Question

Solve the following systems of linear equations by using the method of elimination by equating the coefficients:
\[8x + 5y = 9{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \& {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x + 2y = 4\]

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 \[8x + 5y = 9{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \& {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x + 2y = 4\],
Eliminating the variable \[x\] by taking LCM of the two equations we get,
\[LCM(8,3) = 24\]
Therefore our new equations are,
\[3 \times (8x + 5y = 9{\kern 1pt} ) \Rightarrow 24x + 15y = 27\]--------------\[(1)\]
\[{\kern 1pt} 8 \times (3x + 2y = 4) \Rightarrow 24x + 16y = 32\]--------------\[(2)\]
Now, \[(1)\]from\[(2)\]we get,
\[24x + 16y - 32 - 24x - 15y + 27 = 0\]
\[ \Rightarrow y - 5 = 0\]
\[ \Rightarrow y = 5\]
Now substituting \[y = 5\]in \[3x + 2y = 4\]we get,
\[ 3x + 2.5 = 4 \\
   \Rightarrow 3x = 4 - 10 \\
   \Rightarrow 3x = - 6 \\
   \Rightarrow x = - 2 \]
Therefore our solution for the above equations is $x= - 2, y=5$.

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, constant \[0\] diploma variables can be there.

Note: It is very important that we know how to calculate the LCM of the numbers 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.