Answer
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Hint: A linear equation is any equation that can be written in the form $ax + by + c = 0$ where a and b are real numbers and x and y are variables. This form is sometimes called the standard form of a linear equation with two variables. If the equation contains any fractions use the least common denominator to clear the fraction. We will do this by multiplying both sides of the equation by LCD.
Firstly we add the both equations then we get another linear equation in two variables. Secondly, we subtract the both given equations, we get another linear equation in two variables. Further both equations are solved by simple methods.
Complete step-by-step answer:
We have given two equation
\[ \Rightarrow \]\[152x - 378y{\text{ }} = {\text{ }} - 74\;.{\text{ }}.....{\text{ }}\left( 1 \right)\]
\[ \Rightarrow \]\[ - 378x + 152y{\text{ }} = {\text{ }} - 604\;......{\text{ }}\left( 2 \right)\]
In given equation, the coefficient of variable x and y in
Equation is the coefficient of y and x in 2 equations respectively.
Adding Equation 1 and 2, we get,
\[x = {\text{ }}2\]
\[\begin{array}{*{20}{l}}
{152x{\text{ }} - \;378y{\text{ }} = {\text{ }}-{\text{ }}74} \\
{ - 378x\; + 152y{\text{ }} = {\text{ }}-{\text{ }}604} \\
{ - - - - - - - - - - - - - - - - - - } \\
{ - 226x{\text{ }} - 226y{\text{ }} = {\text{ }} - {\text{ }}678}
\end{array}\]
\[ \Rightarrow \]\[ - 226\left( {x + y{\text{ }}} \right){\text{ }} = {\text{ }} - 678\]
\[ \Rightarrow \]\[x + y{\text{ }} = {\text{ }} - 678/ - 226 = {\text{ }}3\]
\[ \Rightarrow \] \[x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3 \ldots \ldots \ldots \ldots \ldots \ldots .\left( 3 \right)\]
After solving, we get the linear equation marked as 3 equation,
\[Subtracting{\text{ }}equation{\text{ }}1{\text{ }}and{\text{ }}2\]
\[\begin{array}{*{20}{l}}
{152x{\text{ }} - \;378y{\text{ }} = {\text{ }}-{\text{ }}74} \\
{ - 378x\; + 152y{\text{ }} = {\text{ }}-{\text{ }}604} \\
{\left( + \right)\;\;\;\;\;\left( - \right)\;\;\;\;\;\;\;\;\;\left( + \right)} \\
{ - - - - - - - - - - - - - - - } \\
{530x{\text{ }} - {\text{ }}530y{\text{ }} = {\text{ }}530}
\end{array}\]
\[ \Rightarrow \] \[530{\text{ }}\left( {x - {\text{ }}y{\text{ }}} \right){\text{ }} = {\text{ }}530\]
\[ \Rightarrow \] \[x - y{\text{ }} = {\text{ }}\dfrac{{530}}{{530}} = {\text{ }}1\]
\[ \Rightarrow \] \[x{\text{ }} - \;y{\text{ }} = {\text{ }}1 \ldots \ldots \ldots \ldots \ldots \ldots .\left( 4 \right)\]
Here, we use the elimination method.
On adding equation 3 and 4, we get,
\[\begin{array}{*{20}{l}}
{x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3} \\
{x{\text{ }} - \;y{\text{ }} = {\text{ }}1\;\;\;\;\;\;\;\;\;\;} \\
{ - - - - - - } \\
{2x{\text{ }} = {\text{ }}4}
\end{array}\]
\[ \Rightarrow \]\[x{\text{ }} = {\text{ }}\dfrac{4}{2}\]=\[x{\text{ }} = {\text{ }}2\]
On substituting \[x{\text{ }} = {\text{ }}2\] in equation 3
\[ \Rightarrow \]\[x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3\]
\[ \Rightarrow \]\[2{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3\]
\[ \Rightarrow \] ${y{\text{ }} = {\text{ }}3 - 2} $
\[ \Rightarrow \] ${y{\text{ }} = {\text{ }}1} $
Hence, \[x = {\text{ }}2\] and \[y = {\text{ }}1\] is the required solution.
