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Question: How do you solve and draw the graph \(\dfrac{2x}{3} >4x+10\)?...

How do you solve and draw the graph 2x3>4x+10\dfrac{2x}{3} >4x+10?

Explanation

Solution

We try to take points which have x coordinates that satisfies 2x3>4x+10\dfrac{2x}{3}>4x+10. There is no restriction on the y coordinates. Based on the points we try to find the space or region in the 2-D plane which satisfies 2x3>4x+10\dfrac{2x}{3}>4x+10. We multiply with 3 and solve to get the required interval.

Complete step-by-step solution:
The inequation 2x3>4x+10\dfrac{2x}{3}>4x+10 represents the space or region in 2-D plane where the x coordinates of points satisfy 2x3>4x+10\dfrac{2x}{3}>4x+10.
We can solve the inequation treating them as equations for the operations like addition and subtraction. In case of multiplication and division we need to watch out for the negative values as that changes the inequality sign.
We first take some points for the x coordinates where 2x3>4x+10\dfrac{2x}{3}>4x+10.
We multiply with 3 to both sides of the inequality.
3×2x3>3(4x+10) 2x>12x+30 \begin{aligned} & 3\times \dfrac{2x}{3}>3\left( 4x+10 \right) \\\ & \Rightarrow 2x>12x+30 \\\ \end{aligned}
Now we take the variables on one side and get
2x>12x+30 2x12x>30 10x>30 \begin{aligned} & 2x>12x+30 \\\ & \Rightarrow 2x-12x>30 \\\ & \Rightarrow -10x>30 \\\ \end{aligned}
Now we divide with negative number 10-10 and get

& \dfrac{-10x}{-10}<\dfrac{30}{-10} \\\ & \Rightarrow x<-3 \\\ \end{aligned}$$ **The interval is $x <-3$.** ![](https://www.vedantu.com/question-sets/f2c8858e-34e3-4151-a969-5ff6b158326b3073605444269861098.png) **Note:** We can also express the inequality as the interval system where $\dfrac{2x}{3}>4x+10$ defines that $x\in \left( -\infty ,-3 \right)$. The interval for the y coordinates will be anything which can be defined as $y\in \left( -\infty ,\infty \right)$. We also need to remember that the points on the line $\dfrac{2x}{3}>4x+10$ will not be the solution for the inequation.