Grade 10 Math Review Notes

Solving Linear Systems

• Solving a linear system involves determining the values for the variables that satisfy all equations within the system simultaneously. This typically means finding the point where the lines intersect, indicating a common solution that makes all equations true.

• Methods:

  • Graphing: Visualize both lines on a coordinate plane and identify their point of intersection. This method is effective for simple systems but may lack precision for non-integer solutions.

  • Substitution: Solve one equation for one variable in terms of the other, then substitute this expression into the other equation. This reduces the problem to a single-variable equation.

  • Elimination: Manipulate the equations (typically by multiplying by constants) to make the coefficients of one variable identical or opposite. Adding or subtracting the equations can then eliminate that variable.

Graphing Method

1. Rearrange equations into the slope-intercept form, $y = mx + b$, where $m$ represents the slope and $b$ is the y-intercept.

   • $x - y = 5$ transforms to $\Rightarrow -y = -x + 5 \Rightarrow y = x - 5$. The line has a slope $m = 1$ and a y-intercept $b = -5$.

   • $3x + y = 3$ becomes $\Rightarrow y = -3x + 3$. The line has a slope $m = -3$ and a y-intercept $b = 3$.

2. Plot the y-intercept on the y-axis. Use the slope (rise over run) to graphically find additional points on each line. For instance, a slope of 1 (or 1/1) means for every one unit you move to the right, move one unit up.

3. Draw a straight line through the plotted points to represent each equation on the graph.

4. Locate the point where the lines intersect. The coordinates $(x, y)$ of this point provide the solution to the system.

   • In the given example, the lines intersect at the point $(2, -3)$, indicating that the solution is $x = 2$ and $y = -3$.

Elimination Method

1. Organize the equations in the standard form $Ax + By = C$, where $A$, $B$, and $C$ are constants.

2. Adjust the equations by multiplying one or both by suitable constants so that the coefficients of either $x$ or $y$ are numerically equal (but possibly opposite in sign).

3. Combine the equations by either adding or subtracting them to eliminate one variable.

   • If the coefficients of the variable to be eliminated have opposite signs, add the equations.

   • If the coefficients are the same, subtract the equations.

4. After eliminating one variable, solve the resulting equation for the remaining variable.

5. Once you've found the value of one variable, substitute it back into either of the original equations to find the value of the other variable.

• Example:

  • $x - y = 5$

  • $3x + y = 3$

  • Adding the equations eliminates $y$: $4x = 8 \Rightarrow x = 2$.

  • Substitute $x = 2$ into $x - y = 5$: $2 - y = 5 \Rightarrow y = -3$.

  • The solution to the system is $x = 2$ and $y = -3$.

Substitution Method

1. From one of the equations, express one variable in terms of the other. For example, in $x - y = 5$, isolate $x$: $x = 5 + y$.

2. Replace the isolated variable in the other equation with the expression found in step 1. This will yield an equation in only one variable.

   • Substituting $x = 5 + y$ into $3x + y = 3$ gives: $3(5 + y) + y = 3$.

3. Solve the single-variable equation resulting from the substitution.

   • $15 + 3y + y = 3 \Rightarrow 4y = -12 \Rightarrow y = -3$.

4. Substitute the value of the variable found in step 3 back into the rearranged equation from step 1 to solve for the other variable.

   • With $y = -3$, then $x = 5 + (-3) \Rightarrow x = 2$.

   • Therefore, the solution is $x = 2$ and $y = -3$.

Harder Linear Systems: Elimination

• Goal: Transform the given equations by multiplication so that the coefficients of one variable have the same absolute value, which facilitates elimination.

• Determine if the equations should be added or subtracted to eliminate a variable. This depends on whether the matching coefficients have opposite or identical signs.

• Example:

  • $4x + 3y = 13$

  • $5x - 4y = -7$

  • Multiply the first equation by 5: $20x + 15y = 65$.

  • Multiply the second equation by 4: $20x - 16y = -28$.

  • Subtract the second equation from the first: $31y = 93 \Rightarrow y = 3$.

  • Substitute $y = 3$ into $4x + 3y = 13$: $4x + 9 = 13 \Rightarrow x = 1$.

  • Thus, the solution is $x = 1$ and $y = 3$.

Application of Solving Linear Systems

• Define variables to represent the unknowns in the problem. For instance, let $x$ be the number of Adidas shoes and $y$ be the number of Air Jensen shoes.

• Develop two equations that relate these variables based on the information provided in the problem.

• Use one of the methods (graphing, substitution, or elimination) to solve the system of equations.