Note: To remember the process of framing simultaneous linear equations from mathematical problems. To remember how to solve simultaneous equations by the method of comparison and method of cross multiplications. There is yet another visual way of representing quantitative data and its frequencies. This is a polygon.
Firstly we add the both equations then we get another linear equation in two variables. Secondly, we subtract the both given equations, we get another linear equation in two variables. Further both equations are solved by simple methods.
Complete step-by-step answer:
We have given two equation
\[ \Rightarrow \]\[152x - 378y{\text{ }} = {\text{ }} - 74\;.{\text{ }}.....{\text{ }}\left( 1 \right)\]
\[ \Rightarrow \]\[ - 378x + 152y{\text{ }} = {\text{ }} - 604\;......{\text{ }}\left( 2 \right)\]
In given equation, the coefficient of variable x and y in
Equation is the coefficient of y and x in 2 equations respectively.
Adding Equation 1 and 2, we get,
\[x = {\text{ }}2\]
\[\begin{array}{*{20}{l}}
{152x{\text{ }} - \;378y{\text{ }} = {\text{ }}-{\text{ }}74} \\
{ - 378x\; + 152y{\text{ }} = {\text{ }}-{\text{ }}604} \\
{ - - - - - - - - - - - - - - - - - - } \\
{ - 226x{\text{ }} - 226y{\text{ }} = {\text{ }} - {\text{ }}678}
\end{array}\]
\[ \Rightarrow \]\[ - 226\left( {x + y{\text{ }}} \right){\text{ }} = {\text{ }} - 678\]
\[ \Rightarrow \]\[x + y{\text{ }} = {\text{ }} - 678/ - 226 = {\text{ }}3\]
\[ \Rightarrow \] \[x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3 \ldots \ldots \ldots \ldots \ldots \ldots .\left( 3 \right)\]
After solving, we get the linear equation marked as 3 equation,
\[Subtracting{\text{ }}equation{\text{ }}1{\text{ }}and{\text{ }}2\]
\[\begin{array}{*{20}{l}}
{152x{\text{ }} - \;378y{\text{ }} = {\text{ }}-{\text{ }}74} \\
{ - 378x\; + 152y{\text{ }} = {\text{ }}-{\text{ }}604} \\
{\left( + \right)\;\;\;\;\;\left( - \right)\;\;\;\;\;\;\;\;\;\left( + \right)} \\
{ - - - - - - - - - - - - - - - } \\
{530x{\text{ }} - {\text{ }}530y{\text{ }} = {\text{ }}530}
\end{array}\]
\[ \Rightarrow \] \[530{\text{ }}\left( {x - {\text{ }}y{\text{ }}} \right){\text{ }} = {\text{ }}530\]
\[ \Rightarrow \] \[x - y{\text{ }} = {\text{ }}\dfrac{{530}}{{530}} = {\text{ }}1\]
\[ \Rightarrow \] \[x{\text{ }} - \;y{\text{ }} = {\text{ }}1 \ldots \ldots \ldots \ldots \ldots \ldots .\left( 4 \right)\]
Here, we use the elimination method.
On adding equation 3 and 4, we get,
\[\begin{array}{*{20}{l}}
{x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3} \\
{x{\text{ }} - \;y{\text{ }} = {\text{ }}1\;\;\;\;\;\;\;\;\;\;} \\
{ - - - - - - } \\
{2x{\text{ }} = {\text{ }}4}
\end{array}\]
\[ \Rightarrow \]\[x{\text{ }} = {\text{ }}\dfrac{4}{2}\]=\[x{\text{ }} = {\text{ }}2\]
On substituting \[x{\text{ }} = {\text{ }}2\] in equation 3
\[ \Rightarrow \]\[x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3\]
\[ \Rightarrow \]\[2{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}3\]
\[ \Rightarrow \] ${y{\text{ }} = {\text{ }}3 - 2} $
\[ \Rightarrow \] ${y{\text{ }} = {\text{ }}1} $
Hence, \[x = {\text{ }}2\] and \[y = {\text{ }}1\] is the required solution.
Note: To remember the process of framing simultaneous linear equations from mathematical problems. To remember how to solve simultaneous equations by the method of comparison and method of cross multiplications. There is yet another visual way of representing quantitative data and its frequencies. This is a polygon.
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