Linear Inequalities - SAT Math

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$Ax + By = C$. General form where $A$, $B$, and $C$ are constants.

$Ax + By = C$. General form with integer coefficients $A$, $B$, and $C$.

$-5$. Coefficient of $x$ in slope-intercept form.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$2x + y = 5$. Move $x$-term to left: $2x + y = 5$.

$(x, y) = (6, 4)$. Add equations to get $2x = 12$, so $x = 6$, then $y = 4$.

$(x, y) = (4, 3)$. Add equations to eliminate $y$: $4x = 16$, so $x = 4$.

$(x, y) = (-\frac{2}{3}, \frac{5}{3})$. Set equal: $2x + 3 = -x + 1$, solve to get $x = -\frac{2}{3}$.

Add/subtract equations to eliminate a variable. Multiply equations to make coefficients opposites, then combine.

Infinitely many solutions. Second equation is twice the first, creating coincident lines.

Infinitely many solutions. Dependent equations represent the same geometric line.

$(x, y) = (1, 3)$. Set equal: $x + 2 = -2x + 5$, solve to get $x = 1$.

$x = 4$. Substitute $y = 0$ into equation: $5x = 20$.

Infinitely many solutions. Second equation is twice the first, so lines are identical.

Same slope and same intercept. Lines overlap completely when all coefficients are proportional.

$(x, y) = (6, 1)$. Add equations to eliminate $y$: $2x = 12$, so $x = 6$.

$x = 4$. Substitute $y = 0$ into first equation: $3x = 12$.

$(x, y) = (0, 0)$. Add equations: $2x = 0$, so $x = 0$, then $y = 0$.

Plot equations, solution is intersection point. Draw both lines; they meet at one point.

$(x, y) = (2, 1)$. Add equations to eliminate $y$: $3x = 6$, so $x = 2$.

Equations represent the same line. One equation is a multiple of the other.

$y = 3$. Substitute $x = 5$ into first equation directly.

Determinant $\neq 0$. Non-zero determinant ensures coefficient matrix is invertible.

Infinitely many solutions. Second equation is $-2$ times the first, so lines coincide.

Use matrices and row operations or inverses. Convert system to matrix form and solve systematically.

Infinitely many solutions. Second equation is twice the first, creating identical lines.

Infinitely many solutions. Second equation is twice the first, so lines coincide.

Infinitely many solutions. Second equation is twice the first, creating identical lines.

Lines are parallel, same slope, different intercepts. Lines never intersect when they have identical slopes.

$(x, y) = (3, 0)$. Add equations to eliminate $y$: $6x = 18$, so $x = 3$.

Solve one equation for a variable, substitute in the other. Express one variable in terms of another from first equation.

$(x, y) = (3, 2)$. Add equations to eliminate $y$: $2x = 6$, so $x = 3$.

$x = 6$. Substitute $y = 0$ into equation: $2x = 12$.

Yes, they are parallel. Both lines have slope 2, so they're parallel.

1. Set $y = 0$ to get $4x = 16$, so $x = 4$.

1. No vertical change means zero slope.

Undefined. Division by zero makes slope undefined.

$2x + y = 5$. Add $2x$ to both sides to get standard form.

$Ax + By = C$. General form with constants $A$, $B$, and $C$.

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

Slope of the line. The coefficient of $x$ that determines steepness and direction.

-3. The coefficient of $x$ in slope-intercept form.

-7. The constant term in slope-intercept form.

Yes, they are perpendicular. Slopes are 3 and $-\frac{1}{3}$; their product is -1.

The slopes are equal. Lines with identical slopes never intersect.

$p \leq 4$. 'At most' means less than or equal to.

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

The y-intercept is $b$. The constant term where the line crosses the y-axis.

$y = -\frac{2}{3}x + 2$. Isolate $y$ by subtracting $2x$ and dividing by $3$.

$Ax + By = C$. General form with integer coefficients $A$, $B$, and $C$.

The x-intercept is $x = 4$.. Set $y = 0$ and solve for $x$.

The slope is $-5$. The coefficient of $x$ in slope-intercept form.

$y - y_1 = m(x - x_1)$. Uses a known point and slope to define the line.

Slope-intercept form. Shows slope $m$ and y-intercept $b$ directly.

$x = 2$. Substitute $y = 3$ and solve: $2x + 3 = 7$.

$y = 2x - 4$. Isolate $y$ by subtracting $4x$ and dividing by $-2$.

$y = -(x - 2) + 2$. Horizontal shift affects the $x$-term inside parentheses.

$\bigg( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \bigg)$. Average of coordinates for both $x$ and $y$ values.

$(3, 4)$. Use midpoint formula with given coordinates.

$y - y_1 = m(x - x_1)$. Uses a known point $(x_1, y_1)$ and slope $m$.

$y - 2 = 4(x - 1)$. Substitute point coordinates and slope into formula.

Y-intercept of the line. The constant term where the line crosses the y-axis.

Product of slopes is -1. Negative reciprocals create 90-degree intersections.

$y = 3x - 4$. Direct substitution into $y = mx + b$.

$y = -\frac{3}{4}x + 3$. Isolate $y$ by dividing both sides by 4.

