Question

2 (x +2y = -14)
1 (-2x -6y = 40)
2x +4y = -28
-2x -6y = 40
\overline{\square x + \square y = \square}

Explanation:

Step1: Add the two equations

We have the equations \(2x + 4y=-28\) and \(-2x - 6y = 40\). When we add the left - hand sides and the right - hand sides of the two equations respectively:
\((2x+4y)+(-2x - 6y)=2x + 4y-2x-6y\) and \(-28 + 40\)
Simplify the left - hand side: \(2x-2x+4y - 6y=-2y\)
Simplify the right - hand side: \(12\)
So we get \(-2y=12\)? Wait, no, wait. Wait, actually, when we add \(2x + 4y\) and \(-2x-6y\), the \(x\) terms: \(2x-2x = 0\), the \(y\) terms: \(4y-6y=-2y\), and the constant terms: \(-28 + 40 = 12\). But also, if we consider the coefficients of \(x\) and \(y\) in the sum:
The coefficient of \(x\): \(2x-2x=0x\)
The coefficient of \(y\): \(4y-6y=-2y\)
The constant term: \(-28 + 40 = 12\)
But the problem is asking for the form \(\square x+\square y=\square\). Wait, maybe I made a mistake. Wait, let's re - add the two equations:

\(2x+4y=-28\)

\(-2x - 6y=40\)

Adding the left - hand sides: \(2x+4y-2x - 6y=(2x-2x)+(4y-6y)=-2y\)

Adding the right - hand sides: \(-28 + 40 = 12\)

But also, if we consider the sum of the two equations as a linear combination, the coefficient of \(x\) is \(2-2 = 0\), the coefficient of \(y\) is \(4-6=-2\), and the constant term is \(-28 + 40 = 12\). So the equation after adding is \(0x-2y = 12\)

Wait, but maybe the problem is to add the two equations \(2x + 4y=-28\) and \(-2x-6y = 40\) term by term:

For the \(x\) terms: \(2x+(-2x)=0x\)

For the \(y\) terms: \(4y+(-6y)=-2y\)

For the constant terms: \(-28 + 40 = 12\)

So the equation is \(0x-2y = 12\)

Answer:

\(0\) \(x\) \(+\) \((-2)\) \(y\) \(=\) \(12\) (or we can also write it as \(0x-2y = 12\))