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How would I "solve by addition"? I'm not sure how to solve this.

$3x + 2y = 11$ and under that $3x – 2y = 13$

My notes that go along with it are:

In the addition method, you want to add the equations in such way so that one of the variables (letters) drops out. $x$ and $y$ are on the same side. So how do we solve? The number in front of one of the letters must be the same number and opposite in sign.

But I am completely lost. From the start, how do I solve this by addition method?

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  • $\begingroup$ Add the two equations. $2y$ and $-2y$ will cancel out. $\endgroup$ – foobar1209 Mar 17 '14 at 18:58
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Add the 2 equation to obtain $3x+3x+2y-2y=11+13$, the $y$ cancel out and you are left with $6x=24$ from which $x=4$ follows, to find y substitute 4 for $x$ into one of the original equations

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Here's the full solution. We have: \begin{align} 3x+2y=11\\ 3x-2y=13\end{align}

After adding the two equations, we have: \begin{align} 6x=24 \implies \boxed{x=4} \end{align}

Now plug back into either equation to solve for $y$: \begin{align} 3(4)+2y=11\\ 2y=-1 \implies \boxed{y=\dfrac{-1}{2}}\\ \end{align}

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The problem with an equation of two unknowns is that there is no unique solution. With two equations of two unknowns however, constituting a system of equations, we can sometimes find a unique solution. There are several methods, in this case we "solve by addition":

You add, from both equations, the left hand sides together and the right hand sides.

Left: $3x + 2y + 3x - 2y = 6x$ Right: $11 + 13 = 24$

This gives you the new equation $6x = 24$ which you can solve for $x$, you get $x = 4$.

Then plug this solution into one of the equations you have, for example $3x + 2y = 11$. You get $3 \cdot 4 + 2y = 11$. You can solve this equation to get $y = -\frac{1}{2}$.

Hence the solution to your system of equations is $x = 4$ and $y = -\frac{1}{2}$.

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