Algebra Questions

Introduction:

Algebra is a field of mathematics that aids in the representation of problems and circumstances into mathematical expressions. To produce a coherent mathematical expression, it uses variables like x, y, and z, as well as addition, subtraction, multiplication, and division. Trigonometry, calculus, and coordinate geometry are all fields of mathematics that use algebra. In algebra, 2x + 4 = 8 is a simple example of an expression.

Algebra is concerned with symbols, which are connected through operators. It is more than a mathematical idea; it is a skill that we all utilize without even realizing it in our daily lives. Solving equations and finding the solution is less crucial than understanding algebra as a concept.

It is more vital to understand algebra as a concept than it is to solve problems and find the correct solution.

Now we are going to solve a few problems related to algebra in which we will utilize various algebraic functions, identities and formulas for simplification and finding unknown quantities.

Example 1: A tourist bus travels at a constant speed for 480 miles. It would have taken 3 hours longer to traverse the distance if the speed had been 8 miles per hour slower. Determine the tourist bus’s speed.

Solution  – Let the speed of a tourist’s bus be x miles per hour.

Total distance to be covered = 480 miles

Time taken with speed x km per hour = 480/x hrs

Now, the speed has been reduced by 8 miles per hour.

Therefore, the new speed be (x-8) miles per hour.

Time taken with speed (x-8) miles per hour = 480/(x-8) hrs

According to the question, we have

480/(x-8) – 480/x = 3

Taking lcm of the denominators; i.e, x and (x-8) we get

480x – 480(x-8) = 3x(x-8)

480x – 480x + 3840 = 3x2 – 24x

3x2 – 24x -3840=0

x2 – 8x – 1280=0

x2 – 40x + 32x – 1280=0

x(x-40) + 32(x-40) =0

(x-40) (x+32) =0

x-40 = 0

x= 40

Therefore, the required value of x is 40 miles per hour.

Example 2: Give algebraic expressions for the following situations:

(I) 8 multiplied by 4x

(ii) multiplying 9 by 3a

(iii) R is reduced by 2.

(iv) 7 times the value of O

Solution –

(I) 8 x 4x = 32x

 (II) 9 x 3a = 27a

(III) R-2

(IV) 7 x O = 7O

Example 3:  Three numbers add up to 131. The second of the three numbers is seven times as large as the first. The third number is a fraction of a dozen less than the first. What is the best equation to employ to solve the problem?

Solution – Let the first required number be x

 the second number be y

the third number be z

As per the question sum of the 3 numbers is given to be 131

 we get,

x+y+z=131 …(1)

Also, we get that the second number is 7 more than twice the first.

Therefore

y=7+2x…(2)

again from the question we have that the z is 12 less than the x.

Thus,

z=x−12⋯(3)

Replacing in (1) we get

x+2x+7+x−12=131.

So the required equation is

4x−5=131.

For x we get

4x=136

x=136/4

x=34.

Putting in (2) and (3) we get

y=2(34) +7=75.

z=34−12=22.

Example 4: Solve the equation and find the value of x: 

3x + 25 = 31

Solution – 3x + 25 = 31

           3x = 31 – 25 

           3x = 6

           x = 6/3

          x = 2

hence, the required value of x is 2.

Conclusion:

Algebra began with arithmetic-style computations, with letters standing in for numbers. This allows for proof of qualities that are true regardless of the numbers used. In the quadratic equation,

 for example, ax2 + bx + c=0

The quadratic formula can be used to obtain the values of the unknown quantity x that satisfy the equation fast and easily. The quadratic formula can be used to quickly and easily identify the values of the unknown quantity x that satisfy the equation. a,b,c can be any numbers (except 0), and the quadratic formula can be used to quickly and easily determine the values of the unknown quantity x that fulfil the equation. To put it another way, you’re looking for all of the solutions to the problem.

Solving equations, such as the quadratic equation above, has traditionally and now been the first step in learning algebra. Then, more general questions like “does an equation have a solution?” are investigated. “What can be said about the nature of the answers?” and “How many solutions does an equation have?” As a result of these explorations, algebra was expanded to encompass non-numerical objects such as permutations, vectors, matrices, and polynomials. The structural properties of these non-numerical objects were eventually formalised into algebraic structures like as groups, rings, and fields.