Question

Solve the following system of linear equations.
$\begin{align}
  & x-2y=0 \\
 & 3x+4y=20 \\
\end{align}$

Answer 1

Hint: To solve a system of linear equations, we have to multiply suitable values to both the equations so that when we add or subtract the two equations we can eliminate one variable. We are given two equations where the coefficients of x and y do not match. Therefore. we will multiply constants throughout both equations so that we can find x and y. After we eliminate one variable, we can find the other and use it in any of the two equations to find the formerly eliminated variable.

Complete step-by-step answer:
A linear equation is one which contains only the first power of a variable and constants. A system of linear variables has as many equations as the number of unknowns. Here, there are two variables x and y and two linear equations. The given equations are,
$\begin{align}
  & x-2y=0 \\
 & 3x+4y=20 \\
\end{align}$
The coefficients of x and y are not equal in the two equations. Hence we multiply a constant to make one of them equal. We can clearly see that the coefficient of y in the first equation is -2, and the coefficient of y in the second equation is 4. Therefore, we multiply the first equation by 2. We get,
$\begin{align}
  & 2\left( x-2y \right)=2\times 0 \\
 & 2x-4y=0......\left( i \right) \\
 & 3x+4y=20......\left( ii \right) \\
\end{align}$
Now, adding equations (i) and (ii), we get,
$\begin{align}
  & 2x-4y+3x+4y=0+20 \\
 & \Rightarrow 2x+3x=20 \\
 & \Rightarrow 5x=20 \\
 & x=4 \\
\end{align}$
Hence, the value of x is 4. Now, substitute this value of x in any one of equations (i) and (ii), we get,
$\begin{align}
  & 2\left( 4 \right)-4y=0 \\
 & \Rightarrow 4y=8 \\
 & \Rightarrow y=2 \\
\end{align}$
Hence, the value of y is 2. Now, to verify these values we can substitute them in equation (ii), we get,
$\begin{align}
  & 2\left( 4 \right)+4\left( 2 \right)=20 \\
 & \Rightarrow 20=20 \\
\end{align}$
Hence, the values of x and y are correct. The solution to these system of linear equations is,
$\begin{align}
  & x=4 \\
 & y=2 \\
\end{align}$

Note: Another way to solve this question is by using matrices. Enter the coefficients in the matrix in the order x1, y1, x2, y2 and enter the RHS with the constants.
$\left( \begin{matrix}
   1 & -2 \\
   3 & 4 \\
\end{matrix} \right)\left( \begin{matrix}
  & x \\
 & y \\
\end{matrix} \right)=\left( \begin{matrix}
  & 0 \\
 & 20 \\
\end{matrix} \right)$
Now, perform the operation ${{R}_{2}}\leftrightarrow {{R}_{2}}-3{{R}_{1}}$ on both RHS and LHS, we get,
$\begin{align}
  & \left( \begin{matrix}
   1 & -2 \\
   3-\left( 3\times 1 \right) & 4-\left( 3\times -2 \right) \\
\end{matrix} \right)\left( \begin{matrix}
  & x \\
 & y \\
\end{matrix} \right)=\left( \begin{matrix}
  & 0 \\
 & 20-\left( 3\times 0 \right) \\
\end{matrix} \right) \\
 & \left( \begin{matrix}
   1 & -2 \\
   0 & 10 \\
\end{matrix} \right)\left( \begin{matrix}
  & x \\
 & y \\
\end{matrix} \right)=\left( \begin{matrix}
  & 0 \\
 & 20 \\
\end{matrix} \right) \\
 & \Rightarrow 10y=20 \\
 & \Rightarrow y=2 \\
\end{align}$
Using this value, we can find the value of x.
$\begin{align}
  & x-2y=0 \\
 & \Rightarrow x-\left( 2\times 2 \right)=0 \\
 & \Rightarrow x=4 \\
\end{align}$
Hence the values of x and y are 4 and 2 respectively.