Systems of Equations - SAT Math

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

1. Horizontal lines have zero rise over run.

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

$x = 4$. Undefined slope means vertical line.

$\frac{1}{2}$. Parallel lines have identical slopes.

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

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

$y = -2x + 6$. Use point-slope: $y - 4 = -2(x - 1)$, then solve.

$y = 2x - 4$. Divide both sides by -2 to isolate $y$.

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

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

$y - 1 = 3(x - 1)$. Use point $(1,1)$ in point-slope formula.

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

$(2, 0)$. Set $y = 0$: $0 = 4x - 8$, so $x = 2$.

$2$. Set $x = 0$: $5y = 10$, so $y = 2$.

Standard form. Form $Ax + By = C$ with integer coefficients.

$x = 3$. Substitute and solve the system of equations.

$y = 5$. Zero slope creates horizontal line.

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

$b$. Constant term where line crosses y-axis.

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

1. Horizontal lines have zero slope.

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

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

$y = -\frac{1}{3}x + 6$. Same slope, different y-intercept through given point.

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

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

$-\frac{3}{2}$. Set $y = 0$: $0 = 2x + 3$, so $x = -\frac{3}{2}$.

Undefined. Vertical lines have undefined slope.

$-\frac{1}{4}$. Perpendicular slopes are negative reciprocals.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Rise over run: change in y divided by change in x.

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

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

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

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

$y = -\frac{1}{2}x + \frac{7}{2}$. Perpendicular slope $-\frac{1}{2}$, use point-slope form.

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

Yes. Each x-value has exactly one y-value.

Point-slope form. Form $y - y_1 = m(x - x_1)$ with known point.

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

$y = -\frac{2}{3}x + 2$. Solve for $y$ by isolating it on one side.

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

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

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

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

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

$y + 3 = 2(x - 1)$. Substitute the point and slope into point-slope form.

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

1. The constant term is the y-intercept.

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

1. No vertical change means zero slope.

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

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

1. Set $y = 0$ and solve for $x$.

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

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

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

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

$y = \frac{3}{2}x - 3$. Solve for $y$ by isolating it on one side.

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

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

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

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

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

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

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

1. The constant term when $x = 0$.

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

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

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

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

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

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

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

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

$\frac{1}{2}$. Parallel lines have identical slopes.

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

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

$y = mx + b$. 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.

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

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

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

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

Undefined. Division by zero makes slope undefined.

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

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

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

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

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

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

1. The coefficient of $x$ is the slope.

$x \leq 7$. 'At most' means less than or equal to.

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

$\geq$. Standard mathematical symbol for greater than or equal to.

No, $x = -1$ does not satisfy $2x + 3 > 0$. Substituting gives $2(-1) + 3 = 1 > 0$ is false.

$x < 10$. Standard notation for values below 10.

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

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

$x \geq 5$. Distribute 3, add 6 to both sides, then divide by 3.

A horizontal line at $y = 4$ with a dashed line. Uses dashed line since $\neq$ excludes the boundary value.

$4 \leq x$. Reverses the typical order to show 4 as the lesser value.

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

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

$x < 3$. Move $2x$ left and $-3$ right, then divide by 3.

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

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

The solution set is $x > 2$. Subtract 5 from both sides, then divide by 2.

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

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

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

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

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

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

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

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

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

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

Infinite solutions (dependent system). Second equation is twice the first, so they represent the same line.

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

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

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

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

No solution (inconsistent system). Parallel lines with different y-intercepts never intersect.

Lines intersect at one or more points. Consistent means the system has at least one solution.

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

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

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

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

Non-parallel lines (different slopes). Different slopes ensure the lines intersect at exactly one point.

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

$(x, y) = (3, 1)$. Add equations to get $2x = 6$, so $x = 3$; substitute to find $y = 1$.

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

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

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

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

$(x, y) = (-\frac{2}{3}, \frac{5}{3})$. 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.

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

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

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

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

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

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

$x = 6$. Subtract 9 from both sides to isolate $x$.

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

$x = 12$. Multiply both sides by 4 to isolate $x$.

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

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

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

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

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

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

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

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

$x = 3$. Subtract 5 from both sides, then divide by 2.

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

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

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

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

Division. Multiplication and division are inverse operations.

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

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

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

$x = 2$. Subtract 5 from both sides, then multiply by -1.

Subtraction. Addition and subtraction are inverse operations.

$x = -\frac{b}{a}$. Subtract $b$ from both sides, then divide by $a$.

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

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

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

Addition. Add 5 to both sides to isolate x.

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

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

$y - y_1 = m(x - x_1)$ to $y = mx + b$. Expand point-slope and solve for $y$.

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

$x = 4$. Set $y = 0$ and solve for $x$.

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

$3x - y = -5$. Expand and rearrange to $Ax + By = C$ form.

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

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

Slopes are equal. Parallel lines have identical slopes but different intercepts.

$2$. Use slope formula: $\frac{7-3}{4-2} = \frac{4}{2} = 2$.

$m$. Coefficient of $x$ represents the rate of change.

$(0, 2)$. Set $x = 0$: $6y = 12$, so $y = 2$.

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

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

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

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

$y = -\frac{2}{3}x + 2$. Solve for $y$ by isolating it on one side.