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Solve the system of equations 3x+2y=143x+2y =14 and 2x+4y=202x + 4y=20 by substitution method

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Q. Solve the system of equations 3x+2y=143x+2y =14 and 2x+4y=202x + 4y=20 by substitution method
  1. Solve for x: Solve the first equation for x.\newline3x+2y=143x + 2y = 14\newline3x=14−2y3x = 14 - 2y\newlinex=14−2y3x = \frac{14 - 2y}{3}
  2. Substitute x: Substitute xx in the second equation.\newline2x+4y=202x + 4y = 20\newline2(14−2y3)+4y=202\left(\frac{14 - 2y}{3}\right) + 4y = 20\newline28−4y3+4y=20\frac{28 - 4y}{3} + 4y = 20
  3. Clear fraction: Clear the fraction by multiplying through by 33. \newline(28−4y)+12y=60(28 - 4y) + 12y = 60\newline28+8y=6028 + 8y = 60
  4. Solve for y: Solve for y.\newline8y=60−288y = 60 - 28\newline8y=328y = 32\newliney=328y = \frac{32}{8}\newliney=4y = 4
  5. Substitute yy: Substitute yy back into the equation for xx.
    x=(14−2×4)/3x = (14 - 2 \times 4) / 3
    x=(14−8)/3x = (14 - 8) / 3
    x=6/3x = 6 / 3
    x=2x = 2

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