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Solve.\newliney=1y = 1\newline–2x−2y=–16–2x − 2y = –16\newline(_____, _____)

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Q. Solve.\newliney=1y = 1\newline–2x−2y=–16–2x − 2y = –16\newline(_____, _____)
  1. Substitute y=1y = 1: First, we substitute y=1y = 1 into the equation −2x−2y=−16-2x − 2y = -16. This gives us −2x−2(1)=−16-2x − 2(1) = -16. After simplifying, we get −2x−2=−16-2x − 2 = -16.
  2. Add 22 to isolate: Next, we add 22 to both sides of the equation to isolate −2x-2x. This gives us −2x=−16+2-2x = -16 + 2. After simplifying, we get −2x=−14-2x = -14.
  3. Divide by −2-2: Finally, we divide both sides of the equation by −2-2 to solve for xx. This gives us x=−14/−2x = -14 / -2. After simplifying, we get x=7x = 7.
  4. Substitute values for x,yx, y: We found y=1y = 1 and x=7x = 7. Substitute these values in x,yx, y to get the coordinate point.\newlineThis gives us the coordinate point 7,17, 1.

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