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Solve the system by substitution. {:[y=-7x-2],[-2x-2y=40]:}

Solve the system by substitution.\newliney=−7x−2−2x−2y=40 \begin{aligned} y & =-7 x-2 \\ -2 x-2 y & =40 \end{aligned}

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Q. Solve the system by substitution.\newliney=−7x−2−2x−2y=40 \begin{aligned} y & =-7 x-2 \\ -2 x-2 y & =40 \end{aligned}
  1. Identify Equations: First, we need to identify the two equations in the system that we will be working with.\newlineEquation 11: y=−7x−2y = -7x - 2\newlineEquation 22: −2x−2y=40-2x - 2y = 40\newlineWe will use substitution to solve the system, which means we will substitute the expression for yy from Equation 11 into Equation 22.
  2. Substitute yy into Equation 22: Substitute y=−7x−2y = -7x - 2 from Equation 11 into Equation 22.\newlineEquation 22 becomes: −2x−2(−7x−2)=40-2x - 2(-7x - 2) = 40\newlineNow we need to simplify and solve for xx.
  3. Simplify and Solve for x: Simplify the equation by distributing the −2-2 into the parentheses.\newline−2x+14x+4=40-2x + 14x + 4 = 40\newlineCombine like terms.\newline12x+4=4012x + 4 = 40
  4. Isolate x Term: Subtract 44 from both sides of the equation to isolate the term with xx.\newline12x+4−4=40−412x + 4 - 4 = 40 - 4\newline12x=3612x = 36
  5. Solve for x: Divide both sides by 1212 to solve for x.\newline12x12=3612\frac{12x}{12} = \frac{36}{12}\newlinex=3x = 3
  6. Substitute xx into Equation 11: Now that we have the value of xx, we can substitute it back into Equation 11 to find the value of yy.\newliney=−7(3)−2y = -7(3) - 2\newlineCalculate the value of yy.\newliney=−21−2y = -21 - 2\newliney=−23y = -23
  7. Calculate yy: We have found the values of xx and yy that satisfy both equations in the system.\newlinex=3x = 3 and y=−23y = -23\newlineThese values are the solution to the system of equations.

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