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Is (1,7)(1,7) a solution to this system of equations?\newliney=5x+2y = 5x + 2\newliney=7x+3y = 7x + 3\newlineChoices:\newline(A)yes\newline(B)no

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Q. Is (1,7)(1,7) a solution to this system of equations?\newliney=5x+2y = 5x + 2\newliney=7x+3y = 7x + 3\newlineChoices:\newline(A)yes\newline(B)no
  1. Substitute and Verify: First, we will substitute the point (1,7)(1,7) into the first equation and check if it holds true. The first equation is y=5x+2y = 5x + 2. If we substitute x=1x=1 and y=7y=7, we get 7=5×1+27 = 5 \times 1 + 2.
  2. First Equation Check: After performing the calculation, we find that 7=5+27 = 5 + 2, which simplifies to 7=77 = 7, which is true. Therefore, the point (1,7)(1,7) satisfies the first equation.
  3. Second Equation Check: Next, we will substitute the point (1,7)(1,7) into the second equation and check if it holds true. The second equation is y=7x+3y = 7x + 3. If we substitute x=1x=1 and y=7y=7, we get 7=7×1+37 = 7\times1 + 3.
  4. Solution Verification: After performing the calculation, we find that 7=7+37 = 7 + 3, which simplifies to 7=107 = 10, which is not true. Therefore, the point (1,7)(1,7) does not satisfy the second equation.
  5. Final Conclusion: Since the point (1,7)(1,7) does not satisfy both equations, it is not a solution to the system of equations.

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