \) and \( x = \dfrac{20}{9} \) respectively.

Factoring out the Common Factor


Sometimes, when we have fractions in the numerator and denominator, we need to simplify the expression by factoring out the common factor.

To factor out the common factor \(x\) in the numerator and denominator of the expression \(\dfrac{x^3 + 3x}{4x^2+2x}\), we identify the terms that contain \(x\) and divide them by \(x\).

\[ \frac{{x^3 + 3x}}{{4x^2+2x}} = \frac{{x(x^2 + 3)}}{{x(4x + 2)}} \]

Next, we can cancel out the common factor \(x\) in the numerator and denominator, which simplifies the expression:

\[ \frac{{x(x^2 + 3)}}{{x(4x + 2)}} = \frac{{x^2 + 3}}{{4x + 2}} \]

Therefore, after factoring out the common factor \(x\) and canceling it out, we are left with the simplified expression \(\dfrac{{x^2 + 3}}{{4x + 2}}\).

Simplify the expression by factoring out the common \(a^2\):

\[ \text{Simplify: } \frac{{4a^3 + a^2}}{{3a^5 - a^2}} \]

To factor out the common \(a^2\), we can rewrite it as:

\[\frac{{a^2(4a + 1)}}{{a^2(3a^3 - 1)}}\]

Canceling out the common factor \(a^2\), the simplified expression is:

\[\frac{{4a + 1}}{{3a^3 - 1}}\]

Cross-Multiplication


Cross multiplication is an powerful technique for solving equations that involve fractions. To solve the equation \( \dfrac{x}{3} = \dfrac{x+1}{5} \) using cross multiplication, we multiply the numerator of the first fraction by the denominator of the second fraction and vice versa, setting them equal to each other

So we get:
\[ 5x = 3(x+1) \]

Simplifying further, combining like terms and dividing both sides by 2:

\[ 5x = 3x + 3 \] \[ 2x = 3 \] \[ x = \frac{3}{2} \]

Therefore, the solution to the equation is \( x = \frac{3}{2} \).

*** Solve x/x-3 = 3/4 *** Solve the A = B/mC example on page 62