How do you solve for y and graph y+2x=3?

Hint: This type of question is based on the concept of linear equations with two variables. We have to first consider the given equation and subtract 2x from both the sides of the equation to solve y. Then, substitute y=0 and find the value of x. Similarly, substitute x=0 and find the value of y. Therefore, we get the x-intercept and y-intercept of the given linear equation. Plot the x and y intercepts in a graph and joins the two points with a line. Thus, we have plotted the graph for y+2x=3.

Complete step-by-step solution:
According to the question, we are asked to solve for y and graphs of y+2x=3.
We have been given the equation is y+2x=3. ----------(1)
First, we have to consider the equation (1).
Subtract 2x from both the sides of the equation. We get
y+2x-2x=3-2x
We know that the terms with the same magnitude and opposite signs cancel out.
Therefore, y=3-2x
Now, we have to plot a graph for y+2x=3.
Since it is a linear equation with two variables, find the intercepts are enough to plot the graph.
To find the x-intercept, we have to substitute y=0.
0+2x=3
\[\Rightarrow 2x=3\]
Divide the whole expression by 2. We get
\[\Rightarrow \dfrac{2x}{2}=\dfrac{3}{2}\]
Cancelling out the common term 2 from the LHS, we get
\[x=\dfrac{3}{2}\]
Therefore, the x-intercept is \[\dfrac{3}{2}\] and the point is \[\left( \dfrac{3}{2},0 \right)\].
Now, we have to find the y-intercept.
Substitute x=0. We get,
y+0=3
Therefore, y=3.
Hence, the y-intercept is 3 and the point is (0,3).
Plot the points \[\left( \dfrac{3}{2},0 \right)\] and (0,3) in the graph.
Join the two points with a straight line.
We get

Therefore, the value of y in y+2x=3 is 3-2x and the graph of y+2x=3 is obtained.

Note: We should simply draw straight lines between the two points. Since the given equation is a linear equation, the graph will always be a straight line. It is enough to just find the x and y intercepts and plot a graph with those points. Avoid calculation mistakes based on finding the values of the intercepts.