Question

How do you solve $3x+2y=7$ and $x-y+3=0$ using substitution?

Answer 1

Hint: There are two unknowns $x$ and $y$ and also two equations to solve. We are applying the process of substitution and then the reduction. We take the value of the one variable and place that on another equation to solve the variables. We solve the equations equating the coefficients of one variable and omitting the variable. The other variable remains with the constants. Using the binary operation, we find the value of the other variable.

Complete step by step answer:
The given equations $3x+2y=7$ and $x-y+3=0$ are linear equations of two variables.
We know that the number of equations has to be equal to the number of unknowns to solve them.
We take the equations as $3x+2y=7.....(i)$ and $x-y+3=0......(ii)$.
We can also find the value of one variable $y$ with respect to $x$ based on the equation
$x-y+3=0$ where $y=x+3$. We replace the value of $y$ in the second equation of
$3x+2y=7$ and get
\[\begin{align}
  & 3x+2y=7 \\
 & \Rightarrow 3x+2\left( x+3 \right)=7 \\
 & \Rightarrow 3x+2x+6=7 \\
\end{align}\]
We get the equation of $x$ and solve
$\begin{align}
  & 3x+2x+6=7 \\
 & \Rightarrow 5x=7-6=1 \\
 & \Rightarrow x=\dfrac{1}{5} \\
\end{align}$
Putting the value of $x$ we get $y=x+3=\dfrac{1}{5}+3=\dfrac{16}{5}$.

Therefore, the values are $x=\dfrac{1}{5},y=\dfrac{16}{5}$.

Note: Now we solve it through reduction method.
We multiply 2 to the both sides of the second equations and get
$\begin{align}
  & 2\times \left( x-y \right)=2\times \left( -3 \right) \\
 & \Rightarrow 2x-2y=-6 \\
\end{align}$
We add this equation with $3x+2y=7$ to get
$\begin{align}
  & \left( 2x-2y \right)+\left( 3x+2y \right)=-6+7 \\
 & \Rightarrow 3x+2x=1 \\
 & \Rightarrow 5x=1 \\
 & \Rightarrow x=\dfrac{1}{5} \\
\end{align}$
The value of $x$ is $\dfrac{1}{5}$. Now putting the value in the equation $y=x+3$, we get
$y=x+3=\dfrac{1}{5}+3=\dfrac{16}{5}$.
Therefore, the values are $x=\dfrac{1}{5},y=\dfrac{16}{5}$.