
How do you solve and write the following in the interval equation : $7\ge 2x-5$ or $\dfrac{3x-4}{4}>4$ ?
Answer
557.1k+ views
Hint: There are 2 inequalities in the question. To solve this question we will solve both inequalities separately and the final answer will be a union of the 2 separate answers because we can see there is a word or present between 2 inequalities.
Complete step by step answer:
Given 2 inequalities are $7\ge 2x-5$ or $\dfrac{3x-4}{4}>4$
First solve $7\ge 2x-5$ , we can add 5 to both LHS and RHS
$\Rightarrow 12\ge 2x$
Now we can divide both LHS and RHS by 2
$6\ge x$ ; so x is always less than equal to 6, $x\in \left( -\infty ,6 \right]$
Let’s solve $\dfrac{3x-4}{4}>4$ , we can multiply 4 with LHS and RHS
$\Rightarrow 3x-4>16$
Now we can add 4 to both sides
$\Rightarrow 3x>20$
So $x>\dfrac{20}{3}$ so x is always greater than $\dfrac{20}{3}$ , $x\in \left( \dfrac{20}{3},\infty \right)$
So the final answer is $\left( -\infty ,6 \right]\cup \left( \dfrac{20}{3},\infty \right)$
If x is in the range $\left( -\infty ,6 \right]\cup \left( \dfrac{20}{3},\infty \right)$ then it will satisfy inequality $7\ge 2x-5$ or $\dfrac{3x-4}{4}>4$
Note:
While writing the range of the variable, keep in mind that if we put the parenthesis bracket then the boundary value will not come in the range and if we use square brackets then the boundary value will come in the range. Whenever we write $\infty $ or $-\infty $ we always use parentheses brackets. When we multiply or divide some negative number to both LHS and RHS in an inequality then the inequality sign changes. If it is less than, it will change into greater than and if it is greater than it will change into less than.
Complete step by step answer:
Given 2 inequalities are $7\ge 2x-5$ or $\dfrac{3x-4}{4}>4$
First solve $7\ge 2x-5$ , we can add 5 to both LHS and RHS
$\Rightarrow 12\ge 2x$
Now we can divide both LHS and RHS by 2
$6\ge x$ ; so x is always less than equal to 6, $x\in \left( -\infty ,6 \right]$
Let’s solve $\dfrac{3x-4}{4}>4$ , we can multiply 4 with LHS and RHS
$\Rightarrow 3x-4>16$
Now we can add 4 to both sides
$\Rightarrow 3x>20$
So $x>\dfrac{20}{3}$ so x is always greater than $\dfrac{20}{3}$ , $x\in \left( \dfrac{20}{3},\infty \right)$
So the final answer is $\left( -\infty ,6 \right]\cup \left( \dfrac{20}{3},\infty \right)$
If x is in the range $\left( -\infty ,6 \right]\cup \left( \dfrac{20}{3},\infty \right)$ then it will satisfy inequality $7\ge 2x-5$ or $\dfrac{3x-4}{4}>4$
Note:
While writing the range of the variable, keep in mind that if we put the parenthesis bracket then the boundary value will not come in the range and if we use square brackets then the boundary value will come in the range. Whenever we write $\infty $ or $-\infty $ we always use parentheses brackets. When we multiply or divide some negative number to both LHS and RHS in an inequality then the inequality sign changes. If it is less than, it will change into greater than and if it is greater than it will change into less than.
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