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Question #1: Create a system of linear equations with no solutions, one solution...
May 13, 2024
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=2x + y = 2 and x+y=5x + 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(x + y) - (x + y) = 5 - 2, which simplifies to 0=30 = 3, a contradiction
step 4
To create a system with one solution, the equations must be intersecting lines
step 5
Example: x+y=2x + y = 2 and x−y=0x - y = 0. These lines intersect at a single point
step 6
To verify algebraically, solve the system: from x−y=0x - y = 0, we get x=yx = y. Substituting into the first equation, x+x=2x + x = 2, so x=1x = 1 and y=1y = 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=2x + y = 2 and 2x+2y=42x + 2y = 4. These represent the same line
step 9
To verify algebraically, divide the second equation by 2: 2x+2y=42x + 2y = 4 becomes x+y=2x + 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=2x + y = 2, x+y=5x + y = 5. One solution: x+y=2x + y = 2, x−y=0x - y = 0. Infinitely many solutions: x+y=2x + y = 2, 2x+2y=42x + 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=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
step 3
For AB: dAB=(4−1)2+(6−2)2=32+42=5d_{AB} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = 5
step 4
For BC: dBC=(7−4)2+(2−6)2=32+(−4)2=5d_{BC} = \sqrt{(7 - 4)^2 + (2 - 6)^2} = \sqrt{3^2 + (-4)^2} = 5
step 5
For AC: dAC=(7−1)2+(2−2)2=62+02=6d_{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=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}
step 7
For AB: mAB=6−24−1=43m_{AB} = \frac{6 - 2}{4 - 1} = \frac{4}{3}
step 8
For BC: mBC=2−67−4=−43m_{BC} = \frac{2 - 6}{7 - 4} = -\frac{4}{3}
step 9
For AC: mAC=2−27−1=0m_{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 dAB2+dBC2=dAC2d_{AB}^2 + d_{BC}^2 = d_{AC}^2: 52+52=625^2 + 5^2 = 6^2 which simplifies to 25+25=3625 + 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 mAB⋅mBC=−1m_{AB} \cdot m_{BC} = -1: 43⋅−43=−169\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 ∠ABC\angle ABC: cos(∠ABC)=dAB2+dBC2−dAC22⋅dAB⋅dBCcos(\angle ABC) = \frac{d_{AB}^2 + d_{BC}^2 - d_{AC}^2}{2 \cdot d_{AB} \cdot d_{BC}}
step 16
Calculate cos(∠ABC)cos(\angle ABC): cos(∠ABC)=52+52−622⋅5⋅5=25+25−3650=1450=725cos(\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 ∠ABC\angle ABC by taking the arccosine of 725\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 ∠ABC\angle ABC, we use the arccosine of the calculated cosine value
step 19
∠ABC=arccos(725)\angle ABC = arccos\left(\frac{7}{25}\right)
[question b] Answer
The internal angle ∠ABC\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: 64a6−2564a^6 - 25 can be written as (8a3)2−(5)2(8a^3)^2 - (5)^2
step 2
Factor the difference of squares: (8a3)2−(5)2=(8a3−5)(8a3+5)(8a^3)^2 - (5)^2 = (8a^3 - 5)(8a^3 + 5)
1 Answer
(8a3−5)(8a3+5)(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 4x2y−44xy+72y4x^2y - 44xy + 72y, the common factor is 4y4y
step 2
Factor the quadratic: x2−11x+18x^2 - 11x + 18 factors into (x−2)(x−9)(x - 2)(x - 9)
step 3
Combine the factored quadratic with the common factor: 4y(x−2)(x−9)4y(x - 2)(x - 9)
2 Answer
4y(x−2)(x−9)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 nn: 6n2−11ny−10y26n^2 - 11ny - 10y^2
step 2
Find two numbers that multiply to 6×(−10y2)=−60y26 \times (-10y^2) = -60y^2 and add to −11y-11y
step 3
The numbers are 2y2y and −30y-30y. Rewrite the middle term: 6n2+2ny−30ny−10y26n^2 + 2ny - 30ny - 10y^2
step 4
Factor by grouping: (6n2+2ny)+(−30ny−10y2)(6n^2 + 2ny) + (-30ny - 10y^2)
step 5
Factor out the common factors: 2ny(3n+1)−10y(3n+1)2ny(3n + 1) - 10y(3n + 1)
step 6
Factor out the common binomial: (3n+1)(2ny−10y)(3n + 1)(2ny - 10y)
step 7
Factor out the common factor from the second term: (3n+1)(2n−5)y(3n + 1)(2n - 5)y
3 Answer
(3n+1)(2n−5)y(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 n2n^2 and the constant term, and add to the coefficient of the middle term, then using grouping to factor completely.
