Question

question
solve the system of equations $x + 2y = -14$ and $-2x - 6y = 40$ by combining the equations.
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$$\begin{array}{r}(x\\quad\\quad+2y\\quad=-14)\\\\(-2x\\quad-6y\\quad=\\ 40)\\end{array}$$

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$$\begin{array}{r}2x\\quad+4y\\quad=-28\\\\-2x\\quad-6y\\quad=\\ 40\\\\\\hline -2\\ x+\\ -6\\ y=\\ \\square\\end{array}$$

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Explanation:

Step1: Add the two equations

We have the equations \(2x + 4y = -28\) and \(-2x - 6y = 40\). When we add them together, we add the left - hand sides and the right - hand sides separately.
For the \(x\) - terms: \(2x+(-2x)=0\)
For the \(y\) - terms: \(4y+(-6y)=-2y\)
For the constant terms: \(-28 + 40=12\)
But in the given format, we are adding the equations as they are presented in the table. The first equation after multiplication is \(2x + 4y=-28\) and the second is \(-2x-6y = 40\).
Adding the left - hand sides: \((2x + 4y)+(-2x-6y)=2x-2x + 4y-6y=-2y\) (Wait, but the table has a format with coefficients for \(x\) and \(y\) as \(-2\) and \(-6\) which seems to be a miscalculation in the table's initial step, but we follow the addition of the two equations \(2x + 4y=-28\) and \(-2x-6y = 40\))
Adding the right - hand sides: \(-28+40 = 12\)
But if we consider the way the table is set up (maybe a miscalculation in the first multiplication, but let's follow the addition of the two equations as per the problem's table - like setup where the first equation after multiplying by 2 is \(2x + 4y=-28\) and the second is \(-2x-6y = 40\))
When we add \(2x+4y=-28\) and \(-2x - 6y=40\), we get \((2x-2x)+(4y-6y)=-28 + 40\)
Simplifying, we have \(0x-2y = 12\), but in the table, the coefficients for \(x\) and \(y\) are \(-2\) and \(-6\) which is incorrect. However, if we just add the right - hand sides of the two equations \(2x + 4y=-28\) and \(-2x-6y = 40\), the right - hand side sum is \(-28 + 40=12\)

Step2: Calculate the sum of the right - hand sides

We know that when we add two equations \(A = B\) and \(C = D\), we get \(A + C=B + D\).
For the equations \(2x + 4y=-28\) (Equation 1) and \(-2x-6y = 40\) (Equation 2), adding Equation 1 and Equation 2:
\((2x + 4y)+(-2x-6y)=-28 + 40\)
The left - hand side simplifies to \(2x-2x+4y - 6y=-2y\), and the right - hand side is \(12\). But in the context of the table, we are asked to find the value of the right - hand side when we add the two equations (the boxed value).
So, \(-28+40 = 12\)

Answer:

\(12\)