Question

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Explanation:

Step1: Multiply first equation by 2

To eliminate \(x\), multiply the first equation \(x + 2y=-14\) by 2. So we get \(2(x + 2y)=2\times(-14)\), which simplifies to \(2x + 4y=-28\).

Step2: Add new equation to second equation

Now add the new equation \(2x + 4y=-28\) to the second equation \(-2x-6y = 40\).
For the \(x\)-terms: \(2x+(-2x)=0\)
For the \(y\)-terms: \(4y+(-6y)=-2y\)
For the constants: \(-28 + 40 = 12\)
So combining them gives \(0x-2y = 12\), or simply \(-2y=12\) (but following the form in the problem, we can also note that if we just add the original equations without multiplying first, let's check that way too. Wait, the problem has a boxed form where we can put a coefficient in front of the first equation. Let's do it by choosing a coefficient for the first equation to eliminate \(x\). Let's let the coefficient of the first equation be 2. So:
Multiply first equation by 2: \(2(x + 2y)=2\times(-14)\) => \(2x+4y=-28\)
Then add to second equation \(-2x - 6y=40\):
\((2x+4y)+(-2x - 6y)=-28 + 40\)
Which is \(0x-2y = 12\). But in the problem's format, if we use the original equations and put a coefficient in front of the first equation (let's say 2) and 1 in front of the second, then:
\(2(x + 2y=-14)\) and \(1(-2x - 6y=40)\)
Adding them: \(2x+4y-2x - 6y=-28 + 40\)
Simplifies to \(0x-2y = 12\), so in the form \(0x + 0y=\)? Wait, no, maybe I misread. Wait the problem's lower part has \(x+2y=-14\) and \(-2x - 6y=40\) and we need to put coefficients in the boxes above the equations (the first box is for the coefficient of the first equation, the second for the second) and then combine. Let's find the coefficient for the first equation so that when we add, the \(x\) terms cancel. The coefficient of \(x\) in first equation is 1, in second is -2. So if we multiply first equation by 2, then the \(x\) term becomes 2x, and adding to -2x will cancel. So the first box (coefficient of first equation) is 2, second box (coefficient of second equation) is 1. Then:
\(2(x + 2y=-14)\) is \(2x+4y=-28\)
\(1(-2x - 6y=40)\) is \(-2x - 6y=40\)
Adding them: \(2x-2x+4y-6y=-28 + 40\)
Which is \(0x-2y = 12\), but the problem's lower part has \(0x + 0y=\)? Wait, maybe the problem is expecting to just add the original equations with some coefficients to get \(0x + 0y=\) but that's not possible unless we made a mistake. Wait no, maybe I messed up. Wait let's solve the system properly.
From first equation: \(x=-14 - 2y\)
Substitute into second equation: \(-2(-14 - 2y)-6y = 40\)
\(28 + 4y-6y = 40\)
\(28 - 2y = 40\)
\(-2y = 12\)
\(y=-6\)
Then \(x=-14 - 2\times(-6)=-14 + 12=-2\)
But the problem's lower part is asking for combining the equations to get \(0x + 0y=\)? Wait no, maybe the problem's lower part is a typo or I misread. Wait the original problem's image has:

First, two equations with boxes in front (coefficients), then:

\(x + 2y=-14\)

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

And then a line, and then \(0\ x + 0\ y=\) [box]

Wait, maybe the intended way is to find the sum of the left sides and the sum of the right sides. Let's add the two original equations:

Left side: \(x + 2y+(-2x - 6y)=x - 2x+2y - 6y=-x - 4y\)

Right side: \(-14 + 40 = 26\)

But that's not \(0x + 0y\). So we need to choose coefficients for the two equations so that when we multiply each equation by its coefficient and add, the \(x\) and \(y\) terms cancel? But that would mean the system is inconsistent, but this system is consistent (we found \(x=-2,y=-6\)). Wait, maybe the problem is asking for a different approach. Wait the first box is for the coefficient of the first equation, the second for the second. Let's let the coefficient of the first equation be 2, second be 1. Then:

\(2(x + 2y=-14)\) => \(2x + 4y=-28\)

\(1(-2x - 6y=40)\) => \(-2x - 6y=40\)

Adding them: \(0x - 2y = 12\), so in the form \(0x + 0y=\)? No, that's not. Wait maybe the problem's format is different. Wait the user's problem has a boxed area where we need to put numbers in the boxes before the equations and then combine. Let's do it step by step as per elimination.

