Math Guide

Practice seventh grade pre algebra

Introduction:


   Expressions are a central concept in algebra. A variable can take several values. The values are not fixed.  But a constant has a fixed value. We combine variables and constants with arithmetic operations to make algebraic expressions. For these problems, we can use basic operations such as addition, subtraction, multiplication and division. Some simple algebraic expressions are x-10, y-3, 5+z and so on. The above expressions are obtained by combining variables and constants.


Example Problems to practice seventh grade pre algebra :


1. In (5x2 – 7) we first obtain x2, and multiply it by 5 to get 5x2.From 5x2, we subtract 7 to finally arrive at 5x2 – 7.


2. Identify the like and unlike terms

Pq2, – 4pq2 (vii) mn2, 10mn

Solution:

pq2, – 4pq2 – like terms

The variables in the two terms match, but only their coefficients differ.

mn2, 10mn- unlike terms

variables are same


3. Identify the type of the polynomial ab + a +b

Solution:

Here this polynomial contain three terms. So this is called a trinomial.


4.Collect like terms and simplify the algebraic expression:

12m2 – 9m + 5m – 4m2 – 7m + 10

Solution:

Rearranging terms, we have

12m2 – 4m2 + 5m – 9m – 7m + 10

= (12 – 4) m2 + (5 – 9 – 7) m + 10

= 8 m2 + (– 4 – 7) m + 10

= 8 m2 + (–11) m + 10

= 8 m2 – 11m + 10


5. Convert this in the form of algebraic expressions.

(i) Subtraction of z from y. that is y-z

(ii) One-half of the sum of numbers x and y.that is ½(x+y)

(iii) One-fourth of the product of numbers p and q.that is ¼(p * q)

(iv) Five times the sum of a twice a number and three that is 5(2x +3)


6. Multiply the algebraic expressions (2 ab)×(5b)

Solution:

(2 ab)×(5b) = (2ab)×(5×b) = (2ab)×(b×5) (Commutative property of multiplication)

= [(2ab×b)]×5 (Associative property of multiplication)

= [2a×(b×b)]×5

= 2ab2×5

= 5×(2ab2) (Commutative property of multiplication)

= [(5×2)]×ab2 (Associative property of multiplication)

= 10 × ab2

= 10ab2


7. Evalutate the following using special products :

(a) 1022 (b) 492

Solution :

(a) 1022 = (100+2)2

= 1002+2×100×2+22 (using I)

= 10000+400+4

= 10404

(b) 492 = (50-1)2

= 502-2×50×1+12 (using II)

= 2500-100+1

= 2401