Question

Check which of the following are solutions of an equation $x+2y=4$ ?
A. $\left( 0,2 \right)$
B. $\left( 2,0 \right)$
C. $\left( 4,0 \right)$
D. $\left( \sqrt{2},-3\sqrt{2} \right)$
E. $\left( 1,1 \right)$
F. $\left( -2,3 \right)$

Answer 1

Hint:First we will write down the process for checking whether the given points are a solution of the line $x+2y=4$, after writing the condition we will put each of the given points into that condition and if the taken point satisfies the condition then it will be a solution for the given equation otherwise not.

Complete step by step answer:
We are given the following equation: $x+2y=4$, now as we know a given point is called as the solution of the equation if it satisfies the equation, let’s take a point $\left( x,y \right)$ , we will put this point into the equation and if left-hand side equals right-hand side that means it will satisfy the equation and hence, it will be considered as the solution of the equation.
Let’s start by considering the options one by one:
First we have: $\left( 0,2 \right)$, now for this point we have $x=0$ and $y=2$ , we will put these values in the given equation that is: $x+2y=4$, therefore:
$\Rightarrow \left( 0 \right)+2.\left( 2 \right)=4\Rightarrow 4=4$
As we can see the left-hand side is equal to the right hand side, therefore $\left( 0,2 \right)$ is a solution of : $x+2y=4$.
Similarly, next we take: $\left( 2,0 \right)$, now for this point we have $x=2$ and $y=0$ , we will put these values in the given equation that is: $x+2y=4$, therefore:
$\Rightarrow \left( 2 \right)+2.\left( 0 \right)=4\Rightarrow 2=4$
As we can see the left-hand side is not equal to the right-hand side , therefore $\left( 2,0 \right)$ is not a solution of : $x+2y=4$.

Next, we have with us: : $\left( 4,0 \right)$, now for this point we have $x=4$ and $y=0$ , we will put these values in the given equation that is: $x+2y=4$, therefore:
$\Rightarrow \left( 4 \right)+2.\left( 0 \right)=4\Rightarrow 4=4$
As we can see the left-hand side is equal to the right-hand side , therefore $\left( 4,0 \right)$ is a solution of : $x+2y=4$.

Next, we have with us: : $\left( \sqrt{2},-3\sqrt{2} \right)$, now for this point we have $x=\sqrt{2}$ and $y=-3\sqrt{2}$ , we will put these values in the given equation that is: $x+2y=4$, therefore:
$\Rightarrow \left( \sqrt{2} \right)+2.\left( -3\sqrt{2} \right)=4\Rightarrow -5\sqrt{2}=4$
As we can see the left hand side is not equal to the right hand side , therefore$\left( \sqrt{2},-3\sqrt{2} \right)$ is not a solution of : $x+2y=4$.

Next, we have with us: : $\left( 1,1 \right)$, now for this point we have $x=1$ and $y=1$ , we will put these values in the given equation that is: $x+2y=4$, therefore:
$\Rightarrow \left( 1 \right)+2.\left( 1 \right)=4\Rightarrow 3=4$
As we can see the left-hand side is not equal to the right-hand side , therefore $\left( 1,1 \right)$ is not a solution of : $x+2y=4$.

Finally, we have with us: : $\left( -2,3 \right)$, now for this point we have $x=-2$ and $y=3$ , we will put these values in the given equation that is: $x+2y=4$, therefore:
$\Rightarrow \left( -2 \right)+2.\left( 3 \right)=4\Rightarrow 4=4$
As we can see the left hand side is equal to the right hand side , therefore $\left( -2,3 \right)$ is a solution of : $x+2y=4$.
So, in conclusion: A, C, and F are the solutions of the given equation.

Note:
These questions are normally easy and do not require many calculations, one can easily solve them and you are not required to show this much calculations. Students can always plot the given points and draw the given line on the graph, this will give an additional advantage to the student. So, the graph would be as below,

We see that the solutions of the line lie on it and the points which are not the solutions lie outside the line.