Question

Solve 30x < 200 where x is an integer

Answer 1

Hint: First of all, divide both sides of the given inequality by 30. Now, simplify it by canceling the like terms and get the inequation of the type x < a where ‘a’ is any number. Now take all the integral values of x less than ‘a’ as the solution.

Complete step-by-step answer:
In this question, we have to solve 30x < 200 where x is an integer. Let us consider the inequality given in the question.
30x < 200
We know that if we divide an inequality with a positive number, the sign of inequality remains constant. So, by dividing 30 on both sides of the above inequality. We get,
\[\dfrac{30x}{30}<\dfrac{200}{30}\]
By canceling the like terms and simplifying the inequality, we get,
\[x<\dfrac{200}{30}\]
\[x<\dfrac{20}{3}\]
By changing \[\dfrac{20}{3}\] into decimal form to clearly visualize the inequality, we get,
x < 6.66….
Now, as we are given that x is an integer to x can take any integral values less than \[6.\overline{6}\]. So, we get the values of x as,
\[x=6,5,4,3,2,1,0,-1,-2,-3........-\infty \]
So, we can say that x can take any integral value from \[-\infty \] to 6 or we can write \[x\in \left[ -\infty ,6 \right]\text{ where }x\in I\] (I is for integer).

Note: In this type of question, first of all, students must remember that whenever a positive number is multiplied or divided in an inequality then the sign of the inequality remains the same but if a negative number is multiplied or divided in an inequality then the sign of inequality changes. For example, consider 3 > 2, if we multiply 2 both sides, we get 6 > 4 but if we multiply – 2 both sides, we get, - 6 < – 4. In this question, students can also cross-check their answer by taking two values of x and satisfying it in the given inequality as follows:
Let us take x = 0
30(x) < 200
By substituting x = 0 in inequality. We get,
30(0) < 200
0 < 200
which is true. So, our answer is correct. Similarly, we can check for other integral values of x from \[\left[ -\infty ,6 \right]\].