Question

How do you solve the system \[x=5-y\] and \[3y=3x+1\] by substitution method?

Answer 1

Hint: we can solve this question using basic linear equation concepts. We will first find the value of x using one equation. Then we will substitute the x value we got in the second equation and then we solve the equation for y. using the value of y we will find the value x. so that we can arrive at our solution.

Complete step by step solution:
Given equations are
\[x=5-y\]
\[3y=3x+1\]
First we have to find the value of x using any one equation.
We will take the first equation because we can see that it is given as x value so that we can avoid the calculation to find it using the second equation.
So the x value we have is
\[x=5-y\]
Now we have to substitute this x value in the second equation.
By substituting the x value in the second equation we will get
\[\Rightarrow 3y=3\left( 5-y \right)+1\]
Now we have to simplify the equation.
Now we will remove the parenthesis in the equation.
By removing parenthesis we will get
\[\Rightarrow 3y=15-3y+1\]
Now we will further simplify the equation. We will get
\[\Rightarrow 3y=16-3y\]
Now we have to simplify it for finding the value of y.
Now we will add \[3y\]on both sides of the equation.
We will get
\[\Rightarrow 3y+3y=16-3y+3y\]
By simplifying we will get
\[\Rightarrow 3y+3y=16\]
\[\Rightarrow 6y=16\]
Now we have to with 6 on both sides of the equation.
\[\Rightarrow \dfrac{6y}{6}=\dfrac{16}{6}\]
By simplifying we will get
\[\Rightarrow y=\dfrac{8}{3}\]
So the y value we got is \[\dfrac{8}{3}\].
Now using this y value we have to find x value.
Substitute this y value in the first equation to get the value.
By substituting we will get
 \[x=5-\dfrac{8}{3}\]
By simplifying we will get
\[\Rightarrow x=\dfrac{15-8}{3}\]
\[\Rightarrow x=\dfrac{7}{3}\]

So by solving the given equations we will get x and y as \[x=\dfrac{7}{3}\] and \[y=\dfrac{8}{3}\].

Note: We can also do this using a layered approach. But as it is specified to solve using the substitution method we used this method. We can also check the solutions by substituting the values in both equations and check whether they are satisfied or not.