
How do you solve a system of equations by using the elimination method?
Answer
450.9k+ views
Hint:In the given question, we have been asked to solve a system of equations by using elimination method. In order to solve the system of equations by elimination method, we need to either add both the equation or subtract both the equation to get the equation in one variable. Then we solve the equation in one variable in a way we solve the general linear equation.
Complete step by step solution:
Steps to solve the system of equations by elimination method:
Write both the equations.
Multiply any of the equations by a number to get the same leading coefficient.
Either add or subtract the two equations to eliminate the variable.
We will get a linear equation in one variable, solve for that variable.
Substituting the value of that variable in any of the equations.
Solve for the other given variable.
Let assume we have given a system of equation
\[\Rightarrow 4x+y=2\]------ (1)
\[\Rightarrow 2x-y=0\]------ (2)
As we can see that in both the equations y have the same coefficient that is equals to 1, so we add both the equations.
Adding equation (1) and equation (2), we get
\[\Rightarrow 4x+y+2x-y=2+0\]
Combining the like terms, we get
\[\Rightarrow 6x=2\]
Dividing both the sides of the equation by 2, we get
\[\Rightarrow x=\dfrac{2}{6}=\dfrac{1}{3}\]
\[\therefore x=\dfrac{1}{3}\]
Substitute the value of \[x=\dfrac{1}{3}\] in equation (1), we get
\[\Rightarrow 4\times \dfrac{1}{3}+y=2\]
\[\Rightarrow \dfrac{4}{3}+y=2\]
Subtracting \[\dfrac{4}{3}\] from both the sides of the equation, we get
\[\Rightarrow y=2-\dfrac{4}{3}\]
Solving the right side of the equation by taking the LCM, we get
\[\Rightarrow y=\dfrac{6}{3}-\dfrac{4}{3}=\dfrac{2}{3}\]
\[\therefore y=\dfrac{2}{3}\]
Therefore, we get
\[\Rightarrow \left( x,y \right)=\left( \dfrac{1}{3},\dfrac{2}{3} \right)\]
Therefore, the possible value of ‘x’ and ‘y’ is \[\left( x,y \right)=\left( \dfrac{1}{3},\dfrac{2}{3} \right)\].
It is the required solution.
Note: In an elimination method for solving a system of equations we will either add both the equations or subtract both the equations to get the equation in one variable. To eliminate the variable, you will add the two given equations if the coefficients of one variable are opposites and you will subtract the two equations if the coefficient of one variable is exactly the same. If both the situations do not satisfy then we will need to multiply one of the equations by a number that will lead to the same leading coefficient.
Complete step by step solution:
Steps to solve the system of equations by elimination method:
Write both the equations.
Multiply any of the equations by a number to get the same leading coefficient.
Either add or subtract the two equations to eliminate the variable.
We will get a linear equation in one variable, solve for that variable.
Substituting the value of that variable in any of the equations.
Solve for the other given variable.
Let assume we have given a system of equation
\[\Rightarrow 4x+y=2\]------ (1)
\[\Rightarrow 2x-y=0\]------ (2)
As we can see that in both the equations y have the same coefficient that is equals to 1, so we add both the equations.
Adding equation (1) and equation (2), we get
\[\Rightarrow 4x+y+2x-y=2+0\]
Combining the like terms, we get
\[\Rightarrow 6x=2\]
Dividing both the sides of the equation by 2, we get
\[\Rightarrow x=\dfrac{2}{6}=\dfrac{1}{3}\]
\[\therefore x=\dfrac{1}{3}\]
Substitute the value of \[x=\dfrac{1}{3}\] in equation (1), we get
\[\Rightarrow 4\times \dfrac{1}{3}+y=2\]
\[\Rightarrow \dfrac{4}{3}+y=2\]
Subtracting \[\dfrac{4}{3}\] from both the sides of the equation, we get
\[\Rightarrow y=2-\dfrac{4}{3}\]
Solving the right side of the equation by taking the LCM, we get
\[\Rightarrow y=\dfrac{6}{3}-\dfrac{4}{3}=\dfrac{2}{3}\]
\[\therefore y=\dfrac{2}{3}\]
Therefore, we get
\[\Rightarrow \left( x,y \right)=\left( \dfrac{1}{3},\dfrac{2}{3} \right)\]
Therefore, the possible value of ‘x’ and ‘y’ is \[\left( x,y \right)=\left( \dfrac{1}{3},\dfrac{2}{3} \right)\].
It is the required solution.
Note: In an elimination method for solving a system of equations we will either add both the equations or subtract both the equations to get the equation in one variable. To eliminate the variable, you will add the two given equations if the coefficients of one variable are opposites and you will subtract the two equations if the coefficient of one variable is exactly the same. If both the situations do not satisfy then we will need to multiply one of the equations by a number that will lead to the same leading coefficient.
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