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{[3x-xy=10],[y+xy=6]:}

{3x−xy=10y+xy=6 \left\{\begin{array}{l}3 x-x y=10 \\ y+x y=6\end{array}\right.

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Full solution

Q. {3x−xy=10y+xy=6 \left\{\begin{array}{l}3 x-x y=10 \\ y+x y=6\end{array}\right.
  1. Identify Equations: Identify the equations for substitution or elimination.\newlineEquation 11: 3x−xy=103x - xy = 10\newlineEquation 22: y+xy=6y + xy = 6
  2. Isolate y for Substitution: Isolate y in Equation 22 for substitution.\newliney+xy=6y + xy = 6\newliney(1+x)=6y(1 + x) = 6\newliney=61+xy = \frac{6}{1 + x}
  3. Substitute yy in Equation 11: Substitute yy in Equation 11.3x−x(61+x)=103x - x\left(\frac{6}{1 + x}\right) = 103x−6x1+x=103x - \frac{6x}{1 + x} = 10Multiply through by (1+x)(1 + x) to clear the fraction:3x(1+x)−6x=10(1+x)3x(1 + x) - 6x = 10(1 + x)3x+3x2−6x=10+10x3x + 3x^2 - 6x = 10 + 10x
  4. Simplify and Solve for xx: Simplify and solve for xx.3x2−3x−10=10x3x^2 - 3x - 10 = 10x3x2−13x−10=03x^2 - 13x - 10 = 0
  5. Use Quadratic Formula: Use the quadratic formula to solve for xx.x=−b±b2−4ac2ax = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}x=13±(−13)2−4⋅3⋅(−10)6x = \frac{{13 \pm \sqrt{{(-13)^2 - 4\cdot3\cdot(-10)}}}}{{6}}x=13±169+1206x = \frac{{13 \pm \sqrt{{169 + 120}}}}{{6}}x=13±2896x = \frac{{13 \pm \sqrt{{289}}}}{{6}}x=13±176x = \frac{{13 \pm 17}}{{6}}x=306 or −46x = \frac{{30}}{{6}} \text{ or } \frac{{-4}}{{6}}x=5 or −23x = 5 \text{ or } -\frac{2}{3}
  6. Substitute xx back into yy: Substitute xx back into y=61+xy = \frac{6}{1 + x} to find yy. For x=5x = 5: y=61+5=66=1y = \frac{6}{1 + 5} = \frac{6}{6} = 1 For x=−23x = -\frac{2}{3}: y=61−23=613=18y = \frac{6}{1 - \frac{2}{3}} = \frac{6}{\frac{1}{3}} = 18

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