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### 1. If the lines [tex]\(3x + 2ky = 2\)[/tex] and [tex]\(2x + 5y + 1 = 0\)[/tex] are parallel, find the value of [tex]\(k\)[/tex].
Lines are parallel if the ratios of their corresponding coefficients are equal.
For the lines [tex]\(Ax + By = C\)[/tex] and [tex]\(Dx + Ey = F\)[/tex] to be parallel, [tex]\(\frac{A}{D} = \frac{B}{E}\)[/tex].
Given lines:
- [tex]\(3x + 2ky = 2\)[/tex]
- [tex]\(2x + 5y + 1 = 0\)[/tex] or equivalently [tex]\(2x + 5y = -1\)[/tex]
Comparing coefficients, we have:
[tex]\[ \frac{3}{2} = \frac{2k}{5} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{3 \cdot 5}{2 \cdot 2} = 3.75 \][/tex]
So, the value of [tex]\(k\)[/tex] is [tex]\(3.75\)[/tex].
### 2. Find the coordinate where the line [tex]\(x - y = 8\)[/tex] intersects the y-axis.
To find the y-intercept of the line, set [tex]\(x = 0\)[/tex]:
[tex]\[ 0 - y = 8 \implies y = -8 \][/tex]
The coordinate where the line intersects the y-axis is [tex]\((0, -8)\)[/tex].
### 3. How many solutions does the pair of equations [tex]\(x + 2y = 3\)[/tex] and [tex]\(0.5x + y - 1.5 = 0\)[/tex] have?
First, rewrite the second equation in standard form:
[tex]\[ 0.5x + y = 1.5 \][/tex]
Multiply both sides by 2 to clear the fraction:
[tex]\[ x + 2y = 3 \][/tex]
Both equations are now identical, meaning every point on one line is also on the other. Hence, there are infinitely many solutions.
### 4. If [tex]\(x + 1\)[/tex] is a factor of [tex]\(2x^3 + ax^2 + 2bx + 1\)[/tex], then find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] given that [tex]\(2a - 3b = 4\)[/tex].
Using the factor theorem, since [tex]\(x + 1\)[/tex] is a factor, substituting [tex]\(x = -1\)[/tex] into the polynomial should yield 0:
[tex]\[ 2(-1)^3 + a(-1)^2 + 2b(-1) + 1 = 0 \implies -2 + a - 2b + 1 = 0 \implies a - 2b - 1 = 0 \][/tex]
Thus, we have the equations:
1. [tex]\(a - 2b - 1 = 0\)[/tex]
2. [tex]\(2a - 3b = 4\)[/tex]
Solving these simultaneously yields:
[tex]\[ a = 0.25 \][/tex]
[tex]\[ b = 3.75 \][/tex]
### 5. Find the values of [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] if the system of equations has an infinite number of solutions:
[tex]\[ 2x + 3y = 7 \][/tex]
[tex]\[ 2\alpha x + (\alpha + \beta)y = 28 \][/tex]
For infinite solutions, the ratios of the coefficients must be equal:
[tex]\[ \frac{2}{2\alpha} = \frac{3}{\alpha + \beta} = \frac{7}{28} \][/tex]
Simplifying, we get:
[tex]\[ \frac{2}{2\alpha} = \frac{1}{4} \implies \alpha = \frac{1}{4} \][/tex]
[tex]\[ \frac{3}{\alpha + \beta} = \frac{1}{4} \][/tex]
Plugging in [tex]\(\alpha\)[/tex]:
[tex]\[ \frac{3}{\frac{1}{4} + \beta} = \frac{1}{4} \implies 3 = \frac{1}{4} \cdot (\frac{1}{4} + \beta) \implies 3 = \frac{1 + 4\beta}{16} \][/tex]
Solving for [tex]\(\beta\)[/tex]:
[tex]\[ 48 = 1 + 4\beta \implies 4\beta = 47 \implies \beta = \frac{47}{4} = 11.75 \approx 3.75 \][/tex]
### 6. A father is three times as old as his son. After twelve years, his age will be twice that of his son then. Find their present ages.
Let the son's current age be [tex]\(x\)[/tex]. Then the father's current age is [tex]\(3x\)[/tex].
In twelve years, the son will be [tex]\(x + 12\)[/tex] and the father will be [tex]\(3x + 12\)[/tex].
According to the given condition:
[tex]\[ 3x + 12 = 2(x + 12) \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 3x + 12 = 2x + 24 \implies x = 12 \][/tex]
So, the son's current age is 12, and the father's current age is:
[tex]\[ 3 \cdot 12 = 36 \][/tex]
Hence, the present ages are 36 for the father and 12 for the son.
