Solve for: {\text{begin}array l 3y+5x+57-7x=3x-11y-23 } 4y+9x-y-20=5x-11-8x\text{end}array .

Expression: $\left\{\begin{array} { l } 3y+5x+57-7x=3x-11y-23 \\ 4y+9x-y-20=5x-11-8x\end{array} \right.$

Move the variables to the left-hand side and change their signs

$\left\{\begin{array} { l } 3y+5x+57-7x-3x+11y=-23 \\ 4y+9x-y-20=5x-11-8x\end{array} \right.$

Move the constant to the right-hand side and change its sign

$\left\{\begin{array} { l } 3y+5x-7x-3x+11y=-23-57 \\ 4y+9x-y-20=5x-11-8x\end{array} \right.$

Move the variables to the left-hand side and change their signs

$\left\{\begin{array} { l } 3y+5x-7x-3x+11y=-23-57 \\ 4y+9x-y-20-5x+8x=-11\end{array} \right.$

Move the constant to the right-hand side and change its sign

$\left\{\begin{array} { l } 3y+5x-7x-3x+11y=-23-57 \\ 4y+9x-y-5x+8x=-11+20\end{array} \right.$

Collect like terms

$\left\{\begin{array} { l } 14y+5x-7x-3x=-23-57 \\ 4y+9x-y-5x+8x=-11+20\end{array} \right.$

Collect like terms

$\left\{\begin{array} { l } 14y-5x=-23-57 \\ 4y+9x-y-5x+8x=-11+20\end{array} \right.$

Calculate the difference

$\left\{\begin{array} { l } 14y-5x=-80 \\ 4y+9x-y-5x+8x=-11+20\end{array} \right.$

Collect like terms

$\left\{\begin{array} { l } 14y-5x=-80 \\ 3y+9x-5x+8x=-11+20\end{array} \right.$

Collect like terms

$\left\{\begin{array} { l } 14y-5x=-80 \\ 3y+12x=-11+20\end{array} \right.$

Calculate the sum

$\left\{\begin{array} { l } 14y-5x=-80 \\ 3y+12x=9\end{array} \right.$

Use the commutative property to reorder the terms

$\left\{\begin{array} { l } -5x+14y=-80 \\ 3y+12x=9\end{array} \right.$

Use the commutative property to reorder the terms

$\left\{\begin{array} { l } -5x+14y=-80 \\ 12x+3y=9\end{array} \right.$

Move the variable to the right-hand side and change its sign

$\left\{\begin{array} { l } 14y=-80+5x \\ 12x+3y=9\end{array} \right.$

Move the variable to the right-hand side and change its sign

$\left\{\begin{array} { l } 14y=-80+5x \\ 3y=9-12x\end{array} \right.$

Divide both sides of the equation by $14$

$\left\{\begin{array} { l } y=-\frac{ 40 }{ 7 }+\frac{ 5 }{ 14 }x \\ 3y=9-12x\end{array} \right.$

Divide both sides of the equation by $3$

$\left\{\begin{array} { l } y=-\frac{ 40 }{ 7 }+\frac{ 5 }{ 14 }x \\ y=3-4x\end{array} \right.$

Since both expressions $-\frac{ 40 }{ 7 }+\frac{ 5 }{ 14 }x$ and $3-4x$ are equal to $y$, set them equal to each other forming an equation in $x$

$-\frac{ 40 }{ 7 }+\frac{ 5 }{ 14 }x=3-4x$

Solve the equation for $x$

$x=2$

Substitute the given value of $x$ into the equation $y=3-4x$

$y=3-4 \times 2$

Solve the equation for $y$

$y=-5$

The possible solution of the system is the ordered pair $\left( x, y\right)$

$\left( x, y\right)=\left( 2, -5\right)$

Check if the given ordered pair is the solution of the system of equations

$\left\{\begin{array} { l } 3 \times \left( -5 \right)+5 \times 2+57-7 \times 2=3 \times 2-11 \times \left( -5 \right)-23 \\ 4 \times \left( -5 \right)+9 \times 2-\left( -5 \right)-20=5 \times 2-11-8 \times 2\end{array} \right.$

Simplify the equalities

$\left\{\begin{array} { l } 38=38 \\ -17=-17\end{array} \right.$

Since all of the equalities are true, the ordered pair is the solution of the system

$\left( x, y\right)=\left( 2, -5\right)$