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 - 4}{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 \(y\) (from \(y^2\) and \(y\)): \(\frac{y(y + 2)}{y + 1}\)? Wait, no, wait: \(y^2/y = y\)? Wait, no, \(y^2\) is \(y\cdot y\), and denominator has \(y\), so cancel one \(y\), get \(y\) in numerator. Wait, but let's re - do:

Original expression: \(\frac{y^2}{y - 3}\times\frac{y^2 - y - 6}{y^2 + y}=\frac{y^2}{y - 3}\times\frac{(y - 3)(y + 2)}{y(y + 1)}\)

Cancel \(y - 3\) (non - zero, so \(y
eq3\)), cancel one \(y\) from \(y^2\) and \(y\) (so \(y
eq0\)), we get \(\frac{y(y + 2)}{y + 1}=\frac{y^2+2y}{y + 1}\)? Wait, no, wait the options: option c is \(\frac{y + 2}{y + 1}\), option b is \(\frac{y^2+2y}{y + 1}\) with restrictions \(y
eq3,0, - 1\). Wait, let's check the denominator of the original expression: \(y - 3
eq0\Rightarrow y
eq3\), \(y^2 + y=y(y + 1)
eq0\Rightarrow y
eq0,y
eq - 1\). Now, when we simplify \(\frac{y^2}{y - 3}\times\frac{(y - 3)(y + 2)}{y(y + 1)}\), cancel \(y - 3\) ( \(y
eq3\) ), cancel one \(y\) ( \(y
eq0\) ), so we have \(\frac{y(y + 2)}{y + 1}=\frac{y^2+2y}{y + 1}\), and restrictions \(y
eq3,0, - 1\) (since \(y - 3
eq0\), \(y
eq0\) from \(y\) in denominator, \(y + 1
eq0\Rightarrow y
eq - 1\)). Wait, but option b is \(\frac{y^2 + 2y}{y + 1},y
eq3,0, - 1\), option c is \(\frac{y + 2}{y + 1},y
eq3,0, - 1\). Wait, I must have made a mistake. Wait, \(y^2\) is \(y\times y\), and the second fraction's numerator is \((y - 3)(y + 2)\), denominator is \(y(y + 1)\). So multiplying: \(\frac{y\times y\times(y - 3)\times(y + 2)}{(y - 3)\times y\times(y + 1)}\). Cancel \(y - 3\) ( \(y
eq3\) ), cancel one \(y\) ( \(y
eq0\) ), so we get \(\frac{y(y + 2)}{y + 1}\)? But option c has \(\frac{y + 2}{y + 1}\). Wait, maybe I messed up the factoring. Wait, \(y^2 - y - 6\): let's factor again. \(y^2 - y - 6=(y - 3)(y + 2)\), correct. \(y^2 + y=y(y + 1)\), correct. \(y^2\) is \(y\times y\). So when we multiply, numerator: \(y^2\times(y^2 - y - 6)=y^2(y - 3)(y + 2)\), denominator: \((y - 3)(y^2 + y)=(y - 3)y(y + 1)\). So cancel \(y - 3\) ( \(y
eq3\) ), cancel one \(y\) ( \(y
eq0\) ), so we have \(\frac{y(y + 2)}{y + 1}\), which is \(\frac{y^2+2y}{y + 1}\), and restrictions: \(y
eq3\) (from \(y - 3\)), \(y
eq0\) (from \(y\) in denominator), \(y
eq - 1\) (from \(y + 1\) in denominator). So option b is \(\frac{y^2 + 2y}{y + 1},y
eq3,0, - 1\), option c is \(\frac{y + 2}{y + 1},y
eq3,0, - 1\). Wait, maybe I made a mistake in the number of \(y\)s. Wait, \(y^2\) is \(y\times y\), and the denominator has \(y\), so \(y^2/y=y\), so numerator after canceling is \(y(y + 2)\), denominator is \(y + 1\). But let's check the options again. Wait, maybe the original problem has a typo? Wait, the problem is \(\frac{y^2}{y - 3}\cdot\frac{y^2 - y - 6}{y^2 + y}\). Wait, \(y^2 - y - 6=(y - 3)(y + 2)\), \(y^2 + y=y(y + 1)\). So \(\frac{y^2}{y - 3}\times\frac{(y - 3)(y + 2)}{y(y + 1)}=\frac{y(y + 2)}{y + 1}=\frac{y^2 + 2y}{y + 1}\), with \(y
eq3\) (from \(y - 3\)), \(y
eq0\) (from \(y\) in denominator), \(y
eq - 1\) (from \(y + 1\) in denominator). So option b is correct? Wait, but let's check the options again. Option b: \(\frac{y^2 + 2y}{y + 1},y
eq3,0, - 1\), option c: \(\frac{y + 2}{y + 1},y
eq3,0, - 1\). Wait, maybe I made a mistake in the exponent. Wait, \(y^2\) divided by \(y\) is \(y\), so \(y(y + 2)=y^2 + 2y\), so option b is correct.

