
Solve $3x+4y=25,5x-6y=-9$ using the elimination method.
Answer
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Hint: 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 3 and the second equation by 2 and add the two equations. The resulting equation will be in variable x only. Solve the equation for x. Substitute this value of x in the first equation and hence find the value of y. 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
$\begin{align}
& 3x+4y=25\text{ }\left( i \right) \\
& 5x-6y=-9\text{ }\left( ii \right) \\
\end{align}$
Multiplying equation (i) by 3 and equation (ii) by 2 and adding the resulting equation, we get
$\begin{align}
& 3\left( 3x+4y \right)+2\left( 5x-6y \right)=3\times 25+2\times \left( -9 \right) \\
& \Rightarrow 9x+12y+10x-12y=75-18=57 \\
& \Rightarrow 19x=57 \\
\end{align}$
Dividing both sides of the equation by 19, we get
$x=\dfrac{57}{19}=3$
Substituting the value of x in equation (i), we get
$3\left( 3 \right)+4y=25\Rightarrow 9+4y=25$
Subtracting 9 from both sides of the equation, we get
$4y=25-9=16$
Dividing both sides of the equation by 4, we get
$y=\dfrac{16}{4}=4$
Hence the solution of the given system is x=3 and y = 4
Note: Verification:
We have $3x+4y=3\left( 3 \right)+4\left( 4 \right)=9+16=25$ and $5x-6y=5\left( 3 \right)-6\left( 4 \right)=15-24=-9$
Hence, we have
$3x+4y=9$ and \[5x-6y=-9\]
Hence x = 3 and y = 4 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
$\begin{align}
& 3x+4y=25\text{ }\left( i \right) \\
& 5x-6y=-9\text{ }\left( ii \right) \\
\end{align}$
Multiplying equation (i) by 3 and equation (ii) by 2 and adding the resulting equation, we get
$\begin{align}
& 3\left( 3x+4y \right)+2\left( 5x-6y \right)=3\times 25+2\times \left( -9 \right) \\
& \Rightarrow 9x+12y+10x-12y=75-18=57 \\
& \Rightarrow 19x=57 \\
\end{align}$
Dividing both sides of the equation by 19, we get
$x=\dfrac{57}{19}=3$
Substituting the value of x in equation (i), we get
$3\left( 3 \right)+4y=25\Rightarrow 9+4y=25$
Subtracting 9 from both sides of the equation, we get
$4y=25-9=16$
Dividing both sides of the equation by 4, we get
$y=\dfrac{16}{4}=4$
Hence the solution of the given system is x=3 and y = 4
Note: Verification:
We have $3x+4y=3\left( 3 \right)+4\left( 4 \right)=9+16=25$ and $5x-6y=5\left( 3 \right)-6\left( 4 \right)=15-24=-9$
Hence, we have
$3x+4y=9$ and \[5x-6y=-9\]
Hence x = 3 and y = 4 is the solution of the given system of equations.
Hence our answer is verified to be correct.
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