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$\begin{aligned} \dfrac{1}{3}x + 3y &= 4 \ 2x - 4 &= 2y \end{aligned}$

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Find equations of tangent lines using limits

Full solution

Q.

\begin{aligned} \dfrac{1}{3}x + 3y &= 4 \ 2x - 4 &= 2y \end{aligned}

Solve for y: First, solve the second equation for $y$ : $\newline$ $2x - 4 = 2y$ $\newline$ $2y = 2x - 4$ $\newline$ $y = x - 2$
Substitute into first equation: Substitute $y = x - 2$ into the first equation: $\newline$ $\frac{1}{3}x + 3(x - 2) = 4$
Simplify the equation: Simplify the equation: $\newline$ $\frac{1}{3}x + 3x - 6 = 4$
Combine like terms: Combine like terms: $\newline$ $\frac{1}{3}x + 3x = 10$
Convert to common denominator: Convert $3x$ to a fraction with the same denominator: $\newline$ $\frac{1}{3}x + \frac{9}{3}x = 10$
Combine the fractions: Combine the fractions: $\newline$ $\frac{10}{3}x = 10$
Solve for x: Solve for $x$ : $\newline$ $x = 3$
Substitute back into y: Substitute $x = 3$ back into $y = x - 2$ : $\newline$ $y = 3 - 2$ $\newline$ $y = 1$

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