Solve the system of equations. {:[-5y-10 x=45],[-3y+10 x=-5],[x=◻],[y=◻]:}

Identify operation to eliminate variable: Identify the operation to eliminate one of the variables. In this case, we can add the two equations directly to eliminate the 'x' term since the coefficients of 'x' are opposite in both equations.Add equations to eliminate 'x': Add the two equations to eliminate 'x'. (-5y - 10x) + (-3y + 10x) = 45 - 5 -5y - 10x + (-3y) + 10x = 40 -5y - 3y = 40 -8y = 40Solve for 'y': Solve for 'y'. Divide both sides of the equation by -8 to find the value of 'y'. -8y / -8 = 40 / -8 y = -5Substitute 'y' into equation: Substitute y = -5 into one of the original equations to solve for 'x'. We can use the second equation for this purpose. -3y + 10x = -5 -3(-5) + 10x = -5 15 + 10x = -5Solve for 'x': Solve for 'x'. Subtract 15 from both sides of the equation to isolate the term with 'x'. 10x = -5 - 15 10x = -20 x = -20 / 10 x = -2Write the solution as a coordinate point: Write the solution as a coordinate point. The solution is (-2, -5).

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Solve the system of equations.

{:[-5y-10 x=45],[-3y+10 x=-5],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} -5 y-10 x=45 \\ -3 y+10 x=-5 \\ x=\square \\ y=\square \end{array}$

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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} -5 y-10 x=45 \\ -3 y+10 x=-5 \\ x=\square \\ y=\square \end{array}

Identify operation to eliminate variable: Identify the operation to eliminate one of the variables. In this case, we can add the two equations directly to eliminate the ' $x$ ' term since the coefficients of ' $x$ ' are opposite in both equations.
Add equations to eliminate 'x': Add the two equations to eliminate 'x'. $\newline$ $(-5y - 10x) + (-3y + 10x) = 45 - 5$ $\newline$ $-5y - 10x + (-3y) + 10x = 40$ $\newline$ $-5y - 3y = 40$ $\newline$ $-8y = 40$
Solve for 'y': Solve for 'y'. Divide both sides of the equation by $-8$ to find the value of 'y'. $\frac{-8y}{-8} = \frac{40}{-8}$ $y = -5$
Substitute 'y' into equation: Substitute $y = -5$ into one of the original equations to solve for 'x'. We can use the second equation for this purpose. $\newline$ $-3y + 10x = -5$ $\newline$ $-3(-5) + 10x = -5$ $\newline$ $15 + 10x = -5$
Solve for 'x': Solve for 'x'. Subtract $15$ from both sides of the equation to isolate the term with 'x'. $\newline$ $10x = -5 - 15$ $\newline$ $10x = -20$ $\newline$ $x = \frac{-20}{10}$ $\newline$ $x = -2$
Write the solution as a coordinate point: Write the solution as a coordinate point. The solution is $(-2, -5)$ .