Question

How do you solve $ 4x + 1 \leqslant 17 $ and $ 5x - 1 > 19 $ ?

Answer 1

Hint: To solve this problem, we will take one problem at a time. Therefore, when we consider the first interval and solve for x. We get that the solution of x is $ x \leqslant 4 $ . After that, when we consider the second interval and solve for x. We get that the solution of x is $ x > 4 $ . When we analyze both the intervals together, we see that it clearly says that the solution set is $ \left( { - \infty ,\infty } \right) $ .

Complete step-by-step solution:
The given question we have is $ 4x + 1 \leqslant 17 $ and $ 5x - 1 > 19 $
Taking the first interval, we will solve it first and then move on to the second one
Subtracting 1 from both the sides of the inequality, we will get:-
 $
  4x + 1 - 1 \leqslant 17 - 1 \\
   \Rightarrow 4x \leqslant 16 \\
 $
Dividing both the sides by 4, we will get
 $
  4x \leqslant 16 \\
  \dfrac{{4x}}{4} \leqslant \dfrac{{16}}{4} \\
   \Rightarrow x \leqslant \dfrac{{{4^2}}}{4} \\
   \Rightarrow x \leqslant 4 \\
 $
Hence, we can conclude that for interval 1, that is $ 4x + 1 \leqslant 17 $ . The solution will be $ x \leqslant 4 $ .
Taking the second interval now, we will add 1 to both the sides of the inequality which will give us:-
 $
  5x - 1 + 1 > 19 + 1 \\
   \Rightarrow 5x > 20 \\
 $
Now, to remove the 5 as a coefficient from x, we will divide both the sides of the inequality by 5.
 $ 5x > 20 \\
   \Rightarrow \dfrac{{5x}}{5} > \dfrac{{20}}{5} \\
   \Rightarrow x > \dfrac{{5 \times 4}}{5} \\
   \Rightarrow x > 4 \\
 $
Hence, we can conclude that for interval 2 that is $ 5x - 1 > 19 $ . The solution will be $ x > 4 $ .
Now, if we combine see both the intervals, which are
 $ x \leqslant 4 $ and $ x > 4 $
We can conclude that the solutions are saying that. The answer interval of this question is the set of all real numbers from positive 4 to infinity and 4(including -4) to minus infinity.
Which basically means solution is the set of all real numbers
Hence, in mathematical language, we can say that
Solution interval= $ ( - \infty ,4] \cup (4,\infty ) $

Note: Instead of writing the solution as $ ( - \infty ,4] \cup (4,\infty ) $ we can use our mind a little bit and write the answer as $ ( - \infty ,4] \cup (4,\infty ) $ . This is because, all the numbers from minus infinity to 4(including 4) and then after 4 to positive infinity is actually the set of all the real numbers. Hence, the changed form of the answer is the better one to write.