Question

Find the values of x, for given linear inequality $3x+8>2$ when
(i) x is an integer
(ii) x is a real number

Answer 1

Hint: Take +8 to the RHS of the equation and solve it and find values of x. In case one we have to take only integer values of x because x is an integer. In case 2, we can take any values either integers or non-integers.

Complete step-by-step solution -
According to the question, we have to find the possible values of x using the equation $3x+8>2$.
To solve this equation, first of all, take the constant term that is -8 to the RHS of the given equation.

& 3x+8>2 \\\ & 3x>-8+2 \\\ \end{aligned}$$ On solving, we get $$\begin{aligned} & 3x>-6 \\\ & x>-2 \\\ \end{aligned}$$ Here, we have to take only those values of x which is greater than -2. In case (i) we have to find those values of x which are integers. Since x is an integer, we can have x as $$........,-3,-2,-1,0,1,2,3,.........$$ But we have, $$x>-2$$. So, we have values of x = $$-1,0,1,2,3,4,................$$ In case (ii) we have to find those values of x which are real numbers. Since x is a real number greater than -2 i.e, $$x>-2$$ . So, we have values of x = $$\left( -2,\infty \right)$$ . Note: In this question, one can include -2 as values of x because -2 is an integer and a real number also. But, -2 should not be included because we have to take only those values of x which is greater than -2. So, -2 should not be included. In case (ii) one can make a mistake in taking only integer values of x. Integers are also real numbers. But, we cannot take only integer values. We also have to include the non-integer values and integer values in the range $$\left( -2,\infty \right)$$.