• Example:

  • Let $x$ = number of Adidas shoes and $y$ = number of Air Jensen shoes.

  • Equation 1 (total pairs): $x + y = 75$

  • Equation 2 (total sales): $82x + 95y = 6241$

  • Multiply equation 1 by 82: $82x + 82y = 6150$

  • Subtract the modified equation 1 from equation 2: $13y = 91 \Rightarrow y = 7$

  • Substitute $y = 7$ into $x + y = 75$: $x = 68$

  • Conclusion: There are 68 pairs of Adidas shoes and 7 pairs of Air Jensen shoes.

Midpoint and Distance of a Line Segment

• Midpoint Formula: The midpoint between two points $(x<em>1, y</em>1)$ and $(x<em>2, y</em>2)$ is calculated as $\left(\frac{x<em>1 + x</em>2}{2}, \frac{y<em>1 + y</em>2}{2}\right)$.

• Distance Formula: Derived from the Pythagorean theorem, the distance $d$ between two points $(x<em>1, y</em>1)$ and $(x<em>2, y</em>2)$ is given by $d = \sqrt{(x<em>2 - x</em>1)^2 + (y<em>2 - y</em>1)^2}$.

• Example:

  • Given points A(5, -3) and B(-1, 5):

    • Midpoint: $\left(\frac{5 + (-1)}{2}, \frac{-3 + 5}{2}\right) = (2, 1)$

    • Distance: $\sqrt{(-1 - 5)^2 + (5 - (-3))^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Median of a Triangle

• A median of a triangle is a line segment connecting a vertex to the midpoint of the opposite side, dividing that side into two equal parts.

• Steps to find the equation of a median:

  1. Begin by determining the midpoint of the side opposite the vertex from which the median is drawn, applying the midpoint formula.

  2. Calculate the slope of the median using the coordinates of the original vertex and the computed midpoint: $m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}$.

  3. Use the slope-intercept form $(y = mx + b)$ and substitute either the coordinates of the vertex or the midpoint to solve for the y-intercept $b$.

• Example:

  • In a triangle with vertices A(-1, 3), B(1, 5), and C(3, 1), find the equation of the median from vertex C to side AB.

    • The midpoint of AB is $\left(\frac{-1 + 1}{2}, \frac{3 + 5}{2}\right) = (0, 4)$.

    • The slope of the median from C to AB is $\frac{1 - 4}{3 - 0} = -1$.

    • Using the point (0, 4) in the equation $y = mx + b$: $4 = -1(0) + b \Rightarrow b = 4$.

    • Therefore, the equation of the median is $y = -x + 4$.

Right Bisector (Perpendicular Bisector)

• A right bisector is a line that intersects a line segment at its midpoint and is perpendicular to it.

• Steps to determine the equation of a right bisector:

  1. Find the slope of the original line segment using the slope formula: $m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}$.

  2. Calculate the slope of the right bisector. This slope will be the negative reciprocal of the slope of the original line segment.

  3. Find the midpoint of the original line segment using the midpoint formula. This point lies on the right bisector.

  4. Use the slope-intercept form of a line $(y = mx + b)$. Substitute the coordinates of the midpoint for $x$ and $y$, and the slope of the right bisector for $m$, then solve for $b$ to find the y-intercept.

• Example:

  • Given end points A(-2, -4) and B(8, 6):

    • The slope of $AB$ is $\frac{6 - (-4)}{8 - (-2)} = 1$.

    • The slope of the right bisector is $-1$.

    • The midpoint of $AB$ is $\left(\frac{-2 + 8}{2}, \frac{-4 + 6}{2}\right) = (3, 1)$.

    • Using the point (3, 1) in $y = mx + b$: $1 = -1(3) + b \Rightarrow b = 4$.

    • The equation of the right bisector is $y = -x + 4$.

Classifying Triangles

• Scalene: All three sides are of different lengths, and consequently, all angles are different.

• Isosceles: Two sides are equal in length, resulting in the angles opposite these sides also being equal.

• Equilateral: All three sides are of equal length, and each angle measures 60 degrees.

• Right Triangle: One angle is exactly 90 degrees, satisfying the Pythagorean theorem ($a^2 + b^2 = c^2$), where $c$ represents the length of the hypotenuse.

• To classify a triangle:

  1. Calculate the lengths of all three sides using the distance formula.

  2. Compare the lengths of the sides to determine if the triangle is scalene, isosceles, or equilateral.

  3. Apply the Pythagorean theorem to verify if the triangle is a right triangle.

• Example:

  • Consider a triangle with vertices D(-4, -2), E(-2, 6), and F(4, 4):

    • The length of side DE is $\sqrt{(-2 - (-4))^2 + (6 - (-2))^2} = \sqrt{68}$.

