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Solve simcultanecusily \( x+2 y=5 \) and \( 2 y^{2}-x y-4 x^{2}=8 \)

Ask by Savage Brooks. in South Africa
Feb 18,2025

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Answer

The solutions are \(x = -3, y = 4\) and \(x = 0.5, y = 2.25\).

Solution

Solve the system of equations \( x+2y=5;2y^{2}-xy-4x^{2}=8 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+2y=5\\2y^{2}-xy-4x^{2}=8\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=5-2y\\2y^{2}-xy-4x^{2}=8\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2y^{2}-\left(5-2y\right)y-4\left(5-2y\right)^{2}=8\) - step3: Simplify: \(-12y^{2}+75y-100=8\) - step4: Move the expression to the left side: \(-12y^{2}+75y-100-8=0\) - step5: Subtract the numbers: \(-12y^{2}+75y-108=0\) - step6: Factor the expression: \(-3\left(y-4\right)\left(4y-9\right)=0\) - step7: Divide the terms: \(\left(y-4\right)\left(4y-9\right)=0\) - step8: Separate into possible cases: \(\begin{align}&y-4=0\\&4y-9=0\end{align}\) - step9: Solve the equation: \(\begin{align}&y=4\\&y=\frac{9}{4}\end{align}\) - step10: Calculate: \(y=4\cup y=\frac{9}{4}\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=5-2y\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=5-2y\\y=\frac{9}{4}\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-3\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1}{2}\\y=\frac{9}{4}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-3\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1}{2}\\y=\frac{9}{4}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-3,4\right)\cup \left(x,y\right) = \left(\frac{1}{2},\frac{9}{4}\right)\) The solutions to the system of equations \(x+2y=5\) and \(2y^{2}-xy-4x^{2}=8\) are: 1. \(x = -3, y = 4\) 2. \(x = 0.5, y = 2.25\)

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The Deep Dive

To solve the equations \( x + 2y = 5 \) and \( 2y^2 - xy - 4x^2 = 8 \) simultaneously, we can start by expressing \( x \) in terms of \( y \) from the first equation. Rearranging gives us \( x = 5 - 2y \). Next, we substitute \( x \) into the second equation: \[ 2y^2 - (5 - 2y)y - 4(5 - 2y)^2 = 8 \] Expanding and simplifying that will lead us to a quadratic equation in terms of \( y \). Once we find the values for \( y \), we can substitute back to find the corresponding values of \( x \). Let's see how solution values pan out. After simplification, you’ll want to identify the roots of the quadratic and thus conclude with the \( (x, y) \) pairs from those \( y \) values.

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