Linear Equations - ACT Math

$7$. Constant term in slope-intercept form.

$b = -9$. Substitute $(2,1)$: $1 = 5(2) + b$, so $b = -9$.

$4$. Substitute $(2,9)$: $9=k(2)+1$, so $k=4$.

$b = -9$. Substitute $(2,1)$: $1 = 5(2) + b$, so $b = -9$.

$-\frac{1}{2}$. Perpendicular slope is negative reciprocal: $-\frac{1}{2}$.

$y = 2x + 1$. Distribute and simplify: $y - 3 = 2x - 2$, so $y = 2x + 1$.

1. Set $y = 0$: $4x = 8$, so $x = 2$.

1. Use slope formula: $\frac{8-2}{4-1} = \frac{6}{3} = 2$.

$y = -\frac{3}{4}x + 3$. Solve for $y$: $4y = -3x + 12$, so $y = -\frac{3}{4}x + 3$.

$y = -2$. Horizontal lines have constant y-value regardless of x.

$x = 5$. Vertical lines have constant x-value regardless of y.

undefined. Vertical lines have undefined slope (division by zero).

1. Horizontal lines have zero slope (no rise).

1. In $y = mx + b$, the coefficient of $x$ is the slope.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Rise over run between two points.

$x - 2y = -6$. Multiply by 2: $2y = x + 6$, rearrange to $x - 2y = -6$.

1. Convert to slope-intercept: $y = 2x + 4$, slope is 2.

$y = 3x - 4$. Direct application of slope-intercept form $y = mx + b$.

Slope-intercept form. Format matches $y = mx + b$ with explicit slope and y-intercept.

1. Set $y = 0$: $0 = 7x - 14$, so $x = 2$.

$y = 2x + 1$. Distribute and simplify: $y - 3 = 2x - 2$, so $y = 2x + 1$.

$y = -\frac{1}{4}x + 3$. Same slope $-\frac{1}{4}$, different y-intercept through $(0,3)$.

$y = -\frac{1}{3}x + \frac{17}{3}$. Perpendicular slope is $-\frac{1}{3}$; use point-slope form.

1. Set $x = 0$: $3y = 6$, so $y = 2$.

$\frac{3}{4}$. Convert to slope-intercept: $y = \frac{3}{4}x - 3$, slope is $\frac{3}{4}$.

$\frac{3}{2}$. Use slope formula: $\frac{3-0}{2-0} = \frac{3}{2}$.

$y = -5x + 3$. Use point-slope form: $y - (-2) = -5(x - 1)$.

undefined. Same x-coordinate means vertical line with undefined slope.

1. In $y = mx + b$, the constant term is the y-intercept.