Example

Title: Test2

Grade: 8-b Lesson: S1-P2

Explanation: The best way to understand algebra is by looking at some examples. Take turns and read each example for easy understanding.

Examples:

Simplify the following expression: \$ (3u\sqrt(36v^2)) / (12\sqrt(100uv)) \$.

Step 1a

Simplify inside the square roots: \$ \sqrt 36v^2 = 6v \$(since \$ \sqrt 36 = 6 \$)

⇒ \$ \sqrt100uv = 10\sqrtuv \$

So the expression becomes: \$ 3u 6v \div 12(10\sqrtuv) \$

Explanation: We simplified the expressions inside the square roots and then substituted these simplified forms into the given expression.

Step 1b

Simplify the constants: \$ 3 \times 6 = 18 \$

⇒ \$ 12 \times 10 = 120 \$

Substitute these values: \$ 18uv \div 120(\sqrt100uv) \$

Explanation: We simplified the numbers and then used these simplified values in the expression.

Step 1c

Simplify the fraction (reduce the fraction): Simplify \$ 18uv \div 120(\sqrt100uv) \$

Divide both numerator and denominator by 6: \$ 3uv \div 20\sqrtuv \$

Therefore, the simplified expression is \$ 3uv \div 20\sqrtuv \$.

Explanation: After reducing the fraction, we determine the final result.

Calculate the sum of the cube root of 64 and the square root of 25?

Step 2a

The given cube root = 64 and squre root = 25.

The cube root of 64 is \$root(3)64 = 4\$, becuase \$4 times 4 times 4\$ = 64.

The square root of 25 is \$\sqrt(25) = 5\$, beacuse \$5 times 5\$ = 25.

So, the sum of the cube root of 64 and the square root of 25 is 4 + 5 = 9.

Explanation: Find the cube root and square root of a given number, then add them together.


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