
Draw the graph of the equation given below \[x + y = 2\]
Answer
478.5k+ views
Hint: First, identify the given equation. Then rearrange the given equation with our convenience. Assume some \[x\] values to find the \[y\] values on the given equation. We get some points with the given equation. And mention that points value in tabular format. Labeled the points in a graph. And draw a straight line. Now we get the straight line graph on a given equation.
Complete answer:
Given equation is \[x + y = 2\]
First, we rearrange the linear equation \[x + y = 2\]
We change the equation in right-hand side
We change the equation in right-hand side \[y\] terms and left-hand side in \[x\] terms, constant terms only.
So add \[ - x\] term on both sides of the linear equation
\[x + y - x = 2 - x\]
Subtract the terms on both sides of the linear equation,
\[y = 2 - x\]
Now we get the equation is \[y = 2 - x\] .
Assume that,
\[x = - 3\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - ( - 3)\]
\[ = 2 + 3\]
\[ = 5\]
\[x = - 2\] apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - ( - 2)\]
\[ = 2 + 2\]
\[ = 4\]
\[x = - 1\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - ( - 1)\]
\[ = 2 + 1\]
\[ = 3\]
\[x = 0\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - (0)\]
\[ = 2\]
\[x = 1\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - 1\]
\[ = 1\]
\[x = 2\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - 2\]
\[ = 0\]
\[x = 3\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - 3\]
\[ = - 1\]
Now we mention the values in tabular format,
The Points are, \[( - 3,5),( - 2,4),( - 1,3),(0,2),(1,1),(2,0),(3, - 1)\]
Point out the points in the graph and join all points in a single line.
Draw the straight line.
Then we get a graph of \[x + y = 2\]
So that given equation is a straight line graph.
Note: We assume the \[x\] values to find the \[y\] values. Give the first preference to find the \[x\] values in \[ - 4 \leqslant x \leqslant 4\]these intervals. Because \[x \in [ - 4,4]\]in this interval we get the points are easily and simply. Draw a graph correctly labeled the points. Little change to the points labeled that are mis-matched in the graph.
Complete answer:
Given equation is \[x + y = 2\]
First, we rearrange the linear equation \[x + y = 2\]
We change the equation in right-hand side
We change the equation in right-hand side \[y\] terms and left-hand side in \[x\] terms, constant terms only.
So add \[ - x\] term on both sides of the linear equation
\[x + y - x = 2 - x\]
Subtract the terms on both sides of the linear equation,
\[y = 2 - x\]
Now we get the equation is \[y = 2 - x\] .
Assume that,
\[x = - 3\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - ( - 3)\]
\[ = 2 + 3\]
\[ = 5\]
\[x = - 2\] apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - ( - 2)\]
\[ = 2 + 2\]
\[ = 4\]
\[x = - 1\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - ( - 1)\]
\[ = 2 + 1\]
\[ = 3\]
\[x = 0\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - (0)\]
\[ = 2\]
\[x = 1\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - 1\]
\[ = 1\]
\[x = 2\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - 2\]
\[ = 0\]
\[x = 3\]apply the \[x\] value in the equation \[y\] Then,
\[y = 2 - 3\]
\[ = - 1\]
Now we mention the values in tabular format,
| \[x\] | \[ - 3\] | \[ - 2\] | \[ - 1\] | \[0\] | \[1\] | \[2\] | \[3\] |
| \[y\] | \[5\] | \[4\] | \[3\] | \[2\] | \[1\] | \[0\] | \[ - 1\] |
The Points are, \[( - 3,5),( - 2,4),( - 1,3),(0,2),(1,1),(2,0),(3, - 1)\]
Point out the points in the graph and join all points in a single line.
Draw the straight line.
Then we get a graph of \[x + y = 2\]
So that given equation is a straight line graph.
Note: We assume the \[x\] values to find the \[y\] values. Give the first preference to find the \[x\] values in \[ - 4 \leqslant x \leqslant 4\]these intervals. Because \[x \in [ - 4,4]\]in this interval we get the points are easily and simply. Draw a graph correctly labeled the points. Little change to the points labeled that are mis-matched in the graph.
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