(2) \( 2^{y-3 x}=\frac{1}{16} \) \( x^{2}+x y=24 \)

Solve the equation by following steps: - step0: Solve for \(y\): \(2^{y-3x}=\frac{1}{16}\) - step1: Rewrite in exponential form: \(2^{y-3x}=2^{-4}\) - step2: Set the exponents equal: \(y-3x=-4\) - step3: Move the expression to the right side: \(y=-4+3x\) Solve the system of equations \( x^{2}+x y=24; y=3x-4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}+xy=24\\y=3x-4\end{array}\right.\) - step1: Substitute the value of \(y:\) \(x^{2}+x\left(3x-4\right)=24\) - step2: Simplify: \(4x^{2}-4x=24\) - step3: Move the expression to the left side: \(4x^{2}-4x-24=0\) - step4: Factor the expression: \(4\left(x-3\right)\left(x+2\right)=0\) - step5: Divide the terms: \(\left(x-3\right)\left(x+2\right)=0\) - step6: Separate into possible cases: \(\begin{align}&x-3=0\\&x+2=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=3\\&x=-2\end{align}\) - step8: Calculate: \(x=3\cup x=-2\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=3\\y=3x-4\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=3x-4\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=3\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=-10\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=-10\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=5\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=-10\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=5\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-2,-10\right)\cup \left(x,y\right) = \left(3,5\right)\) Let's solve the given system of equations step by step. ### Step 1: Solve the first equation The first equation is: \[ 2^{y-3x} = \frac{1}{16} \] We can rewrite \(\frac{1}{16}\) as \(2^{-4}\). Thus, we have: \[ 2^{y-3x} = 2^{-4} \] Since the bases are the same, we can set the exponents equal to each other: \[ y - 3x = -4 \] Rearranging this gives us: \[ y = 3x - 4 \] ### Step 2: Substitute into the second equation Now, we substitute \(y = 3x - 4\) into the second equation: \[ x^2 + xy = 24 \] Substituting for \(y\): \[ x^2 + x(3x - 4) = 24 \] This simplifies to: \[ x^2 + 3x^2 - 4x = 24 \] Combining like terms: \[ 4x^2 - 4x - 24 = 0 \] ### Step 3: Simplify the equation Dividing the entire equation by 4 gives: \[ x^2 - x - 6 = 0 \] ### Step 4: Factor the quadratic equation Now we can factor this equation: \[ (x - 3)(x + 2) = 0 \] Setting each factor to zero gives us: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] ### Step 5: Find corresponding \(y\) values Now we can find the corresponding \(y\) values for each \(x\): 1. For \(x = 3\): \[ y = 3(3) - 4 = 9 - 4 = 5 \] 2. For \(x = -2\): \[ y = 3(-2) - 4 = -6 - 4 = -10 \] ### Final Solutions The solutions to the system of equations are: \[ (x, y) = (3, 5) \quad \text{and} \quad (x, y) = (-2, -10) \]