Elementary Algebra:
Elementary algebra is a branch of the science of calculation like arithmetic. In strategies to teach elementary algebra, we use the symbols to denote the quantities and progressions. The arithmetic has a single definite value, but strategies algebraic symbols have a general and not merely a single definite value. Let see about the identities of strategies to teach elementary algebra.
The followings are some of the algebraic identities using in the elementary algebra.
Let see the examples of strategies to teach elementary algebra.
1. Expand: (2x + 3y+ 4z)^{2}
Solution:
Comparing the given expression (2x + 3y+ 4z)^{2} with (a +b +c)^{2}.
Then we have,
a= 2x
b= 3y
c=4z
(a +b +c)^{2} = a^{2} +b^{2} +c^{2} +2ab +2bc +2ca
Therefore (2x + 3y+ 4z)^{2} = (2x)^{2} + (3y)^{2} + (4z)^{2} +2(2x) (3y) +2(3y) (4y) + 2(4x) (2x)
= 4x^{2} + 9y^{2} + 16z^{2} +12xy +24yz +16zx
2. Find the cube of 4x-5y.
Solution:
We know that (a -b) ^{3} = a^{3} - b^{3} -3ab (a-b)
Here, a = 4x,
b = 5y
(4x-5y)^{3} = (4x)^{3} - (5y) -3* 4x * 5y (4x – 5y)
= 64x^{3} - 125y^{3} -60xy (4x- 5y).
3. Evaluate: 102³ using suitable identity.
Solution:
102^{3} = (100 +2)^{3}
Use the identity (a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)
Therefore, 102^{3} = (100 + 2)^{3}
= 100^{3} +2^{3} +3* 100* 2(100+ 2)
= 1000000 + 8 + 600(100 +2)
= 1000000 + 8 + 60000 + 1200
= 1061208
4. Evaluate 3x² - y² - z if x = 3, y = -1 and z = -2
Solution:
Substitute the values for the variables.
= 3(3)^{2} - (-1)^{2} - (-2)
= 3(9) - (1) - (-2)
= 27 - (1) + (2)
= 26 + 2
= 28
These are the examples of strategies to teach elementary algebra.