Question

solve the system of linear equations by elimination.\\(2x + 7y = 1\\)\\(2x - 4y = 12\\)\
solution: (\\(\square\\), \\(\square\\))

Explanation:

Step1: Subtract the two equations to eliminate \(x\)

We have the system:
\[

$$\begin{cases} 2x + 7y = 1 \\ 2x - 4y = 12 \end{cases}$$

\]
Subtract the second equation from the first equation:
\((2x + 7y)-(2x - 4y)=1 - 12\)
Simplify the left - hand side: \(2x+7y - 2x + 4y=11y\)
Simplify the right - hand side: \(1-12=-11\)
So we get the equation \(11y=-11\)

Step2: Solve for \(y\)

From \(11y = - 11\), divide both sides by 11:
\(y=\frac{-11}{11}=-1\)

Step3: Substitute \(y = - 1\) into one of the original equations to solve for \(x\)

Substitute \(y=-1\) into the first equation \(2x + 7y = 1\):
\(2x+7\times(-1)=1\)
Simplify: \(2x-7 = 1\)
Add 7 to both sides: \(2x=1 + 7=8\)
Divide both sides by 2: \(x=\frac{8}{2}=4\)

Answer:

\((4,-1)\)