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Solve the system of linear equation of $3x+2y+25=0,2x+y+10=0$

seo-qna
Last updated date: 13th Jun 2024
Total views: 392.7k
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Answer
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Hint: We have a given set of equation:
$\begin{align}
  & 3x+2y+25=0......(1) \\
 & 2x+y+10=0......(2) \\
\end{align}$
To get the values of x and y we need to solve the given set of linear equations. We can solve a set of equations using a substitution method. In the substitution method, we get the value of one variable in terms of another variable from one equation and substitute it in the other equation to get the value of another variable. So, we will rearrange equation 1 and take y in terms of x and substitute it in rearranged equation 2.

Complete step-by-step solution
We can write the set of linear equations as:
$\begin{align}
  & 3x+2y=-25......(3) \\
 & 2x+y=-10......(4) \\
\end{align}$
So, by applying substitution method for the given set of linear equations:
From equation (4), we can write:
$\begin{align}
  & 3x+2y=-25 \\
 & \Rightarrow 2y=-25-3x \\
 & \Rightarrow y=\dfrac{-25-3x}{2}......(5) \\
\end{align}$
Now, we get the value of y in terms of x.
Put the value of y in equation (3), we get:
$\begin{align}
  & \Rightarrow 2x+y=-10 \\
 & \Rightarrow 2x+\left( \dfrac{-25-3x}{2} \right)=-10 \\
 & \Rightarrow 4x-25-3x=-20 \\
 & \Rightarrow x=5......(6) \\
\end{align}$
Put the value of x from equation (6) in equation (5) to get the value of y:
So, we have:
\[\begin{align}
  & y=\dfrac{-25-3\left( 5 \right)}{2} \\
 & =\dfrac{-25-15}{2} \\
 & =\dfrac{-40}{2} \\
 & =-20
\end{align}\]
Hence, x = 5 and y = -20

Note: There is another method to solve a given set of linear equations, i.e. Elimination method. In the elimination method, we multiply or divide both the equations with a number to eliminate one variable in both the equations by adding or subtracting them and get the value of another value. Then we put the value in one of the equations and get the value of the eliminated variable.
For the given set of equations:
$\begin{align}
  & 3x+2y=-25......(1) \\
 & 2x+y=-10......(2) \\
\end{align}$
Multiply equation (1) by 2 and equation (2) by 3, we get:
$\begin{align}
  & 6x+4y=-50......(3) \\
 & 6x+3y=-30......(4) \\
\end{align}$
Subtract equation (4) from equation (3); we get:
$\begin{align}
  & \Rightarrow 4y-3y=-50+30 \\
 & \Rightarrow y=-20......(5) \\
\end{align}$
Now put the value of y from equation (5) in any of the above equation to get the value of x.
Let’s put value of y in equation (1), we get:
$\begin{align}
  & \Rightarrow 3x+2y=-25 \\
 & \Rightarrow 3x+2\left( -20 \right)=-255 \\
 & \Rightarrow 3x-40=-25 \\
 & \Rightarrow 3x=15 \\
 & \Rightarrow x=5 \\
\end{align}$