Question

what is the product in simplest form? state any restrictions on the variable.

1. \\(\frac{y^2}{y - 3} \cdot \frac{y^2 - y - 6}{y^2 + 1y}\\)

a. \\(\frac{y^2 + 2y}{y + 1}, y \
eq 3, -1\\) c. \\(\frac{y + 2}{y + 1}, y \
eq 3, 0, -1\\)
b. \\(\frac{y^2 + 2y}{y + 1}, y \
eq 3, 0, -1\\) d. \\(\frac{y + 2}{y + 1}, y \
eq 3, -1\\)
what is the quotient in simplified form? state any restrictions on the variable.

1. \\(\frac{a + 2}{a - 5} \div \frac{a + 1}{a^2 - 8a + 15}\\)

a. \\(\frac{(a + 2)(a - 3)}{a + 1}, a \
eq 5, -1, 3\\) c. \\(\frac{(a + 2)(a - 3)}{a + 1}, a \
eq 3, -1\\)
b. \\(\frac{(a + 2)(a + 1)}{(a - 5)^2(a - 3)}, a \
eq 5, 3, -1\\) d. \\(\frac{(a + 2)(a + 1)}{(a - 5)^2(a - 3)}, a \
eq 5, 3\\)
simplify the difference.

1. \\(\frac{n^2 - 10n + 24}{n^2 - 13n + 42} - \frac{9}{n - 7}\\)

a. \\(\frac{n - 13}{n - 7}\\) c. \\(n - 13\\)
b. \\(\frac{n - 9}{n - 7}\\) d. \\(\frac{n^2 - 10n + 15}{n^2 - 13n + 42}\\)

1. what is the sum \\(\frac{90x}{x - 3} + \frac{10x + 5}{x - 3}\\)?

a. \\(\frac{100x + 5}{x - 3}\\) c. \\(\frac{100x + 5}{2x - 6}\\)
b. \\(\frac{105x}{x - 3}\\) d. \\(\frac{105x}{2x - 6}\\)
solve the equation. check the solution.

1. \\(\frac{4}{a} + \frac{5}{3a} = 3\\)

a. \\(\frac{17}{9}\\) b. \\(\frac{17}{3}\\) c. \\(\frac{19}{9}\\) d. \\(\frac{3}{4}\\)

1. \\(\frac{-4}{x + 1} = \frac{-1}{x + 5}\\)

a. \\(-\frac{19}{4}\\) b. \\(\frac{1}{3}\\) c. \\(-\frac{19}{3}\\) d. \\(2\\)

Explanation:

Question 57

Step1: Factor numerators/denominators

Factor \(y^2 - y - 6=(y - 3)(y + 2)\) and \(y^2 + y=y(y + 1)\). The expression becomes \(\frac{y^2}{y - 3}\cdot\frac{(y - 3)(y + 2)}{y(y + 1)}\).

Step2: Cancel common factors

Cancel \(y - 3\) and one \(y\) (since \(y^2=y\cdot y\)). We get \(\frac{y(y + 2)}{y + 1}=\frac{y^2 + 2y}{y + 1}\)? Wait, no, wait: \(y^2/y\) is \(y\), so after canceling \(y - 3\) and \(y\), it's \(\frac{y(y + 2)}{y + 1}\)? Wait, no, original numerator first fraction is \(y^2\), second is \((y - 3)(y + 2)\); denominators: \(y - 3\) and \(y(y + 1)\). So cancel \(y - 3\), and \(y^2/y=y\), so we have \(\frac{y(y + 2)}{y + 1}\)? Wait, but let's check restrictions. Denominators: \(y - 3
eq0\Rightarrow y
eq3\); \(y^2 + y=y(y + 1)
eq0\Rightarrow y
eq0,-1\). Wait, but let's re - do the simplification. Wait, \(y^2/(y - 3)\times(y^2 - y - 6)/(y^2 + y)\). Factor \(y^2 - y - 6=(y - 3)(y + 2)\), \(y^2 + y=y(y + 1)\). So multiply numerators: \(y^2\times(y - 3)(y + 2)\), denominators: \((y - 3)\times y(y + 1)\). Cancel \(y - 3\) and one \(y\) (from \(y^2\) and \(y\)), so we get \(\frac{y(y + 2)}{y + 1}=\frac{y^2 + 2y}{y + 1}\)? Wait, no, \(y^2\times(y - 3)(y + 2)=y^2(y - 3)(y + 2)\), \((y - 3)y(y + 1)=y(y - 3)(y + 1)\). So cancel \(y(y - 3)\) from numerator and denominator, we get \(\frac{y(y + 2)}{y + 1}\)? Wait, no, \(y^2/y = y\), so numerator after cancel: \(y(y + 2)\), denominator: \(y + 1\). Wait, but let's check the options. Option b: \(\frac{y^2 + 2y}{y + 1},y
eq3,0,-1\). Wait, but wait, maybe I made a mistake. Wait, let's do it again. \(\frac{y^2}{y - 3}\times\frac{(y - 3)(y + 2)}{y(y + 1)}\). So \(y^2\) and \(y\) cancel to \(y\), \(y - 3\) cancels. So we have \(\frac{y(y + 2)}{y + 1}=\frac{y^2 + 2y}{y + 1}\). Restrictions: \(y - 3
eq0\Rightarrow y
eq3\); \(y(y + 1)
eq0\Rightarrow y
eq0,-1\). So the answer should be option b? Wait, no, wait the options: option b is \(\frac{y^2 + 2y}{y + 1},y
eq3,0,-1\). But wait, let's check the simplification again. Wait, \(y^2/(y - 3)\times(y^2 - y - 6)/(y^2 + y)\). Let's plug in \(y = 2\) (not a restricted value). Left - hand side: \(\frac{4}{2 - 3}\times\frac{4 - 2 - 6}{4 + 2}=\frac{4}{-1}\times\frac{-4}{6}=-\frac{16}{6}=-\frac{8}{3}\). Option b: \(\frac{4 + 4}{2 + 1}=\frac{8}{3}\)? Wait, that's negative? Wait, no, I must have messed up the sign. Wait, \(y^2 - y - 6\) when \(y = 2\): \(4-2 - 6=-4\), \(y^2 + y=4 + 2 = 6\), \(y - 3=2 - 3=-1\). So \(\frac{4}{-1}\times\frac{-4}{6}=\frac{16}{6}=\frac{8}{3}\). Option b: \(\frac{4 + 4}{2 + 1}=\frac{8}{3}\), which matches. Option a: \(\frac{4 + 4}{2 + 1}=\frac{8}{3}\), but restrictions \(y
eq3,-1\) (missing \(y
eq0\)). Option b: restrictions \(y
eq3,0,-1\), which is correct. Wait, but wait, let's re - check the simplification. Wait, maybe I made a mistake in the first simplification. Let's do it algebraically:

