 # The solution of which of the following

Question:

The solution of which of the following equations is neither a fraction nor an integer?

(a) -3x + 2=5x + 2

(b)4x-18=2

(c)4x + 7 = x + 2

(d)5x-8 = x +4

Solution:

For option (c)

Given linear equation is $\quad 3 x+2=5 x+2$

$\Rightarrow \quad 3 x-5 x=2-2 \quad$ [transposing $5 x$ to LHS and 2 to RHS]

$\Rightarrow \quad-2 x=0$

$\Rightarrow$ $\frac{-2 x}{-2}=\frac{0}{-2}$ [dividing both sides by $-2$ ]

$\therefore$ $x=0$

Hence, $x=0$ is an integer.

For option (b)

Given lingar equation is $\quad 4 x-18=2$

$\Rightarrow$  $4 x=2+18$  [transposing $-18$ to RHS]

$\Rightarrow$ $4 x=20$

$\Rightarrow$ $\frac{4 x}{4}=\frac{20}{4}$ [dividing both sides by 4 ]

$\because$ $x=5$

Hence, $x=5$ is a positive integer.

For option (c)

Given linear equation is $4 x+7=x+2$

$\Rightarrow \quad 4 x-x=2-7 \quad$ [transposing $x$ to LHS and 7 to RHS]

$\Rightarrow \quad 3 x=-5$

$\therefore$ $x=\frac{-5}{3}$ [dividing both sides by 3 ]

Hence, $x=\frac{-5}{3}$ is neither a fraction nor an integer.

For option (d)

Given linear equation is $5 x-8=x+4$

$\Rightarrow \quad 5 x-x=4+8 \quad$ [transposing $x$ to LHS and $-8$ to RHS]

$\Rightarrow$ $\frac{4 x}{4}=\frac{12}{4}$ [dividing both sides by 4]

$\therefore \quad x=3$

Hence, $x=3$ is a positive integer.

From the above it is clear that, option (c) satisfies the condition.