Respuesta :
[tex]\bf \begin{cases} \boxed{y}=6x-11\\ \cline{1-1} -2x-3y=-7 \end{cases}\qquad \implies \stackrel{\textit{substituting \underline{y} in the 2nd equation}}{-2x-3\left( \boxed{6x-11} \right)=-7}[/tex]
[tex]\bf -2x-18x+33=-7\implies -20x+33=-7\implies -20x=-40 \\\\\\ x=\cfrac{-40}{-20}\implies \blacktriangleright x=2 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that }}{y=6x-11}\implies y=6(2)-11\implies y=12-11\implies \blacktriangleright y=1 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (2,1)~\hfill[/tex]
Step 1: Plug 6x - 11 in for y in the equation -2x - 3y = -7
-2x - 3(6x-11) = -7
Step 2: Distribute -3 to the numbers inside the parentheses (6x -11)
-2x - 18x + 33 = -7
Step 3: Combine like terms
x's go with x's
-20x + 33 = -7
Step 4: Subtract 33 to both sides
-20x = -40
Step 5: Isolate x by dividing -20 to both sides
x = 2
Step 6: To solve for y plug 2 in for x in the equation y = 6x - 11
y = 6(2) - 11
y = 12 - 11
y = 1
(2, 1)
Check:
1 = 6(2) - 11
1 = 12 - 11
1 = 1
-2(2) - 3(1) = -7
-4 - 3 = -7
-7 = -7
Hope this helped!
