CALCU-Question #1: Create a system of linear equat

Question #1: Create a system of linear equations with no solutions, one solution, and infinitely many solutions. Explain how you determine the number of solutions without solving them. Then, verify the solutions by solving them, both algebraically and graphically (by free-hand).

Solution by Steps

step 1

To create a system with no solutions, the equations must be parallel and have different y-intercepts

step 2

Example:

x + y = 2

and

x + y = 5

. These lines are parallel and will never intersect

step 3

To verify algebraically, subtract the first equation from the second:

(x + y) - (x + y) = 5 - 2

, which simplifies to

0 = 3

, a contradiction

step 4

To create a system with one solution, the equations must be intersecting lines

step 5

Example:

x + y = 2

and

x - y = 0

. These lines intersect at a single point

step 6

To verify algebraically, solve the system: from

x - y = 0

, we get

x = y

. Substituting into the first equation,

x + x = 2

, so

x = 1

and

y = 1

step 7

To create a system with infinitely many solutions, the equations must be the same line or multiples of each other

step 8

Example:

x + y = 2

and

2x + 2y = 4

. These represent the same line

step 9

To verify algebraically, divide the second equation by 2:

2x + 2y = 4

becomes

x + y = 2

, which is the same as the first equation

step 10

To verify graphically, plot the equations on a coordinate plane. For no solutions, the lines will be parallel. For one solution, the lines will intersect at a point. For infinitely many solutions, the lines will coincide

[question 1] Answer

No solutions:

x + y = 2

,

x + y = 5

. One solution:

x + y = 2

,

x - y = 0

. Infinitely many solutions:

x + y = 2

,

2x + 2y = 4

.

Key Concept

Determining the number of solutions in a system of linear equations without solving

Explanation

Systems with no solutions have parallel lines with different y-intercepts. Systems with one solution have intersecting lines. Systems with infinitely many solutions have coinciding lines or are multiples of each other.

Solution by Steps

step 1

Choose three non-collinear points that do not form vertical or horizontal lines. For example, let's take points A(1,2), B(4,6), and C(7,2)

step 2

Calculate the distances between the points using the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

step 3

For AB:

d_{AB} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = 5

step 4

For BC:

d_{BC} = \sqrt{(7 - 4)^2 + (2 - 6)^2} = \sqrt{3^2 + (-4)^2} = 5

step 5

For AC:

d_{AC} = \sqrt{(7 - 1)^2 + (2 - 2)^2} = \sqrt{6^2 + 0^2} = 6

step 6

Calculate the slopes of the sides using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

step 7

For AB:

m_{AB} = \frac{6 - 2}{4 - 1} = \frac{4}{3}

step 8

For BC:

m_{BC} = \frac{2 - 6}{7 - 4} = -\frac{4}{3}

step 9

For AC:

m_{AC} = \frac{2 - 2}{7 - 1} = 0

step 10

Determine if the triangle is right-angled by checking if the Pythagorean theorem holds for the lengths of the sides

step 11

Check if

d_{AB}^2 + d_{BC}^2 = d_{AC}^2

:

5^2 + 5^2 = 6^2

which simplifies to

25 + 25 = 36

, which is not true. Thus, it is not a right-angled triangle by side lengths

step 12

Determine if the triangle is right-angled by checking if the product of the slopes of two perpendicular lines is -1

step 13

Check if

m_{AB} \cdot m_{BC} = -1

:

\frac{4}{3} \cdot -\frac{4}{3} = -\frac{16}{9}

, which is not -1. Thus, it is not a right-angled triangle by slopes

step 14

Use trigonometry to find an internal angle. We can use the law of cosines or the arccosine of the dot product of two vectors

step 15

Using the law of cosines to find angle

\angle ABC

:

cos(\angle ABC) = \frac{d_{AB}^2 + d_{BC}^2 - d_{AC}^2}{2 \cdot d_{AB} \cdot d_{BC}}

step 16

Calculate

cos(\angle ABC)

:

cos(\angle ABC) = \frac{5^2 + 5^2 - 6^2}{2 \cdot 5 \cdot 5} = \frac{25 + 25 - 36}{50} = \frac{14}{50} = \frac{7}{25}

step 17

Find

\angle ABC

by taking the arccosine of

\frac{7}{25}

[question a] Answer

The triangle formed by points A(1,2), B(4,6), and C(7,2) is not a right-angled triangle as determined by both the Pythagorean theorem and the slopes of the sides.

Key Concept

Determining a right-angled triangle using side lengths and slopes

Explanation

A triangle is right-angled if the square of the length of one side is equal to the sum of the squares of the lengths of the other two sides (Pythagorean theorem), or if the product of the slopes of two sides is -1 (perpendicular slopes).

Solution by Steps

step 18

To find the internal angle

\angle ABC

, we use the arccosine of the calculated cosine value

step 19

\angle ABC = arccos\left(\frac{7}{25}\right)

[question b] Answer

The internal angle

\angle ABC

can be found using the law of cosines and the arccosine function.

Key Concept

Finding an internal angle using the law of cosines

Explanation

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles, allowing us to find that angle using the arccosine function.

Fully factor the following expressions, where possible, by selecting the appropriate strategy. 64a6 – 25 4x2y – 44xy + 72y 6n2 – 11ny – 10y2

Generated Graph

Solution by Steps

step 1

Identify the expression as a difference of squares:

64a^6 - 25

can be written as

(8a^3)^2 - (5)^2

step 2

Factor the difference of squares:

(8a^3)^2 - (5)^2 = (8a^3 - 5)(8a^3 + 5)

1 Answer

(8a^3 - 5)(8a^3 + 5)

Key Concept

Factoring a difference of squares

Explanation

The expression is a difference of two squares, which factors into the product of the sum and difference of the square roots of the individual terms.

step 1

Factor out the common factor: In

4x^2y - 44xy + 72y

, the common factor is

4y

step 2

Factor the quadratic:

x^2 - 11x + 18

factors into

(x - 2)(x - 9)

step 3

Combine the factored quadratic with the common factor:

4y(x - 2)(x - 9)

2 Answer

4y(x - 2)(x - 9)

Key Concept

Factoring by grouping

Explanation

First, factor out the common factor from all terms, then factor the resulting quadratic expression.

step 1

Recognize the trinomial as a quadratic in

n

:

6n^2 - 11ny - 10y^2

step 2

Find two numbers that multiply to

6 \times (-10y^2) = -60y^2

and add to

-11y

step 3

The numbers are

2y

and

-30y

. Rewrite the middle term:

6n^2 + 2ny - 30ny - 10y^2

step 4

Factor by grouping:

(6n^2 + 2ny) + (-30ny - 10y^2)

step 5

Factor out the common factors:

2ny(3n + 1) - 10y(3n + 1)

step 6

Factor out the common binomial:

(3n + 1)(2ny - 10y)

step 7

Factor out the common factor from the second term:

(3n + 1)(2n - 5)y

3 Answer

(3n + 1)(2n - 5)y

Key Concept

Factoring a quadratic trinomial

Explanation

The trinomial is factored by finding two numbers that multiply to the product of the coefficient of

n^2

and the constant term, and add to the coefficient of the middle term, then using grouping to factor completely.

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