\begin{aligned} 19-38y&=76x\\\\ 24x&=-6(2y-1)\end{aligned}

Equation 1 Solution: We have the system of equations: 1) $ 19 - 38y = 76x $ 2) $ 24x = -6(2y - 1) $ Let's solve the first equation for x: $ 76x = 19 - 38y $ Divide both sides by 76 to isolate x: $ x = \frac{19 - 38y}{76} $ Simplify the fraction: $ x = \frac{1 - 2y}{4} $Equation 2 Solution: Now let's solve the second equation for x: $ 24x = -6(2y - 1) $ Divide both sides by 24: $ x = \frac{-6(2y - 1)}{24} $ Simplify the fraction by dividing both numerator and denominator by 6: $ x = \frac{-1(2y - 1)}{4} $ $ x = \frac{-2y + 1}{4} $Combine Equations: We have two expressions for x: 1) $ x = \frac{1 - 2y}{4} $ 2) $ x = \frac{-2y + 1}{4} $ Since both expressions are equal to x, we can set them equal to each other: $ \frac{1 - 2y}{4} = \frac{-2y + 1}{4} $ Since the denominators are the same, we can equate the numerators: $ 1 - 2y = -2y + 1 $ This simplifies to: $ 0 = 0 $ This indicates that the two equations are actually the same line, and therefore, there are infinitely many solutions (the system is dependent).

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$\begin{aligned} 19-38y&=76x \newline24x&=-6(2y-1)\end{aligned}$

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Evaluate definite integrals using the chain rule

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\begin{aligned} 19-38y&=76x \newline24x&=-6(2y-1)\end{aligned}

Equation $1$ Solution: We have the system of equations: $\newline$ $1$ ) $19 - 38y = 76x$ $\newline$ $2$ ) $24x = -6(2y - 1)$ $\newline$ Let's solve the first equation for x: $\newline$ $76x = 19 - 38y$ $\newline$ Divide both sides by $76$ to isolate x: $\newline$ $x = \frac{19 - 38y}{76}$ $\newline$ Simplify the fraction: $\newline$ $x = \frac{1 - 2y}{4}$
Equation $2$ Solution: Now let's solve the second equation for x: $\newline$ $24x = -6(2y - 1)$ $\newline$ Divide both sides by $24$ : $\newline$ $x = \frac{-6(2y - 1)}{24}$ $\newline$ Simplify the fraction by dividing both numerator and denominator by $6$ : $\newline$ $x = \frac{-1(2y - 1)}{4}$ $\newline$ $x = \frac{-2y + 1}{4}$
Combine Equations: We have two expressions for x: $\newline$ $1$ ) $x = \frac{1 - 2y}{4}$ $\newline$ $2$ ) $x = \frac{-2y + 1}{4}$ $\newline$ Since both expressions are equal to x, we can set them equal to each other: $\newline$ $\frac{1 - 2y}{4} = \frac{-2y + 1}{4}$ $\newline$ Since the denominators are the same, we can equate the numerators: $\newline$ $1 - 2y = -2y + 1$ $\newline$ This simplifies to: $\newline$ $0 = 0$ $\newline$ This indicates that the two equations are actually the same line, and therefore, there are infinitely many solutions (the system is dependent).