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# Algebra Problems

Algebra is a vast subject. It is a branch of mathematics that deals with variables and algebraic operations. Algebra is concerned with letters, symbols, numerals and operations. It is the foundation of many other branches of mathematics. Algebra is the study of different operations on constants and variables. There are many branches of algebra which are given below:

• Elementary algebra: It deals with elementary operations like - addition, subtraction, multiplication, division etc. of variables and elementary problems related to algebra.
• Pre algebra: Pre algebra prepares a student for higher-level algebra. It is taught in grade fifth to grade eighth.
• Linear algebra: It majorly deals with matrices, vector, vector spaces etc.
• Modern algebra: It is also known as abstract algebra which is based on axioms and theorems of higher-grade algebra. It is the study of groups, rings, fields etc.
• Boolean algebra: Boolean algebra is a special kind of branch of algebra which deals with two values: true and false. It is based on different logical operators.
Following are the examples of few algebraic expressions:
(1) $x^{2}-3xy+5y^{2}$
(2)
$a^{3}-a^{2}+2a-1$

Some basic problems related to algebra are discussed below:
Problem 1: Solve $3x^{2}-5(x+y)+7x^{2}+4y-8x$.
Solution: $3x^{2}-5(x+y)+7x^{2}+4y-8x$
= $3x^{2}-5x-5y+7x^{2}+4y-8x$
= $(3+7)x^{2}-(5+8)x-(5-4)y$
= $10x^{2}-13x-y$

Problem 2: Find the value of x from the following equation:
$3(x-8)=15$
Solution: $3(x-8)=15$
$x-8=$$\frac{15}{3}$
$x-8=5$
$x=5+8$
x = 13

Problem 3: The sum of a two-digit number is 9. The number obtained by reversing the places of the digits is 27 less than the original number. Calculate the original number.
Solution: Let us assume that digit at ones place = x and digit at tens place = y.
The number so obtained will be 10y + x.
According to the first condition given in question, we have-
x + y = 9 .......(1)
The number obtained by reversing the digits is 10x + y.
According to the second condition given in question, we have-
(10y + x) - (10x + y) = 27
9y - 9x = 27
y - x = 3.......(2)
Hence, the original number is 10y + x = 10 $\times$ 6 + 3 = 63