what is the product in simplest form? state any restrictions on the variable.
57. $\frac{y^2}{y - 3} \cdot \frac{y^2 - y - 6}{y^2 + 1y}$
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.
58. $\frac{a + 2}{a - 5} \div \frac{a + 1}{a^2 - 8a + 15}$
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.
59. $\frac{n^2 - 10n + 24}{n^2 - 13n + 42} - \frac{9}{n - 7}$
a. $\frac{n - 13}{n - 7}$ c. $n - 13$
b. $\frac{n - 4}{n - 7}$ d. $\frac{n^2 - 10n + 15}{n^2 - 13n + 42}$
60. 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.
61. $\frac{4}{a} + \frac{5}{3a} = 3$
a. $\frac{17}{9}$ b. $\frac{17}{3}$ c. $\frac{19}{9}$ d. $\frac{3}{4}$
62. $\frac{-4}{x + 1} = \frac{-1}{x + 5}$
a. $-\frac{19}{4}$ b. $\frac{1}{3}$ c. $-\frac{19}{3}$ d. $2$

Question

what is the product in simplest form? state any restrictions on the variable.
57. $\frac{y^2}{y - 3} \cdot \frac{y^2 - y - 6}{y^2 + 1y}$
  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.
58. $\frac{a + 2}{a - 5} \div \frac{a + 1}{a^2 - 8a + 15}$
  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.
59. $\frac{n^2 - 10n + 24}{n^2 - 13n + 42} - \frac{9}{n - 7}$
  a. $\frac{n - 13}{n - 7}$  c. $n - 13$
  b. $\frac{n - 4}{n - 7}$  d. $\frac{n^2 - 10n + 15}{n^2 - 13n + 42}$
60. 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.
61. $\frac{4}{a} + \frac{5}{3a} = 3$
  a. $\frac{17}{9}$  b. $\frac{17}{3}$  c. $\frac{19}{9}$  d. $\frac{3}{4}$
62. $\frac{-4}{x + 1} = \frac{-1}{x + 5}$
  a. $-\frac{19}{4}$  b. $\frac{1}{3}$  c. $-\frac{19}{3}$  d. $2$

Accepted Answer

### Question 57
# Explanation:
## 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 $y$. We get $\frac{y(y + 2)}{y + 1}=\frac{y^2+2y}{y + 1}$? Wait, no, wait: $y^2/y=y$, so $y\cdot(y + 2)/(y + 1)$? Wait, no, original $y^2$ over $y-3$ times $(y - 3)(y + 2)$ over $y(y + 1)$. So $y^2$ is $y\cdot y$, so cancel $y$ (one) and $y - 3$. So we have $y(y + 2)/(y + 1)$? Wait, no, the options have $y + 2$ over $y + 1$ or $y^2+2y$ over $y + 1$. Wait, $y^2+2y=y(y + 2)$, so $\frac{y(y + 2)}{y + 1}$ is $\frac{y^2+2y}{y + 1}$? But wait, let's check restrictions. Denominators: $y - 3
eq0\Rightarrow y
eq3$, $y^2 + y=y(y + 1)
eq0\Rightarrow y
eq0,-1$. So restrictions $y
eq3,0,-1$. Now, wait, maybe I factored wrong. Wait, the first fraction is $y^2/(y - 3)$, second is $(y^2 - y - 6)/(y^2 + y)$. Wait, $y^2 - y - 6$: discriminant $1 + 24 = 25$, roots $(1\pm5)/2$, so $3$ and $-2$. So $y^2 - y - 6=(y - 3)(y + 2)$. Then $y^2 + y=y(y + 1)$. So multiplying: $\frac{y^2}{y - 3}\cdot\frac{(y - 3)(y + 2)}{y(y + 1)}$. Cancel $y - 3$, cancel one $y$ (from $y^2$ and $y$): so we get $\frac{y(y + 2)}{y + 1}=\frac{y^2+2y}{y + 1}$. Now check restrictions: $y - 3
eq0\Rightarrow y
eq3$, $y(y + 1)
eq0\Rightarrow y
eq0,-1$. So the expression is $\frac{y^2+2y}{y + 1}$ with $y
eq3,0,-1$, which is option b? Wait, but option c is $\frac{y + 2}{y + 1}$ with $y
eq3,0,-1$. Wait, did I make a mistake? Wait, $y^2$ is $y\cdot y$, and the second numerator has $(y - 3)(y + 2)$, denominator $y(y + 1)$. So $y^2/(y - 3)\times(y - 3)(y + 2)/(y(y + 1))$: the $y^2$ is $y\times y$, so one $y$ cancels with the $y$ in the denominator, leaving $y$ in the numerator. So $y\times(y + 2)/(y + 1)$, which is $y(y + 2)/(y + 1)= (y^2 + 2y)/(y + 1)$. But wait, the options: 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 messed up the factoring. Wait, no, $y^2$ is $y\times y$, so when we multiply, $y^2\times(y - 3)(y + 2)$ over $(y - 3)\times y(y + 1)$. So cancel $y - 3$, cancel one $y$ (from $y^2$ and $y$): so numerator is $y(y + 2)$, denominator $y + 1$. So that's $y(y + 2)/(y + 1)$, which is $y^2 + 2y$ over $y + 1$. So restrictions: $y - 3
eq0\Rightarrow y
eq3$, $y(y + 1)
eq0\Rightarrow y
eq0,-1$. So option b? Wait, but let me check again. Wait, maybe the original problem has a typo? Wait, the second fraction's denominator is $y^2 + 1y$, which is $y^2 + y$, correct. Numerator $y^2 - y - 6$, correct. So my calculation gives $y(y + 2)/(y + 1)$ with $y
eq3,0,-1$, which is option b. But wait, maybe I made a mistake. Wait, let's plug in $y = 2$ (which is allowed, since $2
eq3,0,-1$). Original expression: $y^2/(y - 3)\cdot(y^2 - y - 6)/(y^2 + y)$. For $y = 2$: $4/(-1)\cdot(4 - 2 - 6)/(4 + 2)=4/(-1)\cdot(-4)/6= ( - 4)\cdot(-4/6)=16/6 = 8/3$. Now option b: $(4 + 4)/(2 + 1)=8/3$, which matches. Option c: $(2 + 2)/(2 + 1)=4/3$, which doesn't. So option b is correct? Wait, but the options: a is $y^2+2y/y + 1$, $y
eq3,-1$ (missing $0$), b is $y^2+2y/y + 1$, $y
eq3,0,-1$, c is $y + 2/y + 1$, $y
eq3,0,-1$, d is $y + 2/y + 1$, $y
eq3,-1$. So my calculation gives $y^2+2y/y + 1$ with $y
eq3,0,-1$, so option b.