    • The length of side EF is $\sqrt{(4 - (-2))^2 + (4 - 6)^2} = \sqrt{40}$.

    • The length of side DF is $\sqrt{(4 - (-4))^2 + (4 - (-2))^2} = \sqrt{100} = 10$.

    • Since all sides have different lengths, the triangle is scalene.

    • To check if it’s a right triangle: Is $\sqrt{68}^2 + \sqrt{40}^2 = 10^2$? No. Therefore, it is not a right triangle.

Equation of a Circle

• The standard equation of a circle centered at the origin is represented as $x^2 + y^2 = r^2$, where $r$ denotes the radius of the circle.

• Example:

  • For the circle described by the equation $x^2 + y^2 = 36$, what is its radius?

    • Since $r^2 = 36$, then $r = 6$.

Finding the Equation of a Circle

• If the circle's center is at the origin, its equation takes the form $x^2 + y^2 = r^2$.

• Substitute the coordinates of a known point $(x, y)$ on the circle into the equation to determine the value of $r^2$.

• Express the complete equation of the circle by including the calculated value of $r^2$.

• Determining a Point’s Position Relative to a Circle:

  • Insert the coordinates of the point into the circle’s equation.

    • If the result is less than $r^2$, the point is located inside the circle.

    • If the result equals $r^2$, the point lies directly on the circle.

    • If the result is greater than $r^2$, the point resides outside the circle.

• Example:

  • Consider a circle that is centered at the origin and passes through the point (-3, 4).

    • By substituting the point into the equation, we find $(−3)^2 + 4^2 = r^2 \Rightarrow r^2 = 25$, leading to the equation $x^2 + y^2 = 25$.

    • To check the position of the point (5, 2): $5^2 + 2^2 = 29$, which is more than 25. Hence, this point is outside the circle.

Shortest Distance from a Point to a Line

1. Determine the equation of the line that is perpendicular to the given line and also passes through the specified point.

2. Calculate the point at which the original line and the derived perpendicular line intersect.

3. Employ the distance formula to compute the distance between the given point and the point of intersection.

• Example:

  • Calculate the shortest distance from the point (-3, 5) to the line $y = \frac{1}{4}x + 10$.

    • The slope of a line perpendicular to the given line is -4 (the negative reciprocal of 1/4).

    • Using the point-slope form to find the equation of the perpendicular line: Applying $y = mx + b$, $5 = -4(-3) + b \Rightarrow b = -7$. Therefore, the equation is $y = -4x - 7$.

    • To find the intersection point:

      • Set the two line equations equal to each other: $\frac{1}{4}x + 10 = -4x - 7$

      • Simplify: $x + 40 = -16x - 28 \Rightarrow 17x = -68 \Rightarrow x = -4$

      • Substitute $x = -4$ into one of the line equations: $y = -4(-4) - 7 \Rightarrow y = 9$.

      • The point of intersection is (-4, 9).

    • Calculate the shortest distance:

      • Apply the distance formula: $\sqrt{(-4 - (-3))^2 + (9 - 5)^2} = \sqrt{(-1)^2 + (4)^2} = \sqrt{17}$.

Characteristics of Quadratic Functions

• Standard Form: Expressed as $y = ax^2 + bx + c$.

  • The constant term $c$ indicates the y-intercept, the point where the parabola crosses the y-axis.

  • The coefficient $a$ determines the parabola's opening direction (positive for upwards, negative for downwards).

• Vertex Form: Expressed as $y = a(x - h)^2 + k$.

  • The vertex of the parabola is given by the point $(h, k)$, which is the maximum or minimum point of the parabola.

• Factored Form: Expressed as $y = a(x - r)(x - s)$.

  • The roots or x-intercepts of the parabola are $r$ and $s$.

• The value of $a$ in all forms affects the parabola by controlling its direction and the degree of vertical stretch or compression.

Working With Vertex Form

Completing the Table of Information

• The vertex is located at the point $(h, k)$.

• The axis of symmetry is a vertical line that divides the parabola into two symmetrical halves. This line passes through the vertex, with the equation $x = h$.

• The vertical stretch or compression is determined by the absolute value of $a$. If $|a| > 1$, the parabola is stretched vertically. If $0 < |a| < 1$, the parabola is compressed vertically.

• The direction of opening is dictated by the sign of $a$: positive indicates upward opening, negative indicates downward opening.

Graphing the Equation

• Create a table of values with the vertex at the center. Choose points on either side of the vertex to map the shape of the parabola accurately.

