{:[x_(1)+(x_(1)*y)=12],[x_(1)+(x_(1)*y^(2))=30]:}

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Bytelearn · Accepted Answer

Factor Equations: Let's denote $ x_1 $ as $ x $ for simplicity. We have the system of equations: $$ \begin{cases} x + xy = 12 \ x + xy^2 = 30 \end{cases} $$ We can factor out $ x $ from both equations: $$ \begin{cases} x(1 + y) = 12 \ x(1 + y^2) = 30 \end{cases} $$Eliminate x: Now, let's divide the second equation by the first equation to eliminate $ x $: $$ \frac{x(1 + y^2)}{x(1 + y)} = \frac{30}{12} $$ Simplifying, we get: $$ \frac{1 + y^2}{1 + y} = \frac{5}{2} $$Cross-Multiply for y: Next, we cross-multiply to solve for $ y $: $$ 2(1 + y^2) = 5(1 + y) $$ Expanding both sides, we get: $$ 2 + 2y^2 = 5 + 5y $$Form Quadratic Equation: Now, let's rearrange the terms to form a quadratic equation: $$ 2y^2 - 5y + 2 - 5 = 0 $$ Simplifying, we get: $$ 2y^2 - 5y - 3 = 0 $$Solve Quadratic Equation: We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's try to factor it: $$ (2y + 1)(y - 3) = 0 $$ Setting each factor equal to zero gives us: $$ 2y + 1 = 0 \quad \text{or} \quad y - 3 = 0 $$Check Solutions: Solving for $ y $, we find two possible solutions: $$ y = -\frac{1}{2} \quad \text{or} \quad y = 3 $$We need to check which value of $ y $ satisfies both original equations. Let's substitute $ y = -\frac{1}{2} $ into the first equation: $$ x + x\left(-\frac{1}{2}\right) = 12 $$ Simplifying, we get: $$ x - \frac{1}{2}x = 12 $$ $$ \frac{1}{2}x = 12 $$Multiplying both sides by 2 to solve for $ x $, we get: $$ x = 24 $$ Now let's check if this $ x $ value works with $ y = -\frac{1}{2} $ in the second equation: $$ 24 + 24\left(-\frac{1}{2}\right)^2 = 30 $$ Simplifying, we get: $$ 24 + 24\left(\frac{1}{4}\right) = 30 $$ $$ 24 + 6 = 30 $$ This is true, so $ y = -\frac{1}{2} $ and $ x = 24 $ is a solution.Now let's check the other possible value for $ y $, which is $ y = 3 $, using the first equation: $$ x + x(3) = 12 $$ Simplifying, we get: $$ 4x = 12 $$ $$ x = 3 $$ Now let's check if this $ x $ value works with $ y = 3 $ in the second equation: $$ 3 + 3(3)^2 = 30 $$ Simplifying, we get: $$ 3 + 3(9) = 30 $$ $$ 3 + 27 = 30 $$ This is also true, so $ y = 3 $ and $ x = 3 $ is another solution.

$\begin{array}{l}x_{1}+\left(x_{1} \cdot y\right)=12 \\ x_{1}+\left(x_{1} \cdot y^{2}\right)=30\end{array}$

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x1+(x1⋅y)=12x1+(x1⋅y2)=30 \begin{array}{l}x_{1}+\left(x_{1} \cdot y\right)=12 \\ x_{1}+\left(x_{1} \cdot y^{2}\right)=30\end{array} x1​+(x1​⋅y)=12x1​+(x1​⋅y2)=30​

$\begin{array}{l}x_{1}+\left(x_{1} \cdot y\right)=12 \\ x_{1}+\left(x_{1} \cdot y^{2}\right)=30\end{array}$