How do you solve \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\]?
Answer
599.7k+ views
Hint: An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value. Here we need to solve for ‘x’ which is a variable. Solving the given inequality is very like solving equations and we do most of the same thing but we must pay attention to the direction of inequality\[( \leqslant , > )\]. We have a simple linear equation type inequality and we can solve this easily.
Complete step-by-step solution:
Given \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\]
We need to solve for ‘x’.
Since we know that the direction of inequality doesn’t change if we multiply the same positive number on both sides. We multiply \[2x + 3\] on both sides, we have,
\[\Rightarrow x - 1 \leqslant 2x + 3\]
Similarly we subtract 3 on both side of the inequality,
\[\Rightarrow x - 3 - 1 \leqslant 2x\]
Similarly we subtract 3 on both side of the inequality,
\[ \Rightarrow - 3 - 1 \leqslant 2x - x\]
\[\Rightarrow - 4 \leqslant x\]
\[ \Rightarrow x \geqslant - 4\]
Thus the solution of \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\] is \[ \Rightarrow x \geqslant - 4\].
We can write it in the interval form. That is \[[ - 4,\infty )\].
Note: If we take a value of ‘w’ in \[[ - 4,\infty )\] and put it in \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\], it satisfies. That is
Let put \[x = - 4\] in \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\],
\[\dfrac{{ - 4 - 1}}{{2\left( { - 4} \right) + 3}} \leqslant 1\]
\[\dfrac{{ - 5}}{{ - 8 + 3}} \leqslant 1\]
\[\dfrac{{ - 5}}{{ - 5}} \leqslant 1\]
\[1 \leqslant 1\], which is true. Hence it satisfies .
We know that \[a \ne b\] says that ‘a’ is not equal to ‘b’. \[a > b\] means that ‘a’ is less than ‘b’. \[a < b\] means that ‘a’ is greater than ‘b’. These two are known as strict inequality. \[a \geqslant b\] means that ‘a’ is less than or equal to ‘b’. \[a \leqslant b\] means that ‘a’ is greater than or equal to ‘b’.
The direction of inequality do not change in these cases:
- Add or subtract a number from both sides.
- Multiply or divide both sides by a positive number.
- Simplify a side.
- The direction of the inequality change in these cases:
- Multiply or divide both sides by a negative number.
- Swapping left and right hand sides.
Complete step-by-step solution:
Given \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\]
We need to solve for ‘x’.
Since we know that the direction of inequality doesn’t change if we multiply the same positive number on both sides. We multiply \[2x + 3\] on both sides, we have,
\[\Rightarrow x - 1 \leqslant 2x + 3\]
Similarly we subtract 3 on both side of the inequality,
\[\Rightarrow x - 3 - 1 \leqslant 2x\]
Similarly we subtract 3 on both side of the inequality,
\[ \Rightarrow - 3 - 1 \leqslant 2x - x\]
\[\Rightarrow - 4 \leqslant x\]
\[ \Rightarrow x \geqslant - 4\]
Thus the solution of \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\] is \[ \Rightarrow x \geqslant - 4\].
We can write it in the interval form. That is \[[ - 4,\infty )\].
Note: If we take a value of ‘w’ in \[[ - 4,\infty )\] and put it in \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\], it satisfies. That is
Let put \[x = - 4\] in \[\dfrac{{x - 1}}{{2x + 3}} \leqslant 1\],
\[\dfrac{{ - 4 - 1}}{{2\left( { - 4} \right) + 3}} \leqslant 1\]
\[\dfrac{{ - 5}}{{ - 8 + 3}} \leqslant 1\]
\[\dfrac{{ - 5}}{{ - 5}} \leqslant 1\]
\[1 \leqslant 1\], which is true. Hence it satisfies .
We know that \[a \ne b\] says that ‘a’ is not equal to ‘b’. \[a > b\] means that ‘a’ is less than ‘b’. \[a < b\] means that ‘a’ is greater than ‘b’. These two are known as strict inequality. \[a \geqslant b\] means that ‘a’ is less than or equal to ‘b’. \[a \leqslant b\] means that ‘a’ is greater than or equal to ‘b’.
The direction of inequality do not change in these cases:
- Add or subtract a number from both sides.
- Multiply or divide both sides by a positive number.
- Simplify a side.
- The direction of the inequality change in these cases:
- Multiply or divide both sides by a negative number.
- Swapping left and right hand sides.
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