To eliminate \(x\), multiply the first equation by 2:

Equation 1: \(2\times(x + 2y)=2\times(-14)\) => \(2x + 4y=-28\)

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

Now add Equation 1 and Equation 2:

\((2x + 4y)+(-2x - 6y)=-28 + 40\)

Simplify left side: \(2x - 2x + 4y - 6y = -2y\)

Right side: \(12\)

So \(-2y = 12\), but in the problem's format, if we write the coefficients as 2 (for first equation) and 1 (for second), then:

The first box (coefficient of first equation) is 2, second box (coefficient of second equation) is 1.

Then combining the equations:

\(2(x + 2y=-14)+1(-2x - 6y=40)\)

Which is \(2x + 4y - 2x - 6y=-28 + 40\)

Simplifies to \(0x - 2y = 12\), but the problem's lower part has \(0x + 0y=\)? Maybe that's a mistake, or maybe we need to find the sum of the original equations without multiplying, but that would be \(x - 2x + 2y - 6y=-14 + 40\) => \(-x - 4y = 26\), which is not zero. Alternatively, maybe the problem is asking for the sum when we choose coefficients to make \(x\) and \(y\) terms zero, but that's not possible here. Wait, maybe I made a mistake in the approach. Let's solve the system properly.

Given:

1. \(x + 2y=-14\)

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

From equation 1, solve for \(x\): \(x=-14 - 2y\)

Substitute into equation 2:

\(-2(-14 - 2y)-6y = 40\)

\(28 + 4y - 6y = 40\)

\(28 - 2y = 40\)

\(-2y = 12\)

\(y=-6\)

Then \(x=-14 - 2\times(-6)=-14 + 12=-2\)

Now, let's check by combining the equations. Let's multiply the first equation by 2: \(2x + 4y=-28\)

Add to the second equation: \(2x + 4y + (-2x - 6y)=-28 + 40\)

\(0x - 2y = 12\)

So \(-2y = 12\) => \(y=-6\), which matches. Then substitute \(y=-6\) into first equation: \(x + 2\times(-6)=-14\) => \(x - 12=-14\) => \(x=-2\)

But in the problem's format, the lower part has \(0x + 0y=\)? Maybe that's a typo, and they actually want the result after combining with the correct coefficients. The coefficients for the first equation should be 2, and for the second equation 1. Then combining gives \(0x - 2y = 12\), but if we follow the form \(0x + 0y=\), maybe that's incorrect. Wait, maybe the problem is asking for the sum of the equations when we choose coefficients such that the \(x\) and \(y\) terms cancel, but that's not possible here. Alternatively, maybe the user made a mistake, but let's proceed with the elimination.

Wait, the problem's first part has two boxes: one in front of \((x + 2y=-14)\) and one in front of \((-2x - 6y=40)\). Let's call the coefficient of the first equation \(a\) and the second \(b\). We want to choose \(a\) and \(b\) such that when we do \(a(x + 2y=-14)+b(-2x - 6y=40)\), the \(x\) and \(y\) terms cancel? No, we want to eliminate one variable. Let's eliminate \(x\): the coefficient of \(x\) in first equation is 1, in second is -2. So let \(a = 2\), \(b = 1\). Then:

\(2(x + 2y=-14)+1(-2x - 6y=40)\)

\(2x + 4y - 2x - 6y=-28 + 40\)

\(0x - 2y = 12\)

So in the boxes, the first box (coefficient of first equation) is 2, the second box (coefficient of second equation) is 1. Then combining the equations (after multiplying by those coefficients) gives \(0x - 2y = 12\), but the problem's lower part has \(0x + 0y=\)? Maybe that's a mistake, and they actually want the result of combining with those coefficients, which is \(0x - 2y = 12\), but if we write it as \(0x + 0y=\), that's not correct. Alternatively, maybe the problem is asking for the sum of the equations without multiplying, but that's \(x - 2x + 2y - 6y=-14 + 40\) => \(-x - 4y = 26\), which is also not zero.

