Question

Solve 5x – 3 < 3x + 1 when
(A). x is an integer
(B). x is a real number.

Answer 1

- Hint: Solve the given expression until you get a simple expression of x. Now for x is an integer’s which comes in its range. Similarly, when x is a real number, find its range.

Complete step-by-step solution -

(a). We have been given the expression, 5x – 3 < 3x + 1. It is given that x is an integer. We know what an integer is, it is a number with number no fractional part i.e. no decimals.
Thus, we can say that here integers are -2, -1, 0, 1, 2, 3…….etc.
Now, we have been given, 5x – 3 < 3x + 1.
Let us modify the above expression,
\[\begin{align}
  & 5x-3<3x+1 \\
 & 5x-3x<1+3 \\
 & 2x<4 \\
 & x<\dfrac{4}{2} \\
 & x<2-(1) \\
\end{align}\]
Since x is an integer, (…….-3, -2, -1, 0, 1, 2, 3,……)
We need to find the values of x which are less than 2.
So, the value of x can be …….-3, -2, -1, 0, 1.
Thus x = {…..-3, -2, -1, 0, 1}.
(b). It is given that x is a real number. The real number includes all the rational numbers, such as integers, fraction and all irrational numbers. We got that x < 2 from equation (2). Since x is a real number which is less than 2, so it can go till \[-\infty \].
\[\therefore x\in \left( -\infty ,2 \right)\]
Thus we solved, 5x – 3 < 3x + 1 when,
x is an integer. x = {…….-3, -2, -1, 0, 1}.
x is a real number, \[x\in \left( -\infty ,2 \right)\].

Note: You should remember the concept of integers and real numbers to solve this type of question. Integers include positive and negative whole numbers. Whereas real numbers include rational numbers, irrational numbers including integers.