How do you solve $2x-3y=8$ and $x+y=11$ using substitution?

Hint: Now to find the solution of the equation we will first consider the equation $x+y=11$ . Now we will try to write y in terms of x and substitute this value of y in the equation $2x-3y=8$ hence we get a linear equation in x which we will solve to find the value of x. Now we will substitute the value of x in the equation and find the value of y. hence we have the solution of the given equation.

Complete step by step solution:
Now consider the given equation $2x-3y=8$ and $x+y=11$ .
Now we want to find the values of x and y such that the values satisfy both the equations simultaneously.
To solve the equation we will use a method of substitution.
Now consider $x+y=11$
$\Rightarrow y=11-x$
Now we have y in terms of x.
Now we will substitute this value of y in equation $2x-3y=8$
Hence we get,
$2x-3\left( 11-x \right)=8$
Now we know that according to distributive property $a\left( b-c \right)=ab-ac$ Hence using this we get,
$\Rightarrow 2x-33+3x=8$
Now we will separate the variables and constants hence transposing 33 on RHS we get,
$\Rightarrow 2x+3x=33+8$
Now on simplifying we get,
$\Rightarrow 5x=41$
Now on dividing the whole equation by 5 we get $x=\dfrac{41}{5}$
Now we know that $y=11-x$ hence substituting the value of x we get,
$\begin{align}
  & \Rightarrow y=11-\dfrac{41}{5} \\
 & \Rightarrow y=\dfrac{55-41}{5} \\
 & \Rightarrow y=\dfrac{14}{5} \\
\end{align}$
Hence the solution of the given equation is $y=\dfrac{14}{5}$ and $x=\dfrac{41}{5}$ .

Note: Now note that we can solve the linear equation by various methods. We can plot the graphs of lines represented by both the equations. To find the graphical representation substitute random values of x in the equation and find corresponding values of y. Hence we have (x, y) is the point on line. Now plot these points on a graph and draw a line passing through them. Hence we have the equation of line. Now the solution of the system of linear equations is the intersection point of the two lines.