Determine the slope and the y-intercept.
-$$13 y + 3 x + 4 = 3 + 3 x$$
A) Slope $\frac { 3 } { 13 } , y$-intercept $\left( 0 , - \frac { 1 } { 13 } \right)$
B) Slope $0 , y$-intercept $\left( 0 , \frac { 1 } { 13 } \right)$
C) Slope $0 , y$-intercept $\left( 0 , - \frac { 1 } { 13 } \right)$
D) Slope $- \frac { 3 } { 13 } , y$-intercept $\left( 0 , - \frac { 3 } { 13 } \right)$

Question

Quizplus · Accepted Answer

First, simplify the equation: $13y + 3x + 4 = 3 + 3x$. Subtract $3x$ from both sides to get $13y + 4 = 3$. Then, subtract 4 from both sides to get $13y = -1$. Divide by 13 to solve for $y$, resulting in $y = -\frac{1}{13}$. This shows that the slope is $0$ (since there's no $x$ term in the final equation, indicating a horizontal line) and the y-intercept is $\left(0, -\frac{1}{13}\right)$.

Determine the Slope and the Y-Intercept 13y+3x+4=3+3x13 y + 3 x + 4 = 3 + 3 x13y+3x+4=3+3x

Determine the Slope and the Y-Intercept $13 y + 3 x + 4 = 3 + 3 x$