Solve the system of equations. {:[-9x-6y=15],[9x-10 y=145],[x=◻],[y=◻]:}

Identify variable to eliminate: Identify the variable to eliminate. In this case, we can eliminate 'x' as the coefficients are the opposite in both equations.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'. (-9x - 6y) + (9x - 10y) = 15 + 145 -9x - 6y + 9x - 10y = 160 -16y = 160 This gives us -16y = 160.Solve for y: Solve for 'y'. Dividing both sides of the equation by -16 gives us y = -10.Substitute y into first equation: Substitute y = -10 into the first equation to solve for 'x'. Substitute y = -10 in -9x - 6y = 15. We get -9x + 60 = 15. Subtract 60 from both sides, we get -9x = -45. Divide by -9, we get x = 5. This gives us x = 5.Write solution as coordinate point: Write the solution as a coordinate point. The solution is (5, -10).

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Solve the system of equations.

{:[-9x-6y=15],[9x-10 y=145],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} -9 x-6 y=15 \\ 9 x-10 y=145 \\ x=\square \\ y=\square \end{array}$

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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} -9 x-6 y=15 \\ 9 x-10 y=145 \\ 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 opposite in both equations.
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'. $(-9x - 6y) + (9x - 10y) = 15 + 145$ $\newline$ $-9x - 6y + 9x - 10y = 160$ $\newline$ $-16y = 160$ This gives us $-16y = 160$ .
Solve for y: Solve for 'y'. Dividing both sides of the equation by $-16$ gives us $y = -10$ .
Substitute y into first equation: Substitute $y = -10$ into the first equation to solve for ' $x$ '. Substitute $y = -10$ in $-9x - 6y = 15$ . We get $-9x + 60 = 15$ . Subtract $60$ from both sides, we get $-9x = -45$ . Divide by $-9$ , we get $x = 5$ . This gives us $x = 5$ .
Write solution as coordinate point: Write the solution as a coordinate point. The solution is $(5, -10)$ .