Question

How do you solve $3x+2y=5$ and $2x+5y=-3$?

Answer 1

Hint: The two given equations $3x+2y=5$ and $2x+5y=-3$ can be solved through substitution of variables. We find the value of one variable from one equation and put it in the other one to solve it. We also use the method of reduction to solve the equations.

Complete step by step solution:
The given equations $3x+2y=5$ and $2x+5y=-3$ 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 $3x+2y=5.....(i)$ and $2x+5y=-3......(ii)$.
We can also find the value of one variable $y$ with respect to $x$ based on the equation
$3x+2y=5$ where $y=\dfrac{5-3x}{2}$. We replace the value of $y$ in the second equation of
$2x+5y=-3$ and get
\[\begin{align}
  & 2x+5y=-3 \\
 & \Rightarrow 2x+5\left( \dfrac{5-3x}{2} \right)=-3 \\
 & \Rightarrow 4x-15x+25=-6 \\
\end{align}\]
We get the equation of $x$ and solve
$\begin{align}
  & 4x-15x+25=-6 \\
 & \Rightarrow -11x=-25-6=-31 \\
 & \Rightarrow x=\dfrac{-31}{-11}=\dfrac{31}{11} \\
\end{align}$
Putting the value of $x$ we get $y=\dfrac{5-3x}{2}=\dfrac{5-3\times \dfrac{31}{11}}{2}=-\dfrac{19}{11}$.

Therefore, the values are $x=\dfrac{31}{11},y=-\dfrac{19}{11}$.

Note: Now we solve it through a reduction method.
We multiply 2 and 3 to the both sides of the first and second equations respectively and get
$\begin{align}
  & 2\times \left( 3x+2y \right)=2\times 5 \\
 & \Rightarrow 6x+4y=10 \\
\end{align}$
For the second equation
$\begin{align}
  & 3\times \left( 2x+5y \right)=3\times \left( -3 \right) \\
 & \Rightarrow 6x+15y=-9 \\
\end{align}$
We subtract these equations to get
$\begin{align}
  & \left( 6x+4y \right)-\left( 6x+15y \right)=10-\left( -9 \right) \\
 & \Rightarrow -11y=19 \\
 & \Rightarrow y=-\dfrac{19}{11} \\
\end{align}$
The value of $y$ is $-\dfrac{19}{11}$. Now putting the value in the equation $2x+5y=-3......(ii)$, we get
$\begin{align}
  & 2x+5y=-3 \\
 & \Rightarrow 2x=\dfrac{-3-5\left( -\dfrac{19}{11} \right)}{2} \\
 & \Rightarrow x=\dfrac{31}{11} \\
\end{align}$.
Therefore, the values are $x=\dfrac{31}{11},y=-\dfrac{19}{11}$.