Question

How do you solve $\dfrac{3}{4}x+\dfrac{1}{4}y>1$?

Answer 1

For finding the region that satisfies the inequality $\dfrac{3}{4}x+\dfrac{1}{4}y>1$, we need to find the equality $3x+y=4$. This gives the line graph. Change of form of the given equation will give the x-intercept and y-intercept of the line $3x+y=4$. We change it to the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$ to find the x intercept, and y intercept of the line as $p$ and $q$ respectively. Then we place the points on the axes and from there we draw the line on the graph. We find the required region based on the origin point’s validity.

Complete step by step solution:
We first simplify the inequation $\dfrac{3}{4}x+\dfrac{1}{4}y>1$ into equation form to get $3x+y=4$
We have to find the x-intercept, and y-intercept of the line $3x+y=4$.
For this we convert the given equation into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the form we get that the x intercept, and y intercept of the line will be$p$ and $q$ respectively. The points will be $\left( p,0 \right),\left( 0,q \right)$.
The given equation is $3x+y=4$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get

& 3x+y=4 \\\ & \Rightarrow \dfrac{3x}{4}+\dfrac{y}{4}=1 \\\ & \Rightarrow \dfrac{x}{{}^{4}/{}_{3}}+\dfrac{y}{4}=1 \\\ \end{aligned}$$ Therefore, the x intercept, and y intercept of the line $3x+y=4$ is $\dfrac{4}{3}$ and 4 respectively. The axes intersecting points are $\left( \dfrac{4}{3},0 \right),\left( 0,4 \right)$. The line $3x+y=4$ divides the region in two parts. We find the required one based on the credibility of the origin point. The origin point $\left( 0,0 \right)$ doesn’t satisfy the inequality $\dfrac{3}{4}x+\dfrac{1}{4}y>1$ as $0+0>1$ is false. The opposite side of that region of the line $3x+y=4$ would be the required region. **Note:** We take the origin point as that helps in multiplications. We can use any point other than the ones that are on the line itself to find out the required region. For example, we take the point $\left( -1,-1 \right)$ instead of $\left( 0,0 \right)$. This again doesn’t satisfy the inequality $\dfrac{3}{4}x+\dfrac{1}{4}y>1$ where $-1<1$. This means we can take any arbitrary point and that will indicate the area accordingly.