Literal Equation

Algebra literals are one of the parts of basic algebra that is mainly used to find the unknown or it concern’s on putting a real life problem into equations and calculating it. A literal coefficient is defined as any variable in an algebraic expression. For example, in "58ab", a and b are literal coefficients.

algebra literals

Introduction to algebra literals:

            Algebra literals are one of the parts of basic algebra that is mainly used to find the unknown or it concern’s on putting a real life problem into equations and calculating it. A literal coefficient is defined as any variable in an algebraic expression. For example, in "58ab", a and b are literal coefficients. An algebraic expression with literals contains constants, operating symbols such as plus and minus signs and variables. In algebra literal, we frequently use letters to represent numbers.

 

 

Problems on Algebra Literals:

 

Example 1:

Solve the algebraic literal equations         K = 2A + 2B    for B.

Solution:

Given literal equation is

K = 2A + 2B   

To find B first separate it so add -2A on both sides. 

K – 2A = 2A + 2B – 2A

Simplify the above equation 

K – 2A = 2B

Divide by 2 on both sides to obtain B 

B = (K- 2A) / 2

Example 2:

 Solve the algebraic literal equation         G = `sqrt (a^2 + b^2)`

For b, where G, a and b are real numbers and G is greater than a and b.

Solution:

Given equation is

G = sqrt (a 2 + b 2)

Squaring on both sides 

G 2 = a 2 + b 2

Add - a 2 on both sides and solve it. 

G 2 - a 2 = a 2 + b 2 - a 2 

G 2 - a 2 = b 2

Taking square root on both sides. 

b = + or - sqrt (G 2 - a 2)

Since b is a positive real number, then b is calculated as 

b = + sqrt (G 2 - a 2)

 

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Example 3:

Solve the algebraic literal equation and substitute y in terms of x in the equation     c x + d y = e, with d not equal to zero.

 Solution:

c x + d y = e

Add - c x on both sides of the equation 

c x + d y - c x = e - c x

Solve it. 

d y = - c x + e

Divide by b on both sides. 

y = - (c / d) x + e / d

 

Practice Problems on algebra literals:

 

Solved Problems for Algebra Literal:

Problem 1:

Solve algebra literal A = L x W for L .

Solution:

The given literal equation is area of rectangle.

A = L x W

Now we have to Divide by  W both sides.

A/W = (LxW)/W

Therefore, we get

L = `A/W` .

Problem 2:

Solve algebra literal A =`1/2` x B x H for B .

Solution:

The given literal equation is area of triangle.

A = `1/2` x B x H

Now we need to multiply 2 both sides.

2 x A = `1/2 ` x 2 x B x H

So we get,

2A = B x H  

Now we have to Divide by H both sides.

`(2A)/H` = `(B*H)/H`

Therefore, we get

B = `(2A)/H` .

1) Solve the algebraic literal equation         G = K M, for M.

Answer: M= G / K

2) Solve the algebraic literal equation         z = f x + k, for x.

Answer: (z – k) / f

3) Solve the algebraic literal equation         A = (9 / 3) S + 32, for S.

Answer: S = (3A – 96) / 9