Question

Of the following, the linear equation in one variable x is :
a) 4/x = x/4
b) 1/x + 1/x-1 = 1
c) x/2 + x/3 = 1/4
d) x + 2x + 3 = 0

Answer 1

$\bf \frac{x}{2} + \frac{x}{3} = \frac{1}{4}$ is a linear equation in one variable.

it can be expressed as 10x-3=0.

Option (c) is correct.

Given:

• a) 4/x = x/4
• b) 1/x + 1/x-1 = 1
• c) x/2 + x/3 = 1/4
• d) x² + 2x + 3 = 0

To find:

• Of the above written,which one is the linear equation in one variable x.

Solution:

Formula/Concept to be used:

Linear equation in one variable is written in the form $\bf ax + b = 0$

where,a and b are real numbers and a≠0.

Case 1:

Check option a) $\frac{4}{x} = \frac{x}{4}$

cross multiply,

${x}^{2} = 16$

or

${x}^{2} - 16 = 0$

Thus, option (a) is not linear equation.

Case 2:

Check for option b)$\frac{1}{x} + \frac{1}{(x - 1)} = 1$

simply,

Take LCM

$\frac{x - 1 + x}{x(x - 1)} = 1$

or

$2x - 1 = {x}^{2} - x$

or

${x}^{2} - 3x + 1 = 0$

Thus,

Thus, option (b) is not linear equation.

Case 3:

Check for option c) $\frac{x}{2} + \frac{x}{3} = \frac{1}{4}$

or

$\frac{3x + 2x}{6} = \frac{1}{4}$

or

$\frac{5x}{3} = \frac{1}{2}$

or

$10x = 3$

or

$\bf \red{10x - 3 = 0}$

Thus,

Thus, option (c) is linear equation in one variable.

Case 4:

Check for option d): ${x}^{2} + 2x + 3 = 0$

it is clearly a quadratic equation.

Thus,

Thus, option (d) is not linear equation.

Thus,

$\bf \frac{x}{2} + \frac{x}{3} = \frac{1}{4}$ is a linear equation in one variable.

It can be expressed as 10x-3=0.

Option (c) is correct.

Learn more:

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Answer 2

Linear equation:

An equation is said to be a linear equation if the degree of that equation is one and if we graph the equation it should give us a straight line.

We are given four options and we need to find which of them is a linear equation.

a.

$4 \div x = x \div 4$

Let's solve this by cross multiplication method.

$4 \div x = x \div 4 \\16 = {x}^{2} \\ {x}^{2} - 16 = 0$

The above equation is a quadratic equation.

Hence option a is wrong. It's not a linear equation.

b.

$(1 \div x) + (1 \div (x - 1)) = 1$

Let's solve this by taking LCM

$(1 \div x) + (1 \div (x - 1)) = 1 \\ (x - 1 + x) \div (x(x - 1)) = 1 \\ (2x - 1) \div ( {x}^{2} - x) = 1 \\ 2x - 1 = {x}^{2} - x \\ {x}^{2} - 3x + 1 = 0$

Clearly, it is a quadratic equation. Hence it is not a linear equation. Option b is wrong.

c.

$(x \div 2) + (x \div 3) = 1 \div 4$

Let's solve it through LCM method

$(x \div 2) + (x \div 3) = 1 \div 4 \\( 3x + 2x ) \div 6 = 1 \div 4 \\ 5x \div 6 = 1 \div 4 \\ 20x = 6 \\ 10x = 3 \\ 10x - 3 = 0$

Clearly it is a linear equation. Hence, option c is correct.

d.

${x}^{2} + 2x + 3 = 0$

Clearly, it is a quadratic equation. Hence, option d is wrong.

Out of the four options, option c is a linear equation. Hence option c is correct.

Note: option d should be given as

${x}^{2} + 2x + 3 = 0$

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