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Solve the following equations using elimination method:
$3x+4y=25$ and $5x-6y=-9$.

Answer
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Hint: In this question, we will first eliminate one of the variables, by making terms of that variable equal using LCM. And then solve the equations to get required values.

Complete step-by-step answer:
In elimination method, we are to first eliminate one of the variables and then solve the equation for the remaining variable and use the value of that variable to find the value of the first variable.
Now, the given equations are,
 $3x+4y=25\cdots \cdots \left( i \right)$
and, $5x-6y=-9\cdots \cdots \left( ii \right)$.
Let us first eliminate $y$. For this, we have to make terms containing $y$ in both the equations equal.
To do this, let us take LCM of coefficients of $y$ from both the equations, which is 4 and 6. So, required LCM will be 12.
Now, to make coefficients of $y$ in both the equation 12, let us first multiply equation $\left( i \right)$ with 3, so we get,
$3\times \left( 3x+4y \right)=3\times 25$
Applying distributive law, we get,
$9x+12y=75\cdots \cdots \left( iii \right)$.
And, now we multiplying equation $\left( ii \right)$ with 2, so we get,
$2\times \left( 5x-6y \right)=2\times \left( -9 \right)$
Applying distributive law, we get,
$\Rightarrow 10x-12y=-18\cdots \cdots \left( iv \right)$
Now, to eliminate $y$, let us add equation $\left( iii \right)$ and $\left( iv \right)$, so we get,
$\begin{align}
  & \left( 9x+12y \right)+\left( 10x-12y \right)=75+\left( -18 \right) \\
 & \Rightarrow 9x+12y+10x-12y=75-18 \\
 & \Rightarrow 9x+10x=75-18 \\
 & \Rightarrow 19x=57 \\
\end{align}$
Dividing 19 from both sides of the equation, we get,
$x=\dfrac{57}{19}=3$.
Now, putting this value of $x$ in equation $\left( i \right)$, we get,
  $\begin{align}
  & 3x+4y=25 \\
 & \Rightarrow 3\times 3+4y=25 \\
 & \Rightarrow 9+4y=25 \\
\end{align}$
Subtracting 9 from both sides of the above equation, we get,
$\begin{align}
  & 4y=25-9 \\
 & \Rightarrow 4y=16 \\
\end{align}$
Dividing 4 from both sides of the equation, we get,
$y=\dfrac{16}{4}=4$.
Hence, in given equations, the value of $x$ is 3 and value of $y$ is 4.

Note: In this type of question, you can eliminate any one of the variables, but choose that variable for which equations will be multiplied by a smaller number so as to keep calculation easier.