Solve the system of equations. {:[10 y-11 x=-4],[-2y+3x=4],[x=◻],[y=◻]:}

Identify Variable to Eliminate: Identify the variable to eliminate. We can choose to eliminate either 'x' or 'y'. In this case, let's 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 2 and the second equation by 10 to get the coefficients of 'y' to be the same (but opposite in sign). First equation: 2*(10y - 11x) = 2*(-4) gives 20y - 22x = -8. Second equation: 10*(-2y + 3x) = 10*4 gives -20y + 30x = 40.Add Equations to Eliminate Variable: Add the new equations to eliminate 'y'. (20y - 22x) + (-20y + 30x) = -8 + 40 20y - 20y - 22x + 30x = 32 8x = 32Solve for x: Solve for 'x'. Divide both sides of the equation by 8 to find the value of 'x'. 8x / 8 = 32 / 8 x = 4Substitute x into Original Equation: Substitute x = 4 into one of the original equations to solve for 'y'. Let's use the first equation 10y - 11x = -4. 10y - 11(4) = -4 10y - 44 = -4Solve for y: Solve for 'y'. Add 44 to both sides of the equation to isolate the term with 'y'. 10y = 40 Divide both sides by 10 to find the value of 'y'. y = 40 / 10 y = 4

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Solve the system of equations.

{:[10 y-11 x=-4],[-2y+3x=4],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} 10 y-11 x=-4 \\ -2 y+3 x=4 \\ x=\square \\ y=\square \end{array}$

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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} 10 y-11 x=-4 \\ -2 y+3 x=4 \\ x=\square \\ y=\square \end{array}

Identify Variable to Eliminate: Identify the variable to eliminate. We can choose to eliminate either ' $x$ ' or ' $y$ '. In this case, let's 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 $2$ and the second equation by $10$ to get the coefficients of ' $y$ ' to be the same (but opposite in sign). $\newline$ First equation: $2(10y - 11x) = 2(-4)$ gives $20y - 22x = -8$ . $\newline$ Second equation: $10(-2y + 3x) = 10 imes 4$ gives $-20y + 30x = 40$ .
Add Equations to Eliminate Variable: Add the new equations to eliminate 'y'. $\newline$ $(20y - 22x) + (-20y + 30x) = -8 + 40$ $\newline$ $20y - 20y - 22x + 30x = 32$ $\newline$ $8x = 32$
Solve for x: Solve for 'x'. Divide both sides of the equation by $8$ to find the value of 'x'. $\newline$ $\frac{8x}{8} = \frac{32}{8}$ $\newline$ $x = 4$
Substitute $x$ into Original Equation: Substitute $x = 4$ into one of the original equations to solve for ' $y$ '. Let's use the first equation $10y - 11x = -4$ .
$10y - 11(4) = -4$
$10y - 44 = -4$
Solve for y: Solve for 'y'. Add $44$ to both sides of the equation to isolate the term with 'y'. $\newline$ $10y = 40$ $\newline$ Divide both sides by $10$ to find the value of 'y'. $\newline$ $y = \frac{40}{10}$ $\newline$ $y = 4$