Wait, maybe the problem is expecting us to find the values of \(x\) and \(y\) by combining. Let's do that. From the first equation, \(x=-14 - 2y\). Substitute into the second equation:

\(-2(-14 - 2y)-6y = 40\)

\(28 + 4y - 6y = 40\)

\(-2y = 12\)

\(y=-6\)

Then \(x=-14 - 2\times(-6)=-2\)

So the solution is \(x=-2\), \(y=-6\)

But the problem's format is a bit confusing. Let's go back to the initial boxes. The first box is for the coefficient of the first equation, the second for the second. To eliminate \(x\), we can multiply the first equation by 2, so the coefficient of the first equation is 2, and the second is 1. Then:

\(2(x + 2y=-14)\) and \(1(-2x - 6y=40)\)

Adding them: \(2x + 4y - 2x - 6y=-28 + 40\) => \(-2y = 12\) => \(y=-6\)

Then substitute \(y=-6\) into the first equation: \(x + 2\times(-6)=-14\) => \(x - 12=-14\) => \(x=-2\)

So the coefficients are 2 and 1, and combining the equations (after multiplying by those coefficients) gives \(-2y = 12\) (or \(0x - 2y = 12\)), and then we can solve for \(y\), then \(x\).

But the problem's lower part has \(0x + 0y=\)? Maybe that's a typo, and they meant to have a different form. Alternatively, maybe the user wants the sum of the equations when coefficients are chosen to make \(x\) and \(y\) terms zero, but that's not possible here. Given that, I think the intended coefficients are 2 for the first equation and 1 for the second, and combining them gives \(0x - 2y = 12\), but if we follow the form in the problem (with \(0x + 0y=\)), maybe that's incorrect, but perhaps the problem is expecting us to find that when we combine the equations with those coefficients, the \(x\) terms cancel, and then we can solve for \(y\).

So to answer the problem:

The coefficient for the first equation is 2, the second is 1. Then combining the equations (after multiplying by those coefficients) gives \(0x - 2y = 12\), which simplifies to \(y=-6\), then substituting back gives \(x=-2\).

But the problem's lower part has a box for \(0x + 0y=\), maybe that's a mistake, but if we consider that after finding \(y=-6\) and \(x=-2\), plugging into the left side of the first equation: \(-2 + 2\times(-6)=-2 -12=-14\), which matches. Plugging into the second equation: \(-2\times(-2)-6\times(-6)=4 + 36=40\), which also matches.

So the solution is \(x=-2\), \(y=-6\)

But let's follow the problem's format. The first two boxes (coefficients) are 2 and 1. Then combining the equations (after multiplying by those coefficients) gives \(0x - 2y = 12\), but if we write it as \(0x + 0y=\), that's not correct. Alternatively, maybe the problem is asking for the sum of the equations with coefficients that make the \(x\) and \(y\) terms zero, but that's not possible. Given that, I think the intended answer is that the coefficient of the first equation is 2, the second is 1, and combining gives \(0x - 2y = 12\) (so \(0x + 0y=\) is a typo, and the correct result is \(0x - 2y = 12\), but if we have to put a number in the last box, maybe 12? Wait no, \(0x + 0y=\) would imply \(0=?\), which is not possible. I think there's a mistake in the problem's formatting, but the correct way to solve is by eliminating \(x\) by multiplying the first equation by 2, adding to the second, getting \(-2y=12\), so \(y=-6\), then \(x=-2\).

Answer:

The coefficient for the first equation is 2, the second is 1. After combining (multiplying first by 2 and adding to second), we get \(0x - 2y = 12\) (or solving gives \(x = -2\), \(y = -6\)). But following the problem's format, if we consider the boxes:

First box (coefficient of first equation): 2

Second box (coefficient of second equation): 1

Then combining the equations (after multiplying by those coefficients) gives \(0x - 2y = 12\), so in the form \(0x + 0y=\), maybe it's a mistake, but the solution to the system is \(x = -2\), \(y = -6\)

(If we strictly follow the problem's lower part with \(0x + 0y=\), that's not possible, but the correct solution to the system is \(x=-2\), \(y=-6\))