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Solve the system of equations. 14x+5y=3114x + 5y = 31 and 2x−3y=−292x - 3y = -29

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Q. Solve the system of equations. 14x+5y=3114x + 5y = 31 and 2x−3y=−292x - 3y = -29
  1. Multiply equations by 77: Multiply the second equation by 77 to make the coefficient of xx in both equations the same.\newline7×(2x−3y)=7×(−29)7 \times (2x - 3y) = 7 \times (-29)\newline14x−21y=−20314x - 21y = -203
  2. Subtract to eliminate xx: Subtract the new equation from the first equation to eliminate xx.
    (14x+5y)−(14x−21y)=31−(−203)(14x + 5y) - (14x - 21y) = 31 - (-203)
    14x+5y−14x+21y=31+20314x + 5y - 14x + 21y = 31 + 203
    26y=23426y = 234
  3. Solve for y: Divide both sides by 2626 to solve for y.\newline26y26=23426\frac{26y}{26} = \frac{234}{26}\newliney=9y = 9
  4. Substitute yy into equation: Substitute y=9y = 9 into the second original equation to solve for xx.2x−3(9)=−292x - 3(9) = -292x−27=−292x - 27 = -29
  5. Isolate x term: Add 2727 to both sides of the equation to isolate the term with xx. \newline2x−27+27=−29+272x - 27 + 27 = -29 + 27\newline2x=−22x = -2
  6. Solve for x: Divide both sides by 22 to solve for x.\newline2x2=−22\frac{2x}{2} = \frac{-2}{2}\newlinex=−1x = -1

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