# Answer: b. $\frac{y^2 + 2y}{y + 1}, y
eq 3, 0, - 1$

### Question 58
# Explanation:
## Step1: Rewrite division as multiplication
$\frac{a + 2}{a - 5}\div\frac{a + 1}{a^2 - 8a + 15}=\frac{a + 2}{a - 5}\cdot\frac{a^2 - 8a + 15}{a + 1}$
## Step2: Factor $a^2 - 8a + 15$
$a^2 - 8a + 15=(a - 3)(a - 5)$
## Step3: Multiply and cancel
$\frac{a + 2}{a - 5}\cdot\frac{(a - 3)(a - 5)}{a + 1}$. Cancel $a - 5$: $\frac{(a + 2)(a - 3)}{a + 1}$
## 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 combined restrictions: $a
eq5,3,-1$
So the expression is $\frac{(a + 2)(a - 3)}{a + 1}$ with $a
eq5,3,-1$, which is option a? Wait, option a: $\frac{(a + 2)(a - 3)}{a + 1}, a
eq5,-1,3$, yes. Let's check with $a = 0$ (allowed, $0
eq5,3,-1$). Original: $(0 + 2)/(0 - 5)\div(0 + 1)/(0 - 0 + 15)= (2/-5)\div(1/15)= (2/-5)\times15/1= - 6$. Option a: $(0 + 2)(0 - 3)/(0 + 1)= (2)(-3)/1= - 6$, matches. Option c: restrictions $a
eq3,-1$ (missing $5$), so no. So option a.

# Answer: a. $\frac{(a + 2)(a - 3)}{a + 1}, a
eq 5, - 1, 3$

### Question 59
# Explanation:
## Step1: Factor denominators
$n^2 - 13n + 42=(n - 6)(n - 7)$, $n^2 - 10n + 24=(n - 4)(n - 6)$
## Step2: Rewrite the expression
$\frac{(n - 4)(n - 6)}{(n - 6)(n - 7)}-\frac{9}{n - 7}$
## Step3: Cancel common factor
Cancel $n - 6$ (note $n
eq6$, but also $n
eq7$ from denominator). We get $\frac{n - 4}{n - 7}-\frac{9}{n - 7}$
## Step4: Subtract the fractions
$\frac{(n - 4)-9}{n - 7}=\frac{n - 13}{n - 7}$
So the simplified form is $\frac{n - 13}{n - 7}$, which is option a.

# Answer: a. $\frac{n - 13}{n - 7}$

### Question 60
# Explanation:
## Step1: Add the fractions (same denominator)
$\frac{90x}{x - 3}+\frac{10x + 5}{x - 3}=\frac{90x + 10x + 5}{x - 3}$
## Step2: Combine like terms
$90x + 10x=100x$, so numerator is $100x + 5$, denominator $x - 3$
So the sum is $\frac{100x + 5}{x - 3}$, which is option a.

# Answer: a. $\frac{100x + 5}{x - 3}$

### Question 61
# Explanation:
## Step1: Combine the fractions
$\frac{4}{a}+\frac{5}{3a}=\frac{12}{3a}+\frac{5}{3a}=\frac{17}{3a}$
## Step2: Set equal to 3 and solve
$\frac{17}{3a}=3\Rightarrow17 = 9a\Rightarrow a=\frac{17}{9}$
Check: $\frac{4}{17/9}+\frac{5}{3*(17/9)}=\frac{36}{17}+\frac{15}{17}=\frac{51}{17}=3$, correct. So option a.

# Answer: a. $\frac{17}{9}$

### Question 62
# Explanation:
## Step1: Cross - multiply
$-4(x + 5)=-1(x + 1)$
## Step2: Expand both sides
$-4x-20=-x - 1$
## Step3: Solve for x
$-4x+x=-1 + 20\Rightarrow-3x = 19\Rightarrow x=-\frac{19}{3}$
Check: Left side: $-4/(-19/3 + 1)=-4/(-16/3)=12/16 = 3/4$. Right side: $-1/(-19/3 + 5)=-1/(-4/3)=3/4$, correct. So option c.

# Answer: c. $-\frac{19}{3}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 57

Step1: Factor numerators/denominators

Step2: Cancel common factors

Step1: Rewrite division as multiplication

Step2: Factor \(a^2 - 8a + 15\)

Step3: Multiply and cancel

Step4: Find restrictions

Step1: Factor denominators

Step2: Rewrite the expression

Step3: Cancel common factor

Step4: Subtract the fractions

Answer:

Question 58