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Solve $3x-8y=-23,-6x+10y=4$ using the elimination method.
Answer
499.2k+ views
Hint: We will solve the above question using the elimination method. In the elimination method, we eliminate one of the variables from the given equations by taking a certain linear combination of both the equations. Multiply the first equation by 2 and add the resulting equation to the second equation. The resulting equation will be in variable y only. Solve the equation for y. Substitute this value of y in the first equation and hence find the value of x. Hence find the solution of the given system of equations. Verify your answer.
Complete step-by-step answer:
Elimination Method:
In the elimination method, we form an equation in one variable from the equations in two variable by taking a certain linear combination of the two equations, i.e. if the equations are ${{L}_{1}}\left( x,y \right)=0$ and ${{L}_{2}}\left( x,y \right)=0$, we form a third equation $a{{L}_{1}}\left( x,y \right)+b{{L}_{2}}\left( x,y \right)=0$. The scalars a and b are so chosen that the resulting expression is an equation in one variable only.
Consider the given system of equations
$\begin{align}
& 3x-8y=-23\text{ }\left( i \right) \\
& -6x+10y=4\text{ }\left( ii \right) \\
\end{align}$
Multiplying equation (i) by 2 and adding the resulting equation to equation (ii), we get
$\begin{align}
& 2\left( 3x-8y \right)-6x+10y=2\times \left( -23 \right)+4 \\
& \Rightarrow 6x-16y-6x+10y=-46+4=-42 \\
& \Rightarrow -6y=-42 \\
\end{align}$
Dividing both sides of the equation by -6, we get
$y=\dfrac{-42}{-6}=7$
Substituting the value of y in equation (ii), we get
$\begin{align}
& -6x+10\left( 7 \right)=4 \\
& \Rightarrow -6x+70=4 \\
\end{align}$
Subtracting 70 from both sides of the equation, we get
$-6x=-66$
Dividing both sides of the equation by -6, we get
$x=\dfrac{-66}{-6}=11$
Hence the solution of the given system is x=11 and y = 7
Note: Verification:
We can verify the correctness of our solution by checking that the value of x and y satisfy both the equations
We have $3x-8y=3\left( 11 \right)-8\left( 7 \right)=33-56=-23$ and $-6x+10y=-6\left( 11 \right)+10\left( 7 \right)=-66+70=4$
Hence, we have
$3x-8y=-23$ and $-6x+10y=4$
Hence x = 11 and y = 7 is the solution of the given system of equations.
Hence our answer is verified to be correct.
Complete step-by-step answer:
Elimination Method:
In the elimination method, we form an equation in one variable from the equations in two variable by taking a certain linear combination of the two equations, i.e. if the equations are ${{L}_{1}}\left( x,y \right)=0$ and ${{L}_{2}}\left( x,y \right)=0$, we form a third equation $a{{L}_{1}}\left( x,y \right)+b{{L}_{2}}\left( x,y \right)=0$. The scalars a and b are so chosen that the resulting expression is an equation in one variable only.
Consider the given system of equations
$\begin{align}
& 3x-8y=-23\text{ }\left( i \right) \\
& -6x+10y=4\text{ }\left( ii \right) \\
\end{align}$
Multiplying equation (i) by 2 and adding the resulting equation to equation (ii), we get
$\begin{align}
& 2\left( 3x-8y \right)-6x+10y=2\times \left( -23 \right)+4 \\
& \Rightarrow 6x-16y-6x+10y=-46+4=-42 \\
& \Rightarrow -6y=-42 \\
\end{align}$
Dividing both sides of the equation by -6, we get
$y=\dfrac{-42}{-6}=7$
Substituting the value of y in equation (ii), we get
$\begin{align}
& -6x+10\left( 7 \right)=4 \\
& \Rightarrow -6x+70=4 \\
\end{align}$
Subtracting 70 from both sides of the equation, we get
$-6x=-66$
Dividing both sides of the equation by -6, we get
$x=\dfrac{-66}{-6}=11$
Hence the solution of the given system is x=11 and y = 7
Note: Verification:
We can verify the correctness of our solution by checking that the value of x and y satisfy both the equations
We have $3x-8y=3\left( 11 \right)-8\left( 7 \right)=33-56=-23$ and $-6x+10y=-6\left( 11 \right)+10\left( 7 \right)=-66+70=4$
Hence, we have
$3x-8y=-23$ and $-6x+10y=4$
Hence x = 11 and y = 7 is the solution of the given system of equations.
Hence our answer is verified to be correct.
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