Question

How do you solve \[2x-3y=1\] and \[5x+4y=14\] using substitution?

Answer 1

Hint: Any two linear equations can be solved to get a common point lying on their graphs. We solve these equations by addition and subtraction methods. We can solve these equations by substituting one variable after making required modifications such that any one of the variables remains in an equation. Then, we substitute the rearranged variable into the second equation and find the value of each variable.

Complete step by step answer:
As per the given question we need to solve \[2x-3y=1\] and \[5x+4y=14\] to get a common point called a solution of these equations.
Let \[2x-3y=1\] \[---\left( 1 \right)\]
      \[5x+4y=14\] \[---\left( 2 \right)\]
Let us consider equation \[\left( 1 \right)\]
Now we add \[-2x\] on both sides to the equation \[\left( 1 \right)\]. Then the equation becomes
\[\begin{align}
  & \Rightarrow -2x+2x-3y=1-2x \\
 & \Rightarrow -3y=1-2x \\
\end{align}\]
Now we divide with -3 on both sides. Then the equation becomes
\[\Rightarrow \dfrac{-3y}{-3}=\dfrac{1-2x}{-3}\]
\[\Rightarrow y=\dfrac{1-2x}{-3}\] \[---\left( 3 \right)\]
Now we substitute equation 3 in equation 2. Then the equation becomes
\[\Rightarrow 5x+4\left( \dfrac{1-2x}{-3} \right)=14\]
Now we multiply with -3 on both sides. By using distributive property, the equation becomes
\[\Rightarrow 5x\times -3+4\left( \dfrac{1-2x}{-3} \right)\times -3=14\times -3\]
\[\Rightarrow -15x+4\left( 1-2x \right)=-42\]
\[\Rightarrow -15x+4-8x=-42\]
\[\Rightarrow -23x=-46\]
\[\Rightarrow x=\dfrac{-46}{-23}\]
\[\Rightarrow x=2\]
Now we substitute the value of x in equation 1. Then the equation becomes
\[\begin{align}
  & \Rightarrow 2x-3y=1 \\
 & \Rightarrow 2\left( 2 \right)-3y=1 \\
\end{align}\] \[\begin{align}
  & \Rightarrow 5x\times -3+4\left( \dfrac{1-2x}{-3} \right)\times -3=14\times -3 \\
 & \Rightarrow -15x+4\left( 1-2x \right)=-42 \\
 & \Rightarrow -15x+4-8x=-42 \\
 & \Rightarrow -23x=-46 \\
 & \Rightarrow x=\dfrac{-46}{-23} \\
\end{align}\]
\[\Rightarrow 4-3y=1\]
Now we add -4 on both sides then we get
\[\Rightarrow 4-4-3y=1-4\]
\[\Rightarrow -3y=-3\]
Now we divide with -3 on both sides then we get
\[\Rightarrow \dfrac{-3y}{-3}=\dfrac{-3}{-3}\]
\[\Rightarrow y=1\]
Therefore, \[x=2\] and \[y=1\] is the required solution.

Note:
In order to solve these types of problems, we need to have knowledge of straight lines and basic arithmetic properties. We can solve for x and y of two equations by adding them after making required modifications such that any one of the variables gets canceled. Then, we get the value of one variable. Using this we get the other variable value.