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Question

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  1. the non - permissible value(s) for the rational expression \\(\frac{4b + 28}{b^2 + 10b + 21}\\) is/are

a. \\(b\
eq - 21\\)
b. \\(b\
eq - 7\\)
c. \\(b\
eq - 3\\)
d. \\(b\
eq - 7, - 3\\)

  1. simplify \\(\frac{3v^2 - 17vz - 90z^2}{14v^2 - 128vz + 18z^2}\\).

a. \\(\frac{3v + 10z}{2(7v - z)}, v\
eq 9z, \frac{1}{7}z\\)
b. \\(\frac{3v + 10z}{2(7v - z)}, v\
eq \frac{1}{7}z\\)
c. \\(\frac{3v - 10z}{2(7v + z)}, v\
eq 9z, \frac{1}{7}z\\)
d. \\(\frac{3v - 10z}{2(7v - z)}, v\
eq \frac{1}{7}z\\)

  1. what is the solution to the equation \\(\frac{4x - 2}{x - 5}=\frac{4}{3}\\)?

a. \\(-\frac{7}{4}\\)
b. \\(-\frac{7}{13}\\)
c. \\(\frac{7}{4}\\)
d. there is no solution to this equation because \\(x\
eq 5\\).

Explanation:

Response
Question 1

Step1: Find denominator's roots

The denominator is \(b^2 + 10b + 21\). Factor it: \(b^2 + 10b + 21=(b + 7)(b + 3)\).

Step2: Determine non - permissible values

Set denominator equal to zero: \((b + 7)(b + 3)=0\). So \(b=-7\) or \(b=-3\). Non - permissible values are \(b
eq - 7,-3\).

Step1: Factor numerator and denominator

Numerator: \(3v^{2}-17vz - 90z^{2}\). We need two numbers that multiply to \(3\times(-90)=-270\) and add to \(-17\). The numbers are \(-27\) and \(10\). So \(3v^{2}-17vz - 90z^{2}=3v^{2}-27vz + 10vz-90z^{2}=3v(v - 9z)+10z(v - 9z)=(3v + 10z)(v - 9z)\).
Denominator: \(14v^{2}-128vz + 18z^{2}=2(7v^{2}-64vz + 9z^{2})\). Factor \(7v^{2}-64vz + 9z^{2}\). We need two numbers that multiply to \(7\times9 = 63\) and add to \(-64\). The numbers are \(-63\) and \(-1\). So \(7v^{2}-64vz + 9z^{2}=7v^{2}-63vz-vz + 9z^{2}=7v(v - 9z)-z(v - 9z)=(7v - z)(v - 9z)\).

Step2: Simplify the rational expression

\(\frac{3v^{2}-17vz - 90z^{2}}{14v^{2}-128vz + 18z^{2}}=\frac{(3v + 10z)(v - 9z)}{2(7v - z)(v - 9z)}\). Cancel out \((v - 9z)\) (note \(v
eq9z\)). Also, denominator \(2(7v - z)(v - 9z)
eq0\), so \(7v - z
eq0\Rightarrow v
eq\frac{1}{7}z\) and \(v
eq9z\). The simplified expression is \(\frac{3v + 10z}{2(7v - z)}\), with \(v
eq9z,\frac{1}{7}z\).

Step1: Cross - multiply

Given \(\frac{4x - 2}{x - 5}=\frac{4}{3}\), cross - multiply: \(3(4x - 2)=4(x - 5)\).

Step2: Expand both sides

\(12x-6 = 4x-20\).

Step3: Solve for x

Subtract \(4x\) from both sides: \(12x-4x-6=4x - 4x-20\Rightarrow8x-6=-20\). Add \(6\) to both sides: \(8x-6 + 6=-20 + 6\Rightarrow8x=-14\). Divide by \(8\): \(x=-\frac{14}{8}=-\frac{7}{4}\). We need to check if \(x = -\frac{7}{4}\) makes the denominator \(x - 5
eq0\). \(x-5=-\frac{7}{4}-5=-\frac{7 + 20}{4}=-\frac{27}{4}
eq0\).

Answer:

D. \(b
eq - 7,-3\)

Question 2