
Solve the following simultaneous equation.
\[x + 7y = 10\] and \[3x - 2y = 7\]
Answer
500.1k+ views
Hint:To solve the following equation together first we will make one of the variable same , then add or subtract the equation which we have got after making the given equation similar , then cancel out one of the variable , which will provide value of other variable and then we will put this value in given equation to find out the value of other variable.
Complete step by step answer:
Given : \[x + 7y = 10........eqn\left( 1 \right)\]
\[3x - 2y = 7.........eqn\left( 2 \right)\]
Now multiplying eqn(1) by \[3\] , we get
\[3x + 21y = 30........eqn\left( 3 \right)\]
Now subtracting eqn (2) from eqn (3) , we get
\[3x + 21y - 3x + 2y = 30 - 7\]
\[\Rightarrow 23y = 23\]
On simplifying we get ,
\[y = 1\] .
Now we will put the value of \[y\] in the eqn (2) to get the value of \[x\] , we get
\[3x - 2 \times \left( 1 \right) = 7\]
On simplifying , we get
\[3x = 7 + 2\]
On solving , we get
\[x = 3\]
Therefore , the values \[x\] and \[y\] are \[3\] and \[1\] respectively.
Note:Alternative Solution:This method is also called a substitution method , since in this method we find out the value of one variable in terms of another variable and then put the value of that variable in the second equation to find the required values of \[x\] and \[y\]. Given:
\[x + 7y = 10........eqn\left( 1 \right)\]
\[\Rightarrow 3x - 2y = 7.........eqn\left( 2 \right)\]
Now from eqn (1) we have,
\[x = 10 - 7y......eqn(4)\]
Now using eqn (4) we will put the value of \[x\] in eqn (2) , we get
\[3\left( {10 - 7y} \right) - 2y = 7\]
On simplifying we get,
\[30 - 21y - 2y = 7\]
On solving we get,
\[23y = 23\]
On solving we get,
\[y = 1\] .
Moreover, if we plot the given equations on a graph, the point of intersection of both the equations will provide the solution of these equations.
Complete step by step answer:
Given : \[x + 7y = 10........eqn\left( 1 \right)\]
\[3x - 2y = 7.........eqn\left( 2 \right)\]
Now multiplying eqn(1) by \[3\] , we get
\[3x + 21y = 30........eqn\left( 3 \right)\]
Now subtracting eqn (2) from eqn (3) , we get
\[3x + 21y - 3x + 2y = 30 - 7\]
\[\Rightarrow 23y = 23\]
On simplifying we get ,
\[y = 1\] .
Now we will put the value of \[y\] in the eqn (2) to get the value of \[x\] , we get
\[3x - 2 \times \left( 1 \right) = 7\]
On simplifying , we get
\[3x = 7 + 2\]
On solving , we get
\[x = 3\]
Therefore , the values \[x\] and \[y\] are \[3\] and \[1\] respectively.
Note:Alternative Solution:This method is also called a substitution method , since in this method we find out the value of one variable in terms of another variable and then put the value of that variable in the second equation to find the required values of \[x\] and \[y\]. Given:
\[x + 7y = 10........eqn\left( 1 \right)\]
\[\Rightarrow 3x - 2y = 7.........eqn\left( 2 \right)\]
Now from eqn (1) we have,
\[x = 10 - 7y......eqn(4)\]
Now using eqn (4) we will put the value of \[x\] in eqn (2) , we get
\[3\left( {10 - 7y} \right) - 2y = 7\]
On simplifying we get,
\[30 - 21y - 2y = 7\]
On solving we get,
\[23y = 23\]
On solving we get,
\[y = 1\] .
Moreover, if we plot the given equations on a graph, the point of intersection of both the equations will provide the solution of these equations.
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