Question 129145
Edit: {{{4y+4y}}} does not always equal {{{0}}}, but {{{4y-4y=0}}} always. That is what we are doing when we take Equation 1 minus Equation 2. To illustrate my point, I will show it step-by-step:

{{{(3x+4y)-(x+4y)=1-7}}}

Simplify the {{{1-7}}} on the right side of the =.
{{{(3x+4y)-(x+4y)=-6}}}

Remove unnecessary brackets (or parentheses, depending on what you want to call them).
{{{3x+4y-(x+4y)=-6}}}

Now we remove the other pair of brackets. However, the sign before the open bracket is a minus sign, so we change the + sign inside the bracket to a -.
{{{3x+4y-x-4y=-6}}}

To understand why we must change the sign, consider this example. Let x be apples and y be oranges. When I perform {{{-(x+4y)}}}, I am taking away 1 apple and 4 oranges. That would have the same effect as taking away 1 apple, {{{-x}}}, and then taking away 4 oranges, {{{-4y}}}, so:
{{{-(x+4y)=-x-4y}}}

We have {{{3x+4y-x-4y=-6}}}.

Rearranging the terms,
{{{3x-x+4y-4y=-6}}}

Simplifying x,
{{{2x+4y-4y=-6}}}

Obviously +4y and -4y cancel each other out, so our final equation is
{{{2x=-6}}}
{{{x=(-6)/2=-3}}}

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We are given two equations:

Equation 1:{{{3x+4y=1}}}
Equation 2:{{{x+4y=7}}}

Subtracting Equation 2 from Equation 1, we obtain:
{{{(3x+4y)-(x+4y)=1-7=-6}}}
{{{2x=-6}}}
{{{x=(-6)/2=-3}}}

Then, substitute {{{x=-3}}} into Equation 2:
{{{-3+4y=7}}}
{{{4y=7+3=10}}}
{{{y=10/4=5/2}}}

Since {{{x=-3}}} and {{{y=5/2}}}, the solutions of ({{{x}}},{{{y}}}) are ({{{-3}}},{{{5/2}}}), and the answer is {{{D}}}.