\( 4 x^{2}-7 x y+4 y^{2}=16 \) and \( y=2 x-2 \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}4x^{2}-7xy+4y^{2}-16=0\\y-2x+2=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}4x^{2}-7xy+4y^{2}-16=0\\y=2x-2\end{array}\right.\) - step2: Substitute the value of \(y:\) \(4x^{2}-7x\left(2x-2\right)+4\left(2x-2\right)^{2}-16=0\) - step3: Simplify: \(6x^{2}-18x=0\) - step4: Factor the expression: \(6x\left(x-3\right)=0\) - step5: Separate into possible cases: \(\begin{align}&6x=0\\&x-3=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=0\\&x=3\end{align}\) - step7: Calculate: \(x=0\cup x=3\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=0\\y=2x-2\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=2x-2\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=0\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=4\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=4\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right) = \left(0,-2\right)\cup \left(x,y\right) = \left(3,4\right)\) Solve the system of equations \( 4 x^{2}-7 x y+4 y^{2}-16=0; y=2 x-2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}4x^{2}-7xy+4y^{2}-16=0\\y=2x-2\end{array}\right.\) - step1: Substitute the value of \(y:\) \(4x^{2}-7x\left(2x-2\right)+4\left(2x-2\right)^{2}-16=0\) - step2: Simplify: \(6x^{2}-18x=0\) - step3: Factor the expression: \(6x\left(x-3\right)=0\) - step4: Separate into possible cases: \(\begin{align}&6x=0\\&x-3=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=0\\&x=3\end{align}\) - step6: Calculate: \(x=0\cup x=3\) - step7: Rearrange the terms: \(\left\{ \begin{array}{l}x=0\\y=2x-2\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=2x-2\end{array}\right.\) - step8: Calculate: \(\left\{ \begin{array}{l}x=0\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=4\end{array}\right.\) - step9: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=4\end{array}\right.\) - step10: Rewrite: \(\left(x,y\right) = \left(0,-2\right)\cup \left(x,y\right) = \left(3,4\right)\) We have the following system of equations: 1. \( 4 x^{2} - 7 x y + 4 y^{2} = 16 \) 2. \( y = 2 x - 2 \) To solve this system, we can substitute the second equation into the first equation. ### Step 1: Substitute \( y \) in the first equation Substituting \( y = 2x - 2 \) into the first equation: \[ 4 x^{2} - 7 x (2x - 2) + 4 (2x - 2)^{2} = 16 \] ### Step 2: Expand and simplify Now, let's expand and simplify the equation: 1. Expand \( -7x(2x - 2) \): \[ -7x(2x - 2) = -14x^{2} + 14x \] 2. Expand \( 4(2x - 2)^{2} \): \[ 4(2x - 2)^{2} = 4(4x^{2} - 8x + 4) = 16x^{2} - 32x + 16 \] Now, substituting these back into the equation gives: \[ 4x^{2} - 14x^{2} + 14x + 16x^{2} - 32x + 16 = 16 \] ### Step 3: Combine like terms Combining like terms results in: \[ (4x^{2} - 14x^{2} + 16x^{2}) + (14x - 32x) + 16 = 16 \] This simplifies to: \[ 6x^{2} - 18x + 16 = 16 \] ### Step 4: Set the equation to zero Subtract 16 from both sides: \[ 6x^{2} - 18x = 0 \] ### Step 5: Factor the equation Factoring out \( 6x \): \[ 6x(x - 3) = 0 \] ### Step 6: Solve for \( x \) Setting each factor to zero gives: 1. \( 6x = 0 \) → \( x = 0 \) 2. \( x - 3 = 0 \) → \( x = 3 \) ### Step 7: Find corresponding \( y \) values Now, substitute these \( x \) values back into the equation \( y = 2x - 2 \): 1. For \( x = 0 \): \[ y = 2(0) - 2 = -2 \] So, one solution is \( (0, -2) \). 2. For \( x = 3 \): \[ y = 2(3) - 2 = 6 - 2 = 4 \] So, another solution is \( (3, 4) \). ### Final Solutions The solutions to the system of equations are: \[ (x, y) = (0, -2) \cup (3, 4) \]