
Solve the pair of linear equations: 152x - 378y = – 74, -378x +152y = – 604
Answer
551.4k+ views
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.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

