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2y−x=0x=y+72y-x=0 \quad x=y+7. If x,yx,y satisfies the given system of equations, what is the value of xx?

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Full solution

Q. 2y−x=0x=y+72y-x=0 \quad x=y+7. If x,yx,y satisfies the given system of equations, what is the value of xx?
  1. Substitution Method: We have two equations: 2y−x=02y - x = 0 and x=y+7x = y + 7. Let's use the substitution method to solve for xx. We can substitute the expression for xx from the second equation into the first equation.
  2. Substitute xx into first equation: Substitute x=y+7x = y + 7 into the first equation 2y−x=02y - x = 0.\newline2y−(y+7)=02y - (y + 7) = 0
  3. Simplify and combine terms: Simplify the equation by distributing the negative sign and combining like terms.\newline2y−y−7=02y - y - 7 = 0\newliney−7=0y - 7 = 0
  4. Add 77 to solve for y: Add 77 to both sides of the equation to solve for yy. \newliney=7y = 7
  5. Substitute yy back into second equation: Now that we have the value of yy, we can substitute it back into the second equation x=y+7x = y + 7 to find the value of xx.\newlinex=7+7x = 7 + 7
  6. Calculate x value: Calculate the value of x.\newlinex=14x = 14

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