
How do you solve $\dfrac{1}{3}x+2>3$ and $-3x>12$ ?
Answer
451.8k+ views
Hint: We have been given two linear inequalities in one variable. Thus, we shall determine the interval in which variable-x lies in both the inequalities so that both of them hold true simultaneously. First of all, we shall transpose the constant term from the left hand side to the right hand side in each inequality. Then we shall divide both sides of the inequality to make the coefficient of variable-x equal to 1.
Complete step by step solution:
Given that $\dfrac{1}{3}x+2>3$ and $-3x>12$.
Firstly we shall solve $\dfrac{1}{3}x+2>3$ and transpose the constant term 2 from the left hand side to the right hand side of the equation.
$\Rightarrow \dfrac{1}{3}x>3-2$
$\Rightarrow \dfrac{1}{3}x>1$
Now, we shall multiply both sides of the inequality by 3 to make the coefficient of variable-x equal to 1.
$\Rightarrow x>3\left( 1 \right)$
$\Rightarrow x>3$
Here, we have obtained our inequality in the simplest form and variable-x is greater than 3.
Therefore, the solution of the inequality $\dfrac{1}{3}x+2>3$ is $x\in \left( 3,\infty \right)$.
Now we shall solve $-3x>12$
We have to divide both sides of the inequality by -3 to make the coefficient of variable-x equal to 1.
$\Rightarrow x>\dfrac{12}{-3}$
$\Rightarrow x>-4$
Here, we have obtained our inequality in the simplest form and variable-x is greater than -4.
Therefore, the solution of the inequality $-3x>12$ is $x\in \left( -4,\infty \right)$.
However, here we have to consider those values of x only which satisfies both the inequalities. Thus, we shall take the intersection of the two results obtained above, that is $x\in \left( 3,\infty \right)$ and $x\in \left( -4,\infty \right)$.
$x\in \left( 3,\infty \right)\cap \left( -4,\infty \right)$ is equal to $x\in \left( 3,\infty \right)$ only.
Therefore, the solution of $\dfrac{1}{3}x+2>3$ and $-3x>12$ is $x\in \left( 3,\infty \right)$.
Note: In this problem, we have taken the intersection of the two individual solutions of the two given inequalities because the word ‘and’ was mentioned which means that the solution set must satisfy both the inequalities. If the word ‘or’ would have been used in the question, then we would have taken the union of the two solution sets of the two individual equations.
Complete step by step solution:
Given that $\dfrac{1}{3}x+2>3$ and $-3x>12$.
Firstly we shall solve $\dfrac{1}{3}x+2>3$ and transpose the constant term 2 from the left hand side to the right hand side of the equation.
$\Rightarrow \dfrac{1}{3}x>3-2$
$\Rightarrow \dfrac{1}{3}x>1$
Now, we shall multiply both sides of the inequality by 3 to make the coefficient of variable-x equal to 1.
$\Rightarrow x>3\left( 1 \right)$
$\Rightarrow x>3$
Here, we have obtained our inequality in the simplest form and variable-x is greater than 3.
Therefore, the solution of the inequality $\dfrac{1}{3}x+2>3$ is $x\in \left( 3,\infty \right)$.
Now we shall solve $-3x>12$
We have to divide both sides of the inequality by -3 to make the coefficient of variable-x equal to 1.
$\Rightarrow x>\dfrac{12}{-3}$
$\Rightarrow x>-4$
Here, we have obtained our inequality in the simplest form and variable-x is greater than -4.
Therefore, the solution of the inequality $-3x>12$ is $x\in \left( -4,\infty \right)$.
However, here we have to consider those values of x only which satisfies both the inequalities. Thus, we shall take the intersection of the two results obtained above, that is $x\in \left( 3,\infty \right)$ and $x\in \left( -4,\infty \right)$.
$x\in \left( 3,\infty \right)\cap \left( -4,\infty \right)$ is equal to $x\in \left( 3,\infty \right)$ only.
Therefore, the solution of $\dfrac{1}{3}x+2>3$ and $-3x>12$ is $x\in \left( 3,\infty \right)$.
Note: In this problem, we have taken the intersection of the two individual solutions of the two given inequalities because the word ‘and’ was mentioned which means that the solution set must satisfy both the inequalities. If the word ‘or’ would have been used in the question, then we would have taken the union of the two solution sets of the two individual equations.
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