Solution by Steps
step 1
To find the equation of the parabola, we use the given vertex (−2,3)(-2, 3) and x-intercepts (−4,0)(-4, 0) and (0,0)(0, 0)
step 2
The factored form of the parabola is y=a(x+4)(x−0)y = a(x + 4)(x - 0)
step 3
To find the value of aa, we use the vertex. Plugging in the vertex coordinates into the equation, we get 3=a(−2+4)(−2−0)3 = a(-2 + 4)(-2 - 0)
step 4
Solving for aa gives 3=2a⋅−23 = 2a \cdot -2, which simplifies to a=−34a = -\frac{3}{4}
step 5
The equation of the parabola in factored form is y=−34(x+4)xy = -\frac{3}{4}(x + 4)x
step 6
To convert the equation to standard form, we expand the factored form to get y=−34x2−3xy = -\frac{3}{4}x^2 - 3x
step 7
To convert the equation to vertex form, we complete the square. The vertex form is y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex
step 8
We rewrite the standard form as y=−34(x2+4x)+3y = -\frac{3}{4}(x^2 + 4x) + 3 and complete the square inside the parentheses
step 9
Adding and subtracting (4/2)2=4(4/2)^2 = 4 inside the parentheses, we get y=−34(x2+4x+4−4)+3y = -\frac{3}{4}(x^2 + 4x + 4 - 4) + 3
step 10
Simplifying, we have y=−34(x+2)2+3+34⋅4y = -\frac{3}{4}(x + 2)^2 + 3 + \frac{3}{4} \cdot 4
step 11
The final vertex form of the equation is y=−34(x+2)2+6y = -\frac{3}{4}(x + 2)^2 + 6
[question number] Answer
The equation of the parabola in factored form is y=−34(x+4)xy = -\frac{3}{4}(x + 4)x, in standard form is y=−34x2−3xy = -\frac{3}{4}x^2 - 3x, and in vertex form is y=−34(x+2)2+6y = -\frac{3}{4}(x + 2)^2 + 6.
Key Concept
Finding the equation of a parabola using vertex and x-intercepts
Explanation
The equation of a parabola can be determined using its vertex and x-intercepts by first writing it in factored form, then finding the value of the leading coefficient using the vertex, and finally converting it to standard and vertex forms.
Write each of the following quadratic expressions in vertex form using the most appropriate strategy. In your reflection, be sure to reflect on and explain why your strategy is better than other possible ways of solving this problem. y = x2 – 6x – 8 y = -2(x + 3)(x – 7)
Generated Graph
Solution by Steps
step 1
To complete the square for the quadratic expression y=x2−6x−8y = x^2 - 6x - 8, we first identify the coefficient of xx, which is −6-6
step 2
We take half of the coefficient of xx, which is −3-3, and square it, obtaining 99
step 3
We add and subtract this square inside the expression to maintain equality, getting y=(x2−6x+9)−9−8y = (x^2 - 6x + 9) - 9 - 8
step 4
We factor the perfect square trinomial and simplify the constants, resulting in y=(x−3)2−17y = (x - 3)^2 - 17
[question 1] Answer
y=(x−3)2−17y = (x - 3)^2 - 17
Key Concept
Completing the square for a quadratic expression
Explanation
Completing the square involves creating a perfect square trinomial from a quadratic expression to find its vertex form.
Solution by Steps
step 1
To complete the square for the quadratic expression y=−2(x2+6x−21)y = -2(x^2 + 6x - 21), we first factor out the coefficient of the quadratic term, which is −2-2
step 2
We then identify the coefficient of xx within the parentheses, which is 66
step 3
Taking half of this coefficient gives us 33, and squaring it yields 99
step 4
We add and subtract 99 inside the parentheses and factor the resulting perfect square trinomial, getting y=−2[(x+3)2−9−21]y = -2[(x + 3)^2 - 9 - 21]
step 5
We distribute the −2-2 and combine the constants to obtain the vertex form, resulting in y=−2(x+3)2+60y = -2(x + 3)^2 + 60
[question 2] Answer
y=−2(x+3)2+60y = -2(x + 3)^2 + 60
Key Concept
Completing the square for a quadratic expression with a leading coefficient
Explanation
When completing the square for a quadratic expression with a leading coefficient other than 1, we factor out the leading coefficient before creating the perfect square trinomial.
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