\(\frac{y^2}{y - 3}\cdot\frac{y^2 - y - 6}{y^2 + y}=\frac{y^2}{y - 3}\cdot\frac{(y - 3)(y + 2)}{y(y + 1)}\)

Cancel \(y - 3\) and one \(y\) (from \(y^2\) and \(y\)):

\(=\frac{y(y + 2)}{y + 1}=\frac{y^2 + 2y}{y + 1}\)

Restrictions: \(y-3
eq0\Rightarrow y
eq3\); \(y(y + 1)
eq0\Rightarrow y
eq0,-1\). So the correct option is b.

Step1: Rewrite division as multiplication

\(\frac{a + 2}{a - 5}\div\frac{a + 1}{a^2 - 8a + 15}=\frac{a + 2}{a - 5}\times\frac{a^2 - 8a + 15}{a + 1}\)

Step2: Factor the quadratic

Factor \(a^2 - 8a + 15=(a - 3)(a - 5)\). So the expression becomes \(\frac{a + 2}{a - 5}\times\frac{(a - 3)(a - 5)}{a + 1}\)

Step3: Cancel common factors

Cancel \(a - 5\) from numerator and denominator. We get \(\frac{(a + 2)(a - 3)}{a + 1}\)

Step4: Find restrictions

Denominators: \(a - 5
eq0\Rightarrow a
eq5\); \(a + 1
eq0\Rightarrow a
eq - 1\); \(a^2 - 8a + 15=(a - 3)(a - 5)
eq0\Rightarrow a
eq3,5\). So combining, \(a
eq5,3,-1\)

Step1: Factor the denominators

Factor \(n^2 - 13n + 42=(n - 6)(n - 7)\) and \(n^2 - 10n + 24=(n - 4)(n - 6)\). So the expression \(\frac{n^2 - 10n + 24}{n^2 - 13n + 42}-\frac{9}{n - 7}=\frac{(n - 4)(n - 6)}{(n - 6)(n - 7)}-\frac{9}{n - 7}\)

Step2: Cancel common factors

Cancel \(n - 6\) from the first fraction. We get \(\frac{n - 4}{n - 7}-\frac{9}{n - 7}\)

Step3: Subtract the fractions

Since the denominators are the same, subtract the numerators: \(\frac{(n - 4)-9}{n - 7}=\frac{n - 13}{n - 7}\)? Wait, no, \((n - 4)-9=n - 13\)? Wait, no, \((n - 4)-9=n-4 - 9=n - 13\)? Wait, but let's check with \(n = 8\) (not a restricted value, \(n
eq6,7\)). First fraction: \(\frac{64-80 + 24}{64-104 + 42}=\frac{8}{2}=4\). Second fraction: \(\frac{9}{8 - 7}=9\). So \(4-9=-5\). Now, option a: \(\frac{8 - 13}{8 - 7}=\frac{-5}{1}=-5\), which matches. Wait, but let's re - do the algebra. \(\frac{(n - 4)(n - 6)}{(n - 6)(n - 7)}-\frac{9}{n - 7}=\frac{n - 4}{n - 7}-\frac{9}{n - 7}=\frac{n - 4-9}{n - 7}=\frac{n - 13}{n - 7}\). So the correct option is a.

Answer:

b. \(\frac{y^2 + 2y}{y + 1},y
eq3,0,-1\)

Question 58