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Solve the system of equations.

{:[13 x+2y=1],[5x-2y=-19],[x=◻],[y=◻]:}

Solve the system of equations.\newline13x+2y=15x2y=19x=y= \begin{array}{l} 13 x+2 y=1 \\ 5 x-2 y=-19 \\ x=\square \\ y=\square \end{array}

Full solution

Q. Solve the system of equations.\newline13x+2y=15x2y=19x=y= \begin{array}{l} 13 x+2 y=1 \\ 5 x-2 y=-19 \\ x=\square \\ y=\square \end{array}
  1. 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 'yy' because the coefficients of 'yy' are opposite in both equations.
  2. Add equations to eliminate 'y': Add the equations to eliminate 'y'. (13x+2y)+(5x2y)=1+(19)(13x + 2y) + (5x - 2y) = 1 + (-19) \newline13x+2y+5x2y=11913x + 2y + 5x - 2y = 1 - 19 \newline18x=1818x = -18
  3. Solve for 'x': Solve for 'x'. Divide both sides of the equation by 1818 to find the value of 'x'.\newline18x18=1818\frac{18x}{18} = \frac{-18}{18}\newlinex=1x = -1
  4. Substitute x=1x = -1 to solve for 'y': Substitute x=1x = -1 into one of the original equations to solve for 'y'. We can use the first equation 13x+2y=113x + 2y = 1.
    13(1)+2y=113(-1) + 2y = 1
    13+2y=1-13 + 2y = 1
  5. Solve for 'y': Solve for 'y'. Add 1313 to both sides of the equation to isolate the term with 'y'.\newline2y=1+132y = 1 + 13\newline2y=142y = 14
  6. Solve for 'y': Solve for 'y'. Add 1313 to both sides of the equation to isolate the term with 'y'.\newline2y=1+132y = 1 + 13\newline2y=142y = 14Divide both sides of the equation by 22 to find the value of 'y'.\newline2y2=142\frac{2y}{2} = \frac{14}{2}\newliney=7y = 7

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