3y+4x=11 3y+2x=13

$$\boxed{\begin{array}{lrcrcrr} (1) &3y & + & 4x &=& 11 &\\ (2)& 3y& + & 2x &=& 13 & \\&--&-&--&-&--& \\(1) - (2)& 0 y&+&2x&=&-2 & \\&&&2x&=&-2 & \quad | \quad :2 \\ &&&\textcolor[rgb]{1,0,0}{x}&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{-1} & \end{array}}$$

\boxed{\begin{array}{lrcrcrr}
(1) &3y & + & 4x &=& 11 &\\
(2)& 3y& + & 2x &=&  13 &
\\&--&-&--&-&--&
\\(1) - (2)& 0 y&+&2x&=&-2 &
\\&&&2x&=&-2 & \quad | \quad :2
\\ &&&\textcolor[rgb]{1,0,0}{x}&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{-1} &
\end{array}}

$$\boxed{\begin{array}{lrcrcrr} (1) &3y & + & 4x &=& 11 & \qquad x=-1 \\&3y&+&4(-1)&=&11& \\&3y&-&4&=&11& \quad | \quad +4 \\&3y&&&=&15&\quad | \quad :3 \\&\textcolor[rgb]{1,0,0}{y}&&&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{5}&\end{array}}$$

\boxed{\begin{array}{lrcrcrr} (1) &3y & + & 4x &=& 11 &  \qquad x=-1
\\&3y&+&4(-1)&=&11&
\\&3y&-&4&=&11& \quad | \quad +4
\\&3y&&&=&15&\quad | \quad :3
\\&\textcolor[rgb]{1,0,0}{y}&&&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{5}&\end{array}}

Notice that if I rearrange the first equation, we have ...... 3y = 11 - 4x

And substituting into the second equation gives us   ........(11 - 4x) + 2x = 13

Simplifying, we have  .........11 - 2x   = 13 

Add 2x to both sides ..........   11 = 13 + 2x

Subtract 13 from both sides  ........11 - 13 = 2x

Simplify ...........  -2 = 2x

Divide by 2 on both sides ............  -1 = x    ...... so there's "x"

To find "y," just substitute -1 into either of the two original equations for "x" .....I'll use the first one

3y + 4(-1) = 11

3y - 4 = 11   add 4 to both sides

3y = 15     divide by 3 on both sides

y = 5

So we have (x , y)  = (-1 , 5)

really very good explaination CPhill and Heureka! thumbs up for both of u !