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Question \#1: Create a system of linear equations with no solutions, one soluti...
May 27, 2024
Solution by Steps
step 1
To create a system of linear equations with no solutions, we need two lines that are parallel but not coincident. For example, consider the equations: y=2x+3 y = 2x + 3 y=2x+5 y = 2x + 5 These lines have the same slope but different y-intercepts, so they will never intersect
step 2
To create a system of linear equations with one solution, we need two lines that intersect at exactly one point. For example, consider the equations: y=2x+3 y = 2x + 3 y=−x+1 y = -x + 1 These lines have different slopes, so they will intersect at exactly one point
step 3
To create a system of linear equations with infinitely many solutions, we need two lines that are coincident. For example, consider the equations: y=2x+3 y = 2x + 3 2y=4x+6 2y = 4x + 6 These lines are actually the same line, so they will intersect at every point on the line
step 4
To verify the solutions algebraically, we solve each system of equations. For the system with no solutions: y=2x+3 y = 2x + 3 y=2x+5 y = 2x + 5 Subtracting the first equation from the second: 0=2 0 = 2 This is a contradiction, so there are no solutions
step 5
For the system with one solution: y=2x+3 y = 2x + 3 y=−x+1 y = -x + 1 Setting the equations equal to each other: 2x+3=−x+1 2x + 3 = -x + 1 3x=−2 3x = -2 x=−23 x = -\frac{2}{3} Substituting back into the first equation: y=2(−23)+3 y = 2(-\frac{2}{3}) + 3 y=−43+3 y = -\frac{4}{3} + 3 y=53 y = \frac{5}{3} So the solution is (−23,53) \left( -\frac{2}{3}, \frac{5}{3} \right)
step 6
For the system with infinitely many solutions: y=2x+3 y = 2x + 3 2y=4x+6 2y = 4x + 6 Dividing the second equation by 2: y=2x+3 y = 2x + 3 This is the same as the first equation, so there are infinitely many solutions
step 7
To verify the solutions graphically, we plot each system of equations on a graph. For the system with no solutions, the lines will be parallel and never intersect. For the system with one solution, the lines will intersect at exactly one point. For the system with infinitely many solutions, the lines will be coincident and overlap completely
Answer
No solutions: y=2x+3 y = 2x + 3 and y=2x+5 y = 2x + 5
One solution: y=2x+3 y = 2x + 3 and y=−x+1 y = -x + 1
Infinitely many solutions: y=2x+3 y = 2x + 3 and 2y=4x+6 2y = 4x + 6
Key Concept
System of Linear Equations
Explanation
A system of linear equations can have no solutions (parallel lines), one solution (intersecting lines), or infinitely many solutions (coincident lines). The number of solutions can be determined by comparing the slopes and y-intercepts of the lines.
Generated Graph
Solution by Steps
step 1
To factor the expression 64a6−2564a^6 - 25, we recognize it as a difference of squares: 64a6−25=(8a3)2−5264a^6 - 25 = (8a^3)^2 - 5^2
step 2
Using the difference of squares formula, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b), we get: (8a3−5)(8a3+5)(8a^3 - 5)(8a^3 + 5)
step 3
Therefore, the fully factored form of 64a6−2564a^6 - 25 is (8a3−5)(8a3+5)(8a^3 - 5)(8a^3 + 5)
step 4
To factor the expression 4x2y−44xy+72y4x^2y - 44xy + 72y, we first factor out the common term 4y4y: 4y(x2−11x+18)4y(x^2 - 11x + 18)
step 5
Next, we factor the quadratic x2−11x+18x^2 - 11x + 18. We look for two numbers that multiply to 1818 and add to −11-11, which are −2-2 and −9-9
step 6
Thus, x2−11x+18=(x−2)(x−9)x^2 - 11x + 18 = (x - 2)(x - 9)
step 7
Therefore, the fully factored form of 4x2y−44xy+72y4x^2y - 44xy + 72y is 4y(x−2)(x−9)4y(x - 2)(x - 9)
step 8
To factor the expression 6n2−11ny−10y26n^2 - 11ny - 10y^2, we look for two numbers that multiply to 6⋅(−10)=−606 \cdot (-10) = -60 and add to −11-11. These numbers are −15-15 and 44
step 9
We rewrite the middle term using these numbers: 6n2−15ny+4ny−10y26n^2 - 15ny + 4ny - 10y^2
step 10
We factor by grouping: 3n(2n−5y)+2y(2n−5y)3n(2n - 5y) + 2y(2n - 5y)
step 11
We factor out the common binomial factor: (3n+2y)(2n−5y)(3n + 2y)(2n - 5y)
step 12
Therefore, the fully factored form of 6n2−11ny−10y26n^2 - 11ny - 10y^2 is (3n+2y)(2n−5y)(3n + 2y)(2n - 5y)
[3] Answer
A
Key Concept
Factoring Polynomials
Explanation
Factoring polynomials involves recognizing patterns such as the difference of squares, common factors, and factoring by grouping.