Example 11

• Vertex Form: $y = -2(x + 4)^2 + 3$

• Vertex: (-4, 3)

• Axis of Symmetry: $x = -4$

• Stretch/Compression: Vertical stretch by a factor of 2.

• Direction of Opening: Down

• Domain: All real numbers

• Range: $y \leq 3$

Finding the Vertex Form Equation From a Graph

• The vertex form is given by $y = a(x - h)^2 + k$

• To define the equation, you need the vertex and one additional point $(x, y)$ from the graph.

Example 12

• Vertex: (2, -6)

• Y intercept: (0, -4)

• Substitute these values into the vertex form equation: $y=-4$, $x=0$, $h=2$, $k=-6$

• Calculate $a$: $-4 = a (0 - 2)^2 - 6 \Rightarrow a = \frac{1}{2}$

• Therefore, the equation is: $y = \frac{1}{2}(x - 2)^2 - 6$

Describing Transformations in Words

• Begin by identifying the vertex in the form $y = a(x - h)^2 + k$; in this case, (h, k) = (negative five, negative four).

• Recognize that the vertex of the base function $x^2$ is at the origin, (0, 0).

• Transformations from $x^2$:

  • Horizontal shift left by 5 units.

  • Vertical shift down by 4 units.

  • Vertical compression by a factor of 1/3.

  • Vertical reflection (opens down).

Quadratic in Factored Form

• Represented as $y = a(x - r)(x - s)$

• The x-intercepts are located at $r$ and $s$.

• The x-coordinate of the vertex is situated midway between the x-intercepts. It is calculated as the average of $r$ and $s$: $\frac{r+s}{2}$

Example 14

• $y = -4(x - 2)(x - 4)$

• x-intercepts: 2, 4

• The x-coordinate of the vertex = (2 + 4) / 2 = 3

• To find the y-coordinate of the vertex, substitute x = 3 into the equation: $y = -4(3 - 2)(3 - 4) \Rightarrow y = 4$. Thus, the vertex is (3, 4).

• Axis of symmetry: $x = 3$

Example 15

• Given x-intercepts of -8 and 2, and the curve passes through point (0, -8).

• Substitute the intercepts and the point into the equation form: $y = a(x - r)(x - s)$. Then, solve for $a$.

• $-8 = a(0 + 8)(0 - 2) \Rightarrow a = \frac{1}{2}$

• The complete equation is: $y = \frac{1}{2}(x + 8)(x - 2)$

Factoring Quadratics

• First, determine if there is a common factor that can be factored out from all terms.

• Product and Sum Factoring (Short Method): This is efficient when the coefficient of $x^2$ is 1.

  1. Identify two numbers that multiply to give the constant term (c) and add up to the coefficient of the $x$ term (b).

  2. Express the factors using these numbers: (x + number 1)(x + number 2).

• Long Factoring Method:

  1. Find two numbers such that their product equals $a \times c$ and their sum equals $b$.

  2. Rewrite the $bx$ term using these two numbers to split the middle term into two parts.

  3. Factor by grouping to simplify the expression.

Example 16

• Factoring Quadratics

a) $x^2 + 10x + 24$

• Product = 24, Sum = 10: The numbers are 6 and 4

• Factored form: (x + 6)(x + 4)

b) $z^2 + x - 12$

• Product = -12, Sum = 1: The numbers are 4 and -3

• Factored form: (x + 4)(x - 3)

c) $2x^2 + 22x + 48$

• Factor out the 2: 2($x^2 + 11x + 24$)

• Focus on the quadratic inside the brackets.

• The numbers are 8 and 3.

• Final Factored Form: 2(x + 8)(x + 3)

d) $2x^2 + 7x - 15$

• Product = 2 × -15 = -30, Sum = 7: The numbers are 10 and -3

• Split the middle term: $2x^2 + 10x - 3x - 15$

• Factor by grouping: 2x(x + 5) - 3(x + 5)

• Factored form: (x + 5)(2x - 3)

e) $3x^2 + 23x - 8$

• Need a product of $a \times c$, which is 3 × -8 = -24, and a sum equal to b, which is 23.

• The numbers are 24 and -1.

• Split the middle term into 24x minus 1x.

• Factor out a common factor: 3x(x + 8) - 1(x + 8)

• Final Factored equation: (x + 8)(3x - 1)

f) $x^2 - 16$

• Recognize as a Difference of Squares

• Factored form: (x - 4)(x + 4)

g) $4x^2 - 25$

• Difference of Squares: $(2x)^2 - (5)^2$

• Factored form: (2x - 5)(2x + 5)

h) $x^2 +6x + 9$

• Coefficient is one, look for numbers with a product of nine and a sum of six. The numbers are three and three.

