Question

How do you solve the following system: $ x-3y=20 $ , $ 7x+15y=32 $ ?

Answer 1

Hint: In the problem we have given two equations in two variables. For this type of equation, we will consider the coefficients of each variable. Now we will calculate the LCM of either coefficient of variable $ x $ or coefficients of $ y $. Now we will multiple both the equations with the calculated LCM of coefficients. Now we can observe that the coefficients of any one variable in both the equations are the same, so we can add or subtract based on the coefficients to obtain an equation in a single variable. From this equation, we can calculate the value of one variable and from the given equations we can calculate the value of another variable.

Complete step by step answer:
Given equations, $ x-3y=20 $ , $ 7x+15y=32 $ .
The coefficients of the variables in the above equations are
The coefficient of $ x $ in the first equation is $ 1 $ .
The coefficient of $ x $ in the second equation is $ 7 $ .
The coefficient of $ y $ in the first equation is $ -3 $ .
The coefficient of $ y $ in the second equation is $ 15 $ .
Considering the coefficients of $ x $ in both the equations. We know that the LCM of $ 1 $ and $ 7 $ is $ 7 $ .
Multiplying the first equation with $ 7 $ on both sides of the equation, then we will get
 $ 7\left( x-3y \right)=7\times 20 $
Applying distribution law of multiplication in the above equation, then we will get
 $ \Rightarrow 7x-21y=140 $
Subtracting second equation from the above equation, then we will get
 $ \begin{align}
  & 7x-21y-\left( 7x+15y \right)=140-32 \\
 & \Rightarrow 7x-21y-7x-15y=108 \\
 & \Rightarrow -36y=108 \\
 & \Rightarrow y=\dfrac{-108}{36} \\
 & \Rightarrow y=-3 \\
\end{align} $
Substituting the value of $ y $ in the first equation, then we will get
 $ \begin{align}
  & x-3\left( -3 \right)=20 \\
 & \Rightarrow x+9=20 \\
 & \Rightarrow x=20-9 \\
 & \Rightarrow x=11 \\
\end{align} $
Hence, we have the solution for the given equations as $ x=11 $ , $ y=-3 $ .

Note:
 We check whether the obtained result is correct or not by substituting the solution in any one of the equations. Substituting $ x=11 $ , $ y=-3 $ in second equation, then we will get
 $ \begin{align}
  & 7x+15y=32 \\
 & \Rightarrow 7\left( 11 \right)+15\left( -3 \right)=32 \\
 & \Rightarrow 77-45=32 \\
 & \Rightarrow 32=32 \\
\end{align} $
In the above equation we got LHS $ = $ RHS. Hence the obtained result is correct.