Question

Solve the given pair of linear equations:
\[
  11(x - 5) + 10(y - 2) + 54 = 0 \\
  7(2x - 1) + 9(3y - 1) = 25 \]

Answer 1

Hint: We will use the method of elimination by equating the coefficients of either \[x\] or \[y\] to find the desired solution that satisfies both the equations. Eliminate either of the variables and substitute it in one of the equations to get the other variable.

Complete step-by-step solution:
We are given the equations \[11(x - 5) + 10(y - 2) + 54 = 0\] and \[7(2x - 1) + 9(3y - 1) = 25\] i.e.
\[11x - 55 + 10y - 20 + 54 = 0\]
\[ \Rightarrow 11x + 10y - 21 = 0\]
\[14x - 7 + 27y - 9 - 25 = 0\]
\[ \Rightarrow 14x + 27y - 41 = 0\]
Now let us take the LCM of \[11,14\] to proceed further \[LCM(11,14) = 154\],
\[14 \times (11x + 10y - 21 = 0)\]
\[ \Rightarrow 154x + 140y - 294 = 0\]---------- \[(1)\]
\[11 \times (14x + 27y - 41 = 0)\]
\[ \Rightarrow 154x + 297y - 451 = 0\]---------- \[(2)\]
Now let us subtract equation \[(1)\] from equation \[(2)\] to get,
\[ \Rightarrow 154x + 297y - 451 - (154x + 140y - 294) = 0\]
\[ \Rightarrow 157y - 157 = 0 \\
   \Rightarrow 157y = 157 \\
   \Rightarrow y = 1 \]
Now let us substitute the value of \[y = 1\] in the equation \[11x + 10y - 21 = 0\] to get,
\[
  11x + 10.1 - 21 = 0 \\
   \Rightarrow 11x - 11 = 0 \\
   \Rightarrow 11x = 11 \\
   \Rightarrow x = 1 \]
Therefore, we get the required solution for the above equations i.e. \[(1,1)\]
Additional information: A linear equation is an equation that is written for two exceptional variables. This equation may be a linear aggregate of those variables, and a steady state can be present. Surprisingly, while any linear equation is plotted on a graph, it will necessarily produce an instantaneous line - as a result they are called: Linear equations. Linear Equations are an extensive sort of equations altogether. There may be linear equations in one variable, linear equations in two variables, and so on. In every equation, one issue stays regular: The highest and the best diploma of all variables in the equation should be \[1\]. Other than that, constants \[0\] diploma variables can be there.

Note: It is very important that we know how to calculate the LCM of the numbers if needed and then take one variable from either of the variables to calculate the LCM of the coefficients and then multiply each of them to eliminate either of the variables and then calculate the value of the other variable using substitution method.