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 \(a^2 - 8a + 15\)

\(a^2 - 8a + 15=(a - 3)(a - 5)\)

Step3: Multiply and cancel

\(\frac{a + 2}{a - 5}\times\frac{(a - 3)(a - 5)}{a + 1}=\frac{(a + 2)(a - 3)}{a + 1}\) (cancel \(a - 5\), \(a
eq5\))

Step4: Find restrictions

Denominator of original: \(a - 5
eq0\Rightarrow a
eq5\), \(a + 1
eq0\Rightarrow a
eq - 1\), and from \(a^2 - 8a + 15=(a - 3)(a - 5)\), \(a - 3
eq0\) (because we canceled \(a - 5\), but \(a - 3\) was in the numerator, but wait, no: when we have \(\frac{(a - 3)(a - 5)}{a + 1}\), the \(a - 5\) is canceled with \(a - 5\) in the first fraction, but the original denominator of the second fraction is \(a + 1\), and the first fraction's denominator is \(a - 5\), and the factored numerator has \(a - 3\). So restrictions: \(a - 5
eq0\Rightarrow a
eq5\), \(a + 1
eq0\Rightarrow a
eq - 1\), \(a - 3
eq0\Rightarrow a
eq3\) (because if \(a = 3\), then \(a^2 - 8a + 15 = 0\), but in our case, when we rewrite division as multiplication, we have \(\frac{a + 2}{a - 5}\times\frac{(a - 3)(a - 5)}{a + 1}\), so \(a - 3\) is in the numerator, but we still need \(a - 3
eq0\)? Wait, no, because if \(a = 3\), then the numerator of the second fraction is \(0\), but the expression is defined as long as the denominators are non - zero and the divisor is non - zero. The divisor is \(\frac{a + 1}{a^2 - 8a + 15}\), so \(a^2 - 8a + 15
eq0\Rightarrow a
eq3,a
eq5\), and \(a + 1
eq0\Rightarrow a
eq - 1\), and the first fraction's denominator \(a - 5
eq0\Rightarrow a
eq5\). So the restrictions are \(a
eq5, - 1,3\), and the simplified form is \(\frac{(a + 2)(a - 3)}{a + 1}\), which is option a.

Step1: Factor denominators and numerators

\(n^2 - 10n + 24=(n - 4)(n - 6)\), \(n^2 - 13n + 42=(n - 6)(n - 7)\)

Step2: Rewrite the expression

\(\frac{(n - 4)(n - 6)}{(n - 6)(n - 7)}-\frac{9}{n - 7}\)

Step3: Cancel common factor

Cancel \(n - 6\) ( \(n
eq6\) ), get \(\frac{n - 4}{n - 7}-\frac{9}{n - 7}\)

Step4: Subtract the fractions

Since denominators are the same, \(\frac{(n - 4)-9}{n - 7}=\frac{n - 13}{n - 7}\)? Wait, no: \((n - 4)-9=n - 13\)? Wait, \(n - 4-9=n - 13\), but let's check option a: \(\frac{n - 13}{n - 7}\), option b: \(\frac{n - 4}{n - 7}\). Wait, no, wait: \(\frac{(n - 4)(n - 6)}{(n - 6)(n - 7)}=\frac{n - 4}{n - 7}\) ( \(n
eq6\) ), then \(\frac{n - 4}{n - 7}-\frac{9}{n - 7}=\frac{n - 4-9}{n - 7}=\frac{n - 13}{n - 7}\), which is option a.

Answer:

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

Question 58