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Which equation has the solution x=3 ? 9x-9=126 4x-2=-10 7x-6=15 2x-2=1

Which equation has the solution x=3 x=3 ?\newline9x−9=126 9 x-9=126 \newline4x−2=−10 4 x-2=-10 \newline7x−6=15 7 x-6=15 \newline2x−2=1 2 x-2=1

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Solve a system of equations in three variables using eliminationright chevron icon

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Q. Which equation has the solution x=3 x=3 ?\newline9x−9=126 9 x-9=126 \newline4x−2=−10 4 x-2=-10 \newline7x−6=15 7 x-6=15 \newline2x−2=1 2 x-2=1
  1. Find Equation Solution: We need to find the equation that has x=3x=3 as its solution. To do this, we will substitute x=3x=3 into each equation and see which one results in a true statement.
  2. Substitute x=3x=3: Substitute x=3x=3 into the first equation: 9x−9=1269x-9=126.\newlineCalculation: 9(3)−9=27−9=189(3) - 9 = 27 - 9 = 18.\newlineCheck if 1818 equals 126126: 18≠12618 \neq 126.
  3. Check First Equation: Since 1818 does not equal 126126, the first equation does not have x=3x=3 as its solution. Now, substitute x=3x=3 into the second equation: 4x−2=−104x-2=-10.\newlineCalculation: 4(3)−2=12−2=104(3) - 2 = 12 - 2 = 10.\newlineCheck if 1010 equals −10-10: 10≠−1010 \neq -10.
  4. Check Second Equation: Since 1010 does not equal −10-10, the second equation does not have x=3x=3 as its solution. Now, substitute x=3x=3 into the third equation: 7x−6=157x-6=15.\newlineCalculation: 7(3)−6=21−6=157(3) - 6 = 21 - 6 = 15.\newlineCheck if 1515 equals 1515: 15=1515 = 15.
  5. Check Third Equation: Since 1515 equals 1515, the third equation has x=3x=3 as its solution. We do not need to check the fourth equation because we have already found the correct equation.

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