Solve the system of equations. {:[-12 x-5y=40],[12 x-11 y=88],[x=◻],[y=◻]:}

Identify variable to eliminate: Identify the variable to eliminate. In this case, we can eliminate 'x' as the coefficients are the same in both equations but with opposite signs.Identify operation to eliminate variable: Identify the operation to eliminate the variable. Here, we add the equations as the coefficients are opposite.Add equations to eliminate 'x': Add the equations to eliminate 'x'. (-12x - 5y) + (12x - 11y) = 40 + 88 -12x - 5y + 12x - 11y = 128 -16y = 128 This gives us -16y = 128.Solve for 'y': Solve for 'y'. Dividing both sides of the equation by -16 gives us y = -8.Substitute y = -8 into first equation to solve for 'x': Substitute y = -8 into the first equation to solve for 'x'. Substitute y = -8 in -12x - 5y = 40. We get -12x + 40 = 40. Subtract 40 from both sides, we get -12x = 0. Divide by -12, we get x = 0. This gives us x = 0.

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Solve the system of equations.

{:[-12 x-5y=40],[12 x-11 y=88],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} -12 x-5 y=40 \\ 12 x-11 y=88 \\ x=\square \\ y=\square \end{array}$

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Algebra 2

Solve a system of equations using elimination

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Q. Solve the system of equations.

\newline

\begin{array}{l} -12 x-5 y=40 \\ 12 x-11 y=88 \\ x=\square \\ y=\square \end{array}

Identify variable to eliminate: Identify the variable to eliminate. In this case, we can eliminate ' $x$ ' as the coefficients are the same in both equations but with opposite signs.
Identify operation to eliminate variable: Identify the operation to eliminate the variable. Here, we add the equations as the coefficients are opposite.
Add equations to eliminate 'x': Add the equations to eliminate 'x'. $(-12x - 5y) + (12x - 11y) = 40 + 88$ $\newline$ $-12x - 5y + 12x - 11y = 128$ $\newline$ $-16y = 128$ This gives us $-16y = 128$ .
Solve for 'y': Solve for 'y'. Dividing both sides of the equation by $-16$ gives us $y = -8$ .
Substitute $y = -8$ into first equation to solve for 'x': Substitute $y = -8$ into the first equation to solve for 'x'. Substitute $y = -8$ in $-12x - 5y = 40$ . We get $-12x + 40 = 40$ . Subtract $40$ from both sides, we get $-12x = 0$ . Divide by $-12$ , we get $x = 0$ . This gives us $x = 0$ .