$x = 5$. Vertical lines have constant $x$-values.

$4x - y = 3$. Move $y$ to left side and arrange coefficients properly.

$y = 7$. Horizontal lines have constant $y$-values.

$-\frac{2}{3}$. The coefficient of $x$ in slope-intercept form.

1. Set $y = 0$ to get $2x = 10$, so $x = 5$.

$y = 2x + 8$. Vertical shift changes only the y-intercept.

-3. The coefficient of $x$ is the slope.

1. Set $y = 0$ and solve: $0 = 2x - 4$, so $x = 2$.

1. The constant term is the y-intercept.

$y - y_1 = m(x - x_1)$. Uses a known point $(x_1, y_1)$ and slope $m$.

1. $m = \frac{6-2}{3-1} = \frac{4}{2} = 2$

1. Divide by 3: $y = 4x + 2$, so y-intercept is 2.

1. Use $m = \frac{7-3}{4-2} = \frac{4}{2} = 2$.

$\frac{3}{2}$. Rearrange to get $y = \frac{3}{2}x - 3$.

1. Set $x = 0$ to get $3y = 15$, so $y = 5$.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Change in $y$ divided by change in $x$.

$x > 10$. 'More than' means strictly greater than.

$x \neq 2$. Subtract 2, divide by 5 to find excluded value.

$\geq$. Combines greater than and equal to in one symbol.

$x < 6$. Distribute 3, add 12, then divide by 3.

$x \neq 3$. Solve $x = 3$ from the equation $8 - x = 5$.

$x > -2$. Subtract 4, divide by 2 to solve inequality.

$x \neq -2$. Divide both sides by -5 to isolate $x$.

$x \neq 2$. Solve $2x = 4$ to find the excluded value.

$\leq$. Combines less than and equal to in one symbol.

$x \neq 2$. Subtract 7, divide by -3 to get $x = 2$.

$x > 3$. Subtract 9, divide by -2, flip inequality sign.

Negative numbers. All values to the left of zero on number line.

$x > 4$. Distribute 6, add 12, then divide by 6.

$x \neq 6$. Multiply both sides by 3 to isolate $x$.

$>$. Symbol for strict inequality without equality.

$x \leq 3$. Subtract 7, divide by 3 to isolate $x$.

$y < 8$. 'Less than' means strictly smaller than 8.

$x > 7$. Subtract 3 from both sides to isolate $x$.

$x \neq 4$. Add 3, divide by 7 to find excluded value.

Closed circle at -2, arrow to the left. Closed circle includes -2 in the solution set.

$x \geq 5$. Add 5 to both sides to isolate $x$.

$x < 2$. Subtract 8, divide by 4 to solve inequality.

$x \neq -7$. Distribute -3, subtract 12, divide by -3.

$x \leq 2$. Subtract 3, divide by 2 to solve inequality.

$x \geq 5$. 'At least' means greater than or equal to.

$x \neq 4$. Add 11, divide by -4 to get $x = 4$.

Open circle at 3, arrow to the right. Open circle shows 3 is not included.

$x \neq -4$. Subtract 5, divide by -2 to get $x = -4$.

Multiply or divide both sides by a negative number. This operation changes the direction of the inequality.

$x \text{ } \text{≤} \text{ } 3$. Add 4 to both sides, then divide by 5.

$x \neq -3$. Divide both sides by $-3$, keeping the not-equal sign.

$x > 2$. Subtract 3 from both sides, then divide by 2.

$x < 4$. Distribute 3, add 6 to both sides, then divide by 3.

$y \text{ } \text{≤} \text{ } -3$. Subtract 5, divide by $-2$, and flip the inequality sign.

$x$ is less than or equal to zero. The variable can be zero or any negative value.

$x < -4$. Subtract 4 from both sides, then divide by $-1$ and flip sign.

$x = 4$. Subtract 3 from both sides, then divide by 2.

Multiplication. Multiply both sides by 3 to eliminate the fraction.

Addition. Add 5 to both sides to isolate x.

Distribute to get $3x - 6 = 12$. Apply the distributive property to remove parentheses.

$x = 7$. Subtract 5 from both sides to isolate x.

$x = 2$. Add 4, then divide by 3.

For any number $a$, $a + (-a) = 0$. Any number plus its opposite equals zero.

$x = 3$. Subtract 7 from both sides, then multiply by -1.

$x = 4$. Divide both sides by 5.

$x = 3$. Subtract 1, then divide by 3.

An equation of the form $ax + b = 0$. One variable raised to the first power.

Use inverse operations. Apply the opposite operation to both sides.

$x = 27$. Multiply both sides by 3.

$x = 7$. Subtract 8 from both sides.

$(ab)c = a(bc)$. Grouping factors doesn't change the product.

$x = 2$. Add 5 to both sides.

$x = 4$. Subtract 2 from both sides.

A statement that two expressions are equal. Shows two expressions have the same value.

$a + b = b + a$. Order of addends doesn't affect the sum.

$x = 2$. Subtract 5 from both sides, then divide by -2.

For any number $a$, $a \times 1 = a$. Multiplying by 1 leaves any number unchanged.

$x = 2$. Subtract 3, then divide by 2.