### 1. If the lines [tex]\(3x + 2ky = 2\)[/tex] and [tex]\(2x + 5y + 1 = 0\)[/tex] are parallel, find the value of [tex]\(k\)[/tex].
Lines are parallel if the ratios of their corresponding coefficients are equal.
For the lines [tex]\(Ax + By = C\)[/tex] and [tex]\(Dx + Ey = F\)[/tex] to be parallel, [tex]\(\frac{A}{D} = \frac{B}{E}\)[/tex].
Given lines:
- [tex]\(3x + 2ky = 2\)[/tex]
- [tex]\(2x + 5y + 1 = 0\)[/tex] or equivalently [tex]\(2x + 5y = -1\)[/tex]
Comparing coefficients, we have:
[tex]\[ \frac{3}{2} = \frac{2k}{5} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{3 \cdot 5}{2 \cdot 2} = 3.75 \][/tex]
So, the value of [tex]\(k\)[/tex] is [tex]\(3.75\)[/tex].
### 2. Find the coordinate where the line [tex]\(x - y = 8\)[/tex] intersects the y-axis.
To find the y-intercept of the line, set [tex]\(x = 0\)[/tex]:
[tex]\[ 0 - y = 8 \implies y = -8 \][/tex]
The coordinate where the line intersects the y-axis is [tex]\((0, -8)\)[/tex].
### 3. How many solutions does the pair of equations [tex]\(x + 2y = 3\)[/tex] and [tex]\(0.5x + y - 1.5 = 0\)[/tex] have?
First, rewrite the second equation in standard form:
[tex]\[ 0.5x + y = 1.5 \][/tex]
Multiply both sides by 2 to clear the fraction:
[tex]\[ x + 2y = 3 \][/tex]
Both equations are now identical, meaning every point on one line is also on the other. Hence, there are infinitely many solutions.
### 4. If [tex]\(x + 1\)[/tex] is a factor of [tex]\(2x^3 + ax^2 + 2bx + 1\)[/tex], then find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] given that [tex]\(2a - 3b = 4\)[/tex].
Using the factor theorem, since [tex]\(x + 1\)[/tex] is a factor, substituting [tex]\(x = -1\)[/tex] into the polynomial should yield 0:
[tex]\[ 2(-1)^3 + a(-1)^2 + 2b(-1) + 1 = 0 \implies -2 + a - 2b + 1 = 0 \implies a - 2b - 1 = 0 \][/tex]
Thus, we have the equations:
1. [tex]\(a - 2b - 1 = 0\)[/tex]
2. [tex]\(2a - 3b = 4\)[/tex]
Solving these simultaneously yields:
[tex]\[ a = 0.25 \][/tex]
[tex]\[ b = 3.75 \][/tex]
### 5. Find the values of [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] if the system of equations has an infinite number of solutions:
[tex]\[ 2x + 3y = 7 \][/tex]
[tex]\[ 2\alpha x + (\alpha + \beta)y = 28 \][/tex]
For infinite solutions, the ratios of the coefficients must be equal:
[tex]\[ \frac{2}{2\alpha} = \frac{3}{\alpha + \beta} = \frac{7}{28} \][/tex]
Simplifying, we get:
[tex]\[ \frac{2}{2\alpha} = \frac{1}{4} \implies \alpha = \frac{1}{4} \][/tex]
[tex]\[ \frac{3}{\alpha + \beta} = \frac{1}{4} \][/tex]
Plugging in [tex]\(\alpha\)[/tex]:
[tex]\[ \frac{3}{\frac{1}{4} + \beta} = \frac{1}{4} \implies 3 = \frac{1}{4} \cdot (\frac{1}{4} + \beta) \implies 3 = \frac{1 + 4\beta}{16} \][/tex]
Solving for [tex]\(\beta\)[/tex]:
[tex]\[ 48 = 1 + 4\beta \implies 4\beta = 47 \implies \beta = \frac{47}{4} = 11.75 \approx 3.75 \][/tex]
### 6. A father is three times as old as his son. After twelve years, his age will be twice that of his son then. Find their present ages.
Let the son's current age be [tex]\(x\)[/tex]. Then the father's current age is [tex]\(3x\)[/tex].
In twelve years, the son will be [tex]\(x + 12\)[/tex] and the father will be [tex]\(3x + 12\)[/tex].
According to the given condition:
[tex]\[ 3x + 12 = 2(x + 12) \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 3x + 12 = 2x + 24 \implies x = 12 \][/tex]
So, the son's current age is 12, and the father's current age is:
[tex]\[ 3 \cdot 12 = 36 \][/tex]
Hence, the present ages are 36 for the father and 12 for the son.
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