Solve the system of equations. {:[3x-11 y=-1],[2x-5y=-3],[x=◻],[y=◻]:}

Identify Variable to Eliminate: Identify the variable to eliminate. We will eliminate 'y' by finding a common multiple for the coefficients of 'y' in both equations.Multiply Equations by Common Multiple: Multiply the first equation by 5 and the second equation by 11 to get the coefficients of 'y' to be the same. First equation: (3x - 11y) * 5 = -1 * 5 gives us 15x - 55y = -5. Second equation: (2x - 5y) * 11 = -3 * 11 gives us 22x - 55y = -33.Subtract Equations to Eliminate Variable: Subtract the second equation from the first equation to eliminate 'y'. (15x - 55y) - (22x - 55y) = -5 - (-33) 15x - 22x = -5 + 33 -7x = 28Solve for x: Solve for 'x'. Divide both sides of the equation by -7 to find the value of 'x'. -7x / -7 = 28 / -7 x = -4Substitute x into Equation to Solve for y: Substitute x = -4 into one of the original equations to solve for 'y'. We will use the first equation 3x - 11y = -1. 3(-4) - 11y = -1 -12 - 11y = -1Isolate y Term: Add 12 to both sides of the equation to isolate the term with 'y'. -11y = -1 + 12 -11y = 11Solve for y: Divide both sides of the equation by -11 to solve for 'y'. -11y / -11 = 11 / -11 y = -1Write Solution as Coordinate Point: Write the solution as a coordinate point. The solution is (-4, -1).

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Solve the system of equations.

{:[3x-11 y=-1],[2x-5y=-3],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} 3 x-11 y=-1 \\ 2 x-5 y=-3 \\ x=\square \\ y=\square \end{array}$

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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} 3 x-11 y=-1 \\ 2 x-5 y=-3 \\ x=\square \\ y=\square \end{array}

Identify Variable to Eliminate: Identify the variable to eliminate. We will eliminate ' extit{y}' by finding a common multiple for the coefficients of ' extit{y}' in both equations.
Multiply Equations by Common Multiple: Multiply the first equation by $5$ and the second equation by $11$ to get the coefficients of 'y' to be the same. $\newline$ First equation: $(3x - 11y) \times 5 = -1 \times 5$ gives us $15x - 55y = -5$ . $\newline$ Second equation: $(2x - 5y) \times 11 = -3 \times 11$ gives us $22x - 55y = -33$ .
Subtract Equations to Eliminate Variable: Subtract the second equation from the first equation to eliminate 'y'. $\newline$ $(15x - 55y) - (22x - 55y) = -5 - (-33)$ $\newline$ $15x - 22x = -5 + 33$ $\newline$ $-7x = 28$
Solve for x: Solve for 'x'. Divide both sides of the equation by $-7$ to find the value of 'x'. $\frac{-7x}{-7} = \frac{28}{-7}$ $x = -4$
Substitute $x$ into Equation to Solve for $y$ : Substitute $x = -4$ into one of the original equations to solve for ' $y$ '. We will use the first equation $3x - 11y = -1$ .
$3(-4) - 11y = -1$
$-12 - 11y = -1$
Isolate y Term: Add $12$ to both sides of the equation to isolate the term with 'y'. $\newline$ $-11y = -1 + 12$ $\newline$ $-11y = 11$
Solve for y: Divide both sides of the equation by $-11$ to solve for ' $y$ '. $\newline$ $-11y / -11 = 11 / -11$ $\newline$ $y = -1$
Write Solution as Coordinate Point: Write the solution as a coordinate point. The solution is $(-4, -1)$ .