\( \begin{array}{ll}1.8 & \text { Solve for } x \text { and } y \text { simultaneously } \\ 1.8 .1 & 2 x^{2}-3 x y=-4 \text { and } 4^{x+y}=2^{y+4} \\ 1.8 .2 & x^{2}-2 x y-3 y^{2}=0 \text { and } y-x=2 \\ 1.8 .3 & 3^{2 x}=3^{y-1} \text { and } 2 x+2 y=4 \\ 1.8 .4 & y-4=2 x \text { and } 2 x^{2}-3 x y+y^{2}=4\end{array} \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y-4=2x\\2x^{2}-3xy+y^{2}=4\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=2x+4\\2x^{2}-3xy+y^{2}=4\end{array}\right.\) - step2: Substitute the value of \(y:\) \(2x^{2}-3x\left(2x+4\right)+\left(2x+4\right)^{2}=4\) - step3: Simplify: \(4x+16=4\) - step4: Move the constant to the right side: \(4x=4-16\) - step5: Subtract the numbers: \(4x=-12\) - step6: Divide both sides: \(\frac{4x}{4}=\frac{-12}{4}\) - step7: Divide the numbers: \(x=-3\) - step8: Substitute the value of \(x:\) \(y=2\left(-3\right)+4\) - step9: Calculate: \(y=-2\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-3\\y=-2\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-3\\y=-2\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-3,-2\right)\) Solve the system of equations \( x^{2}-2 x y-3 y^{2}=0; y-x=2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}-2xy-3y^{2}=0\\y-x=2\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x^{2}-2xy-3y^{2}=0\\x=-2+y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(-2+y\right)^{2}-2\left(-2+y\right)y-3y^{2}=0\) - step3: Simplify: \(4-4y^{2}=0\) - step4: Move the constant to the right side: \(-4y^{2}=0-4\) - step5: Remove 0: \(-4y^{2}=-4\) - step6: Change the signs: \(4y^{2}=4\) - step7: Divide both sides: \(\frac{4y^{2}}{4}=\frac{4}{4}\) - step8: Divide the numbers: \(y^{2}=1\) - step9: Simplify the expression: \(y=\pm \sqrt{1}\) - step10: Simplify: \(y=\pm 1\) - step11: Separate into possible cases: \(y=1\cup y=-1\) - step12: Rearrange the terms: \(\left\{ \begin{array}{l}x=-2+y\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-2+y\\y=-1\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=-1\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=-1\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-1,1\right)\cup \left(x,y\right) = \left(-3,-1\right)\) Solve the system of equations \( 3^{2 x}=3^{y-1}; 2 x+2 y=4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3^{2x}=3^{y-1}\\2x+2y=4\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}2x=y-1\\2x+2y=4\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}y=2x+1\\2x+2y=4\end{array}\right.\) - step3: Substitute the value of \(y:\) \(2x+2\left(2x+1\right)=4\) - step4: Simplify: \(6x+2=4\) - step5: Move the constant to the right side: \(6x=4-2\) - step6: Subtract the numbers: \(6x=2\) - step7: Divide both sides: \(\frac{6x}{6}=\frac{2}{6}\) - step8: Divide the numbers: \(x=\frac{1}{3}\) - step9: Substitute the value of \(x:\) \(y=2\times \frac{1}{3}+1\) - step10: Substitute back: \(y=2\times 3^{-1}+1\) - step11: Calculate: \(y=\frac{5}{3}\) - step12: Calculate: \(\left\{ \begin{array}{l}x=\frac{1}{3}\\y=\frac{5}{3}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=\frac{1}{3}\\y=\frac{5}{3}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(\frac{1}{3},\frac{5}{3}\right)\) Solve the system of equations \( 2 x^{2}-3 x y=-4; 4^{x+y}=2^{y+4} \). by following steps: - step0: : \(\left\{ \begin{array}{l}2x^{2}-3xy=-4\\4^{x+y}=2^{y+4}\end{array}\right.\) - step1: Calculate: \(\textrm{Undefined}\) Here are the solutions for the given systems of equations: ### 1.8.1 **Equations:** 1. \( 2 x^{2} - 3 x y = -4 \) 2. \( 4^{x+y} = 2^{y+4} \) **Solution:** The system is **undefined**, meaning there are no solutions that satisfy both equations simultaneously. --- ### 1.8.2 **Equations:** 1. \( x^{2} - 2 x y - 3 y^{2} = 0 \) 2. \( y - x = 2 \) **Solution:** The solutions are: - \( (x, y) = (-1, 1) \) - \( (x, y) = (-3, -1) \) --- ### 1.8.3 **Equations:** 1. \( 3^{2 x} = 3^{y-1} \) 2. \( 2 x + 2 y = 4 \) **Solution:** The solutions are: - \( (x, y) = \left(\frac{1}{3}, \frac{5}{3}\right) \) - Alternatively, in decimal form: \( (x, y) = (0.3\dot{3}, 1.6\dot{6}) \) --- ### 1.8.4 **Equations:** 1. \( y - 4 = 2 x \) 2. \( 2 x^{2} - 3 x y + y^{2} = 4 \) **Solution:** The solution is: - \( (x, y) = (-3, -2) \) --- If you have any further questions or need additional assistance, feel free to ask!