Solution by Steps
step 1
Identify the vertex and x-intercepts from the given information. The vertex is at (−2.5,3.5)(-2.5, 3.5) and the x-intercepts are (−1,0)(-1, 0) and (−4,0)(-4, 0)
step 2
Use the factored form of a parabola equation: y=a(x−x1)(x−x2)y = a(x - x_1)(x - x_2), where x1x_1 and x2x_2 are the x-intercepts. Substituting the x-intercepts, we get y=a(x+1)(x+4)y = a(x + 1)(x + 4)
step 3
Substitute the vertex (−2.5,3.5)(-2.5, 3.5) into the equation to find the value of aa. 3.5=a(−2.5+1)(−2.5+4)3.5 = a(-2.5 + 1)(-2.5 + 4)
step 4
Simplify the equation: 3.5=a(−1.5)(1.5)⇒3.5=a(−2.25)⇒a=3.5−2.25⇒a=−1493.5 = a(-1.5)(1.5) \Rightarrow 3.5 = a(-2.25) \Rightarrow a = \frac{3.5}{-2.25} \Rightarrow a = -\frac{14}{9}
step 5
Substitute aa back into the factored form equation: y=−149(x+1)(x+4)y = -\frac{14}{9}(x + 1)(x + 4)
step 6
Convert the factored form to standard form by expanding: y=−149(x2+5x+4)⇒y=−149x2−709x−569y = -\frac{14}{9}(x^2 + 5x + 4) \Rightarrow y = -\frac{14}{9}x^2 - \frac{70}{9}x - \frac{56}{9}
step 7
Convert the standard form to vertex form using the vertex (−2.5,3.5)(-2.5, 3.5). The vertex form is y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex. Substituting a=−149a = -\frac{14}{9}, h=−2.5h = -2.5, and k=3.5k = 3.5, we get y=−149(x+2.5)2+3.5y = -\frac{14}{9}(x + 2.5)^2 + 3.5
Answer
The equation of the parabola in factored form is y=−149(x+1)(x+4)y = -\frac{14}{9}(x + 1)(x + 4).
The equation of the parabola in standard form is y=−149x2−709x−569y = -\frac{14}{9}x^2 - \frac{70}{9}x - \frac{56}{9}.
The equation of the parabola in vertex form is y=−149(x+2.5)2+3.5y = -\frac{14}{9}(x + 2.5)^2 + 3.5.
Key Concept
Equation of a Parabola
Explanation
The equation of a parabola can be expressed in different forms: factored form, standard form, and vertex form. Each form provides different insights into the properties of the parabola, such as its vertex, x-intercepts, and direction of opening.
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 need to rewrite the quadratic part x2−6xx^2 - 6x in a form that allows us to complete the square
step 2
We take half of the coefficient of xx, which is −6-6, divide it by 22 to get −3-3, and then square it to get 99
step 3
Add and subtract 99 inside the expression: y=x2−6x+9−9−8y = x^2 - 6x + 9 - 9 - 8
step 4
Combine the perfect square trinomial and the constants: y=(x−3)2−17y = (x - 3)^2 - 17
Part (b): y=−2(x+3)(x−7)y = -2(x + 3)(x - 7)
step 1
To convert y=−2(x+3)(x−7)y = -2(x + 3)(x - 7) to vertex form, we first expand the expression
step 2
Expand the binomials: y=−2(x2−4x−21)y = -2(x^2 - 4x - 21)
step 3
Distribute −2-2: y=−2x2+8x+42y = -2x^2 + 8x + 42
step 4
To complete the square, factor out −2-2 from the quadratic and linear terms: y=−2(x2−4x)+42y = -2(x^2 - 4x) + 42
step 5
Take half of the coefficient of xx, which is −4-4, divide it by 22 to get −2-2, and then square it to get 44
step 6
Add and subtract 44 inside the parentheses: y=−2(x2−4x+4−4)+42y = -2(x^2 - 4x + 4 - 4) + 42
step 7
Combine the perfect square trinomial and the constants: y=−2((x−2)2−4)+42y = -2((x - 2)^2 - 4) + 42
step 8
Distribute −2-2 and simplify: y=−2(x−2)2+50y = -2(x - 2)^2 + 50
Answer
The vertex form of y=x2−6x−8y = x^2 - 6x - 8 is y=(x−3)2−17y = (x - 3)^2 - 17.
The vertex form of y=−2(x+3)(x−7)y = -2(x + 3)(x - 7) is y=−2(x−2)2+50y = -2(x - 2)^2 + 50.
Key Concept
Completing the Square
Explanation
Completing the square is a method used to convert a quadratic expression into vertex form, which makes it easier to identify the vertex of the parabola.
Solution by Steps
step 1
To determine the value of kk for which the quadratic equation kx2−2kx+6kx^2 - 2kx + 6 has one xx-intercept, we need to find the condition for the quadratic to have exactly one real root. This occurs when the discriminant is zero
step 2
The discriminant Δ\Delta of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is given by Δ=b2−4ac\Delta = b^2 - 4ac. For the given equation kx2−2kx+6=0kx^2 - 2kx + 6 = 0, we have a=ka = k, b=−2kb = -2k, and c=6c = 6
step 3
Substitute aa, bb, and cc into the discriminant formula: Δ=(−2k)2−4(k)(6) \Delta = (-2k)^2 - 4(k)(6) Δ=4k2−24k \Delta = 4k^2 - 24k
step 4
Set the discriminant equal to zero to find the value of kk: 4k2−24k=0 4k^2 - 24k = 0 4k(k−6)=0 4k(k - 6) = 0
step 5
Solve for kk: 4k=0ork−6=0 4k = 0 \quad \text{or} \quad k - 6 = 0 k=0ork=6 k = 0 \quad \text{or} \quad k = 6
step 6
Since k=0k = 0 would make the equation not a quadratic, we discard k=0k = 0. Therefore, the value of kk for which the quadratic equation has one xx-intercept is k=6k = 6
Answer
The value of kk is 66.
Key Concept
Discriminant of a quadratic equation
Explanation
The discriminant of a quadratic equation determines the nature of its roots. For the quadratic to have exactly one real root (one xx-intercept), the discriminant must be zero.
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