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Khan Academy Solve in removal method. {:[3x+8y=15],[2x-8y=10],[x=◻],[y=◻]:}

Khan Academy\newlineSolve in removal method.\newline3x+8y=152x−8y=10x=□y=□ \begin{array}{l} 3 x+8 y=15 \\ 2 x-8 y=10 \\ x=\square \\ y=\square \end{array}

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Q. Khan Academy\newlineSolve in removal method.\newline3x+8y=152x−8y=10x=□y=□ \begin{array}{l} 3 x+8 y=15 \\ 2 x-8 y=10 \\ x=\square \\ y=\square \end{array}
  1. Write Equations: Write down the system of equations. 3x+8y=153x + 8y = 15 2x−8y=102x - 8y = 10
  2. Eliminate y: Add the two equations to eliminate y. (3x+8y)+(2x−8y)=15+10(3x + 8y) + (2x - 8y) = 15 + 10 3x+2x=253x + 2x = 25 5x=255x = 25
  3. Solve for x: Solve for xx.\newlinex=255x = \frac{25}{5}\newlinex=5x = 5
  4. Substitute xx and Solve for yy: Substitute x=5x = 5 into the first equation to solve for yy. 3(5)+8y=153(5) + 8y = 15 15+8y=1515 + 8y = 15 8y=15−158y = 15 - 15 8y=08y = 0 y=0y = 0
  5. Check Solution: Check the solution in the second equation. 2(5)−8(0)=102(5) - 8(0) = 10 10=1010 = 10 (True)

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