• Factored form: $(x + 3)^2$

Expanding Quadratics Into Standard Form

• Use the double distributive property (FOIL method):

  • Multiply the First terms, Outer terms, Inner terms, and Last terms.

• Example:

a) (4x - 1)(x + 7)

• Expand: $4x^2 + 28x - x - 7$

• Collect like terms: $4x^2 + 27x - 7$

b) $(x - 5)^2$ Remember to expand completely; it’s not just $(x)^2 - (5)^2$

• Expand: (x - 5)(x - 5) = $x^2 - 5x -5x+25$

• Collect like terms: $x^2 - 10x + 25$

Standard Form into Vertex Form

• Steps to convert from standard form to vertex form:

  • Start by enclosing the first two terms in brackets.

  • Add and subtract a specific number inside these brackets. This number is determined by halving the coefficient of the $x$ term and then squaring it.

  • This process creates a perfect square trinomial, which can then be easily factored into vertex form.

Example

a) $x^2 + 6x + 11$

1. Put the first two terms in brackets: $(x^2 + 6x) + 11$

2. Add and subtract the required number: $(x^2 + 6x + 9 - 9 )+ 11$ (Here, (6/2)^2 = 9).

3. Regroup and simplify to achieve vertex form.

Solving Quadratics

• Set the quadratic equation equal to zero.

• Factor the quadratic expression.

• Set each factor equal to zero and solve for $x$.

• If the quadratic expression cannot be factored, use the quadratic formula.

• Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Discriminant: $b^2 - 4ac$

  • If $b^2 - 4ac > 0$, there are two distinct real solutions.

  • If $b^2 - 4ac = 0$, there is exactly one real solution (a repeated root).

  • If $b^2 - 4ac < 0$, there are no real solutions, indicating complex roots.
    Factor example:
    $x^2 = 36$
    This can be rearranged and factored into $(x-6)(x+6) = 0$

a) $x^2 = 36$

• Rearrange: $x^2 - 36 = 0$.

• Factor: (x - 6)(x + 6) = 0

• Solve: $x = 6, x = -6$

e) Consider an unfactorable example where you need to apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-7 \pm \sqrt{7^2 - 4<em>1</em>5}}{2}$
$x = \frac{-7 \pm \sqrt{49 - 20}}{2}$
$x = \frac{-7 \pm \sqrt{29}}{2}$
$x = \frac{-7 + \sqrt{29}}{2}; \frac{-7 - \sqrt{29}}{2}$
Approximate values:
The solutions are approximately -0.807 and -6.193.

f) This time, apply the quadratic equation where factoring is not possible:

Given the equation, if simplifying leads to $16 - 56$, we get a negative answer (negative 40).
Since you cannot square root a negative number within real numbers, it leads to no real solutions.

Sketching And labeling

• Find the y-intercept by setting $x = 0$ in the equation.

• Find the x-intercept(s) by setting $y = 0$ and solving for $x$.

• The vertex's x-coordinate can often be found using $x = -b/(2a)$. Substitute this value back into the original equation to find the y-coordinate of the vertex.

Application Of Quadratics

Consider an object launched upwards, where its height in feet is described by a quadratic equation.

• To find when the object lands on the ground, set the height equation to zero and solve for time.

• To determine the maximum height of the object, find the vertex of the quadratic equation. The y-coordinate of the vertex represents the maximum height achieved by the object.

The Law Triangle

• Right-angled triangles have equivalent ratios between their sides, which are used in trigonometry.

• The mnemonic SOHCAHTOA is used to remember these ratios specifically for right-angled triangles.

SOHCAHTOA

• Used exclusively for right-angled triangles to relate angles and sides.

• Recall the equations: Sine of an angle equals Opposite side over Hypotenuse; Cosine of an angle equals Adjacent side over Hypotenuse; and Tangent of an angle equals Opposite side over Adjacent.

Two laws that work for calculating sides and angles of non-right-angled triangles (oblique triangles):

• The Sine Law and Cosine Law are used in oblique triangles to find unknown sides or angles.

• While Sine Law and Cosine Law could also be applied to right triangles, SOHCAHTOA is generally quicker and easier for right triangles.

• The choice between Sine Law and Cosine Law depends on what information is given and what needs to be found. SOHCAHTOA is not applicable for non-right triangles.
  Given two angles and one side of a triangle, the Sine Law is used to find the other sides.
  The Sine Law states that the ratio of the length of a side to the sine of its opposite angle is constant for all sides and angles in a triangle.