Question

Solveing systems of equations by substitution y=6x-11 -2x-3y=-7

Answer 1

`\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}`

`\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`

Answer 2

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!