4. determine whether each of the following values of the variable given is a solution to the equation shown. justify your answer.
(a) $9 - x = x + 1$, $x = 4$ (b) $\frac{n}{2} + 1 = n - 6$, $n = 14$ (c) $2y - 6 = y + 7$, $y = 8$

using your math
5. liam is playing a game where he thinks of a number and then comes up with a riddle involving it. for the number he is thinking of, he knows that if he triples the number and then subtracts 11, the result will be equal to 10.
(a) write an equation for the number, $n$, that liam is thinking about. (b) is the number $n = 8$? check to see if it is a solution to the equation you wrote in (a).

reviewing your math
6. which of the following is equivalent to the expression $10x + 35$?
(1) $5(5x + 7)$ (3) $10(x + 25)$
(2) $7(3x + 5)$ (4) $5(2x + 7)$

7. find each of the following in simplest form.
(a) $\frac{5}{3} + \frac{2}{9}$ (b) $\frac{10}{7} \div \frac{15}{28}$

Question

4. determine whether each of the following values of the variable given is a solution to the equation shown. justify your answer.
(a) $9 - x = x + 1$, $x = 4$  (b) $\frac{n}{2} + 1 = n - 6$, $n = 14$  (c) $2y - 6 = y + 7$, $y = 8$

using your math
5. liam is playing a game where he thinks of a number and then comes up with a riddle involving it. for the number he is thinking of, he knows that if he triples the number and then subtracts 11, the result will be equal to 10.
(a) write an equation for the number, $n$, that liam is thinking about.  (b) is the number $n = 8$? check to see if it is a solution to the equation you wrote in (a).

reviewing your math
6. which of the following is equivalent to the expression $10x + 35$?
(1) $5(5x + 7)$  (3) $10(x + 25)$
(2) $7(3x + 5)$  (4) $5(2x + 7)$

7. find each of the following in simplest form.
(a) $\frac{5}{3} + \frac{2}{9}$  (b) $\frac{10}{7} \div \frac{15}{28}$

Accepted Answer

### Problem 4(a)
# Explanation:
## Step 1: Substitute $ x = 4 $ into the left - hand side (LHS) of the equation $ 9 - x=x + 1 $
The LHS is $ 9-x $. When $ x = 4 $, we have $ 9-4=5 $.
## Step 2: Substitute $ x = 4 $ into the right - hand side (RHS) of the equation
The RHS is $ x + 1 $. When $ x = 4 $, we have $ 4 + 1=5 $.
## Step 3: Compare LHS and RHS
Since LHS $= 5$ and RHS $= 5$, LHS = RHS.

# Answer: $ x = 4 $ is a solution to the equation $ 9 - x=x + 1 $

### Problem 4(b)
# Explanation:
## Step 1: Substitute $ n = 14 $ into the LHS of the equation $ \frac{n}{2}+1=n - 6 $
The LHS is $ \frac{n}{2}+1 $. When $ n = 14 $, we have $ \frac{14}{2}+1=7 + 1=8 $.
## Step 2: Substitute $ n = 14 $ into the RHS of the equation
The RHS is $ n-6 $. When $ n = 14 $, we have $ 14-6 = 8 $.
## Step 3: Compare LHS and RHS
Since LHS $= 8$ and RHS $= 8$, LHS = RHS.

# Answer: $ n = 14 $ is a solution to the equation $ \frac{n}{2}+1=n - 6 $

### Problem 4(c)
# Explanation:
## Step 1: Substitute $ y = 8 $ into the LHS of the equation $ 2y-6=y + 7 $
The LHS is $ 2y-6 $. When $ y = 8 $, we have $ 2\times8-6=16 - 6 = 10 $.
## Step 2: Substitute $ y = 8 $ into the RHS of the equation
The RHS is $ y + 7 $. When $ y = 8 $, we have $ 8+7 = 15 $.
## Step 3: Compare LHS and RHS
Since $ 10
eq15 $, LHS $
eq$ RHS.

# Answer: $ y = 8 $ is not a solution to the equation $ 2y-6=y + 7 $

### Problem 5(a)
# Explanation:
Tripling the number $ n $ gives $ 3n $, subtracting 11 from it gives $ 3n-11 $, and this result is equal to 10. So the equation is $ 3n-11 = 10 $.

# Answer: The equation is $ 3n-11=10 $

### Problem 5(b)
# Explanation:
## Step 1: Substitute $ n = 8 $ into the LHS of the equation $ 3n-11 = 10 $
The LHS is $ 3n-11 $. When $ n = 8 $, we have $ 3\times8-11=24 - 11 = 13 $.
## Step 2: Compare LHS and RHS
The RHS of the equation $ 3n - 11=10 $ is 10. Since $ 13
eq10 $, LHS $
eq$ RHS.

# Answer: $ n = 8 $ is not a solution to the equation $ 3n-11 = 10 $

### Problem 6
# Explanation:
We need to factor out the greatest common factor (GCF) from $ 10x + 35 $. The GCF of 10 and 35 is 5.
Factor out 5 from $ 10x+35 $: $ 10x+35=5\times(2x + 7) $ (because $ 5\times2x=10x $ and $ 5\times7 = 35 $).
Let's check the other options:
- Option (1): $ 5(5x + 7)=25x+35
eq10x + 35 $
- Option (2): $ 7(3x + 5)=21x+35
eq10x + 35 $
- Option (3): $ 10(x + 25)=10x+250
eq10x + 35 $

# Answer: (4) $ 5(2x + 7) $

### Problem 7(a)
# Explanation:
To add $ \frac{5}{3}+\frac{2}{9} $, we first find a common denominator. The common denominator of 3 and 9 is 9.
Rewrite $ \frac{5}{3} $ with denominator 9: $ \frac{5}{3}=\frac{5\times3}{3\times3}=\frac{15}{9} $
Then $ \frac{15}{9}+\frac{2}{9}=\frac{15 + 2}{9}=\frac{17}{9} $

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

### Problem 7(b)
# Explanation:
To divide $ \frac{10}{7}\div\frac{15}{28} $, we use the rule of dividing fractions: $ \frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c} $ (where $ b
eq0,c
eq0,d
eq0 $)
So $ \frac{10}{7}\div\frac{15}{28}=\frac{10}{7}\times\frac{28}{15} $
Simplify: $ \frac{10\times28}{7\times15}=\frac{10\times4}{15}=\frac{40}{15}=\frac{8}{3} $ (we can also simplify before multiplying: $ \frac{10}{7}\times\frac{28}{15}=\frac{10\times4}{15}=\frac{8}{3} $ since $ 28\div7 = 4 $ and $ 10$ and $ 15$ have a GCF of 5, $ 10\div5 = 2 $, $ 15\div5=3 $, but $ 2\times4 = 8 $ and the denominator is 3)

# Answer: $ \frac{8}{3} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 4(a)

Step 1: Substitute \( x = 4 \) into the left - hand side (LHS) of the equation \( 9 - x=x + 1 \)

Step 2: Substitute \( x = 4 \) into the right - hand side (RHS) of the equation

Step 3: Compare LHS and RHS

Step 1: Substitute \( n = 14 \) into the LHS of the equation \( \frac{n}{2}+1=n - 6 \)

Step 2: Substitute \( n = 14 \) into the RHS of the equation

Step 3: Compare LHS and RHS

Step 1: Substitute \( y = 8 \) into the LHS of the equation \( 2y-6=y + 7 \)

Step 2: Substitute \( y = 8 \) into the RHS of the equation

Step 3: Compare LHS and RHS

Answer:

Problem 4(b)