How do you solve \[2x+11\le 5x-10\]?
Answer
571.5k+ views
Hint: This type of problem is based on the concept of inequality. First, we have to consider the whole function. We need to subtract the whole equation by 5x so then we get x terms in the left-hand side of the equation. Then, we need to add the whole equation with -11. And do some necessary calculations. And then, divide the whole equation by -3 using the property of inequality. Solve x considering the inequality.
Complete step by step solution:
According to the question, we are asked to solve the inequality \[2x+11\le 5x-10\].
We have been given the inequality is \[2x+11\le 5x-10\]. -----(1)
We first have to subtract the whole inequality by 5x.
\[\Rightarrow 2x+11-5x\le 5x-10-5x\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 5x, we get
\[\Rightarrow 2x+11-5x\le -10\]
On grouping the x terms in the left-hand side of the inequality and solving them, we get
\[-3x+11\le -10\]
Now, we need to subtract the whole equation by 11.
\[\Rightarrow -3x+11-11\le -10-11\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 11, we get
\[-3x\le -10-11\]
On further simplification, we get
\[-3x\le -21\]
We find that 3 are common in both the sides of the inequality.
To cancel 3, we have to divide the obtained inequality by 3.
\[\Rightarrow \dfrac{-3x}{3}\le \dfrac{-21}{3}\]
We can express the inequality as
\[\dfrac{-3x}{3}\le \dfrac{-3\times 7}{3}\]
We find that 3 are common in both the numerator and denominator of LHS and RHS of the inequality. Cancelling 3, we get
\[-x\le -7\]
Now, we have to find the value of x.
But we know that, if \[-x\le -a\], then \[x\ge a\].
Here, a=7.
Using this rule of inequality, we get
\[x\ge 7\]
We can also write the value of x as \[\left[ 7,\infty \right)\].
Therefore, the value of x in the given inequality \[2x+11\le 5x-10\] is \[x\ge 7\], that is \[x\in \left[ 7,\infty \right)\].
Note: Whenever you get this type of problem, we should always try to make the necessary changes in the given inequality to get the variable in the left-hand side of the equation. We should avoid calculation mistakes based on sign conventions. We should always use the rule of inequality wherever required. We should not write the solution of the equation as \[\left( 7,\infty \right)\] since 7 is also included.
Complete step by step solution:
According to the question, we are asked to solve the inequality \[2x+11\le 5x-10\].
We have been given the inequality is \[2x+11\le 5x-10\]. -----(1)
We first have to subtract the whole inequality by 5x.
\[\Rightarrow 2x+11-5x\le 5x-10-5x\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 5x, we get
\[\Rightarrow 2x+11-5x\le -10\]
On grouping the x terms in the left-hand side of the inequality and solving them, we get
\[-3x+11\le -10\]
Now, we need to subtract the whole equation by 11.
\[\Rightarrow -3x+11-11\le -10-11\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 11, we get
\[-3x\le -10-11\]
On further simplification, we get
\[-3x\le -21\]
We find that 3 are common in both the sides of the inequality.
To cancel 3, we have to divide the obtained inequality by 3.
\[\Rightarrow \dfrac{-3x}{3}\le \dfrac{-21}{3}\]
We can express the inequality as
\[\dfrac{-3x}{3}\le \dfrac{-3\times 7}{3}\]
We find that 3 are common in both the numerator and denominator of LHS and RHS of the inequality. Cancelling 3, we get
\[-x\le -7\]
Now, we have to find the value of x.
But we know that, if \[-x\le -a\], then \[x\ge a\].
Here, a=7.
Using this rule of inequality, we get
\[x\ge 7\]
We can also write the value of x as \[\left[ 7,\infty \right)\].
Therefore, the value of x in the given inequality \[2x+11\le 5x-10\] is \[x\ge 7\], that is \[x\in \left[ 7,\infty \right)\].
Note: Whenever you get this type of problem, we should always try to make the necessary changes in the given inequality to get the variable in the left-hand side of the equation. We should avoid calculation mistakes based on sign conventions. We should always use the rule of inequality wherever required. We should not write the solution of the equation as \[\left( 7,\infty \right)\] since 7 is also included.
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