Find the solution of the system of equations. {:[12 x-10 y=2],[-6x+7y=-11]:} (◻,◻)

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Find the solution of the system of equations.  {:[12 x-10 y=2],[-6x+7y=-11]:}  (◻,◻)

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Label Equations: Let's label the equations for easier reference: Equation (1): 12x - 10y = 2 Equation (2): -6x + 7y = -11 We will use the method of elimination to solve this system of equations. First, we need to make the coefficients of either x or y the same in both equations. We can do this by multiplying equation (2) by 2 to match the coefficient of x in equation (1).Multiply Equation: Multiply equation (2) by 2: 2(-6x + 7y) = 2(-11) -12x + 14y = -22 Now we have a new equation (3): Equation (3): -12x + 14y = -22Add Equations: Add equation (1) and equation (3) together to eliminate x: (12x - 10y) + (-12x + 14y) = 2 + (-22) The x terms cancel out, and we are left with: -10y + 14y = -20 This simplifies to: 4y = -20Solve for y: Divide both sides of the equation by 4 to solve for y: 4y / 4 = -20 / 4 y = -5 We have found the value of y.Substitute and Solve: Now that we have the value of y, we can substitute it back into one of the original equations to find x. We will use equation (1): 12x - 10(-5) = 2 12x + 50 = 2Solve for x: Subtract 50 from both sides of the equation to solve for x: 12x + 50 - 50 = 2 - 50 12x = -48Divide both sides of the equation by 12 to solve for x: 12x / 12 = -48 / 12 x = -4 We have found the value of x.

Find the solution of the system of equations. $\newline$ $\begin{array}{l} 12 x-10 y=2 \\ -6 x+7 y=-11 \end{array}$ $\newline$ $(\square, \square)$

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Find the solution of the system of equations.\newline12x−10y=2−6x+7y=−11 \begin{array}{l} 12 x-10 y=2 \\ -6 x+7 y=-11 \end{array} 12x−10y=2−6x+7y=−11​\newline(□,□) (\square, \square) (□,□)

Find the solution of the system of equations. $\newline$ $\begin{array}{l} 12 x-10 y=2 \\ -6 x+7 y=-11 \end{array}$ $\newline$ $(\square, \square)$