How do you solve the system using the elimination method for $3a+5b-7=0$ and $a-2b-4=0$?

Hint: Mark the equations as (1) and (2). Multiply equation (1) with 1 and equation (2) with 3 and mark the new equations as (3) and (4) respectively. Subtract equation (4) from equation (3) to get the value of ‘b’, Put the value of ‘b’ in any of the equations to get the value of ‘a’.

Complete step by step solution:
Elimination method for solving systems: In elimination method we have to either add or subtract the equations if the coefficients of one variable are the same. If the coefficients of one variable are different, then we have to multiply one equation with a required constant to make the coefficients equal and then we have to add or subtract accordingly.
Now, considering our equations
$3a+5b-7=0$……….(1)
$a-2b-4=0$……….(2)
To make the coefficients of ‘a’ equal, we have to multiply the equation (1) by 1 and the equation (2) by 3
$eq(1)\times 1\Rightarrow 3a+5b-7=0$……….(3)
$eq(2)\times 3\Rightarrow 3a-6b-12=0$……….(4)
Subtracting equation (4) from equation (3), we get
$\begin{align}
  & eq(3)-eq(4) \\
 & \Rightarrow \left( 3a+5b-7 \right)-\left( 3a-6b-12 \right)=0-0 \\
 & \Rightarrow 3a+5b-7-3a+6b+12=0 \\
 & \Rightarrow 11b+5=0 \\
 & \Rightarrow 11b=-5 \\
 & \Rightarrow b=\dfrac{-5}{11} \\
\end{align}$
Putting the value of ‘b’ in equation (2), we get
$\begin{align}
  & a-2b-4=0 \\
 & \Rightarrow a-2\times \left( -\dfrac{5}{11} \right)-4=0 \\
 & \Rightarrow a-\left( -\dfrac{10}{11} \right)-4=0 \\
 & \Rightarrow a+\dfrac{10}{11}=4 \\
 & \Rightarrow a=4-\dfrac{10}{11} \\
 & \Rightarrow a=\dfrac{44-10}{11} \\
 & \Rightarrow a=\dfrac{34}{11} \\
\end{align}$

Hence, the solution of the system $3a+5b-7=0$ and $a-2b-4=0$ is $\left( a,b \right)=\left( -\dfrac{5}{11},\dfrac{34}{11} \right)$.

Note: Coefficients of one variable should be made equal by multiplying suitable constants. After getting one variable, put the value of that variable in any of the equations, to obtain the value of the other variable.