Sure, let's break this problem down step by step.
### Problem 08: Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
We are given the equation [tex]\((x+5, 2y+3) = (9, 6)\)[/tex].
This gives us two separate equations:
1. [tex]\( x + 5 = 9 \)[/tex]
2. [tex]\( 2y + 3 = 6 \)[/tex]
First, solve for [tex]\( x \)[/tex] from the first equation:
[tex]\[ x + 5 = 9 \][/tex]
Subtract 5 from both sides:
[tex]\[ x = 9 - 5 \][/tex]
[tex]\[ x = 4 \][/tex]
Next, solve for [tex]\( y \)[/tex] from the second equation:
[tex]\[ 2y + 3 = 6 \][/tex]
Subtract 3 from both sides:
[tex]\[ 2y = 6 - 3 \][/tex]
[tex]\[ 2y = 3 \][/tex]
Divide both sides by 2:
[tex]\[ y = \frac{3}{2} \][/tex]
[tex]\[ y = 1.5 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are:
[tex]\[ x = 4 \][/tex]
[tex]\[ y = 1.5 \][/tex]
### Problem 09: Express [tex]\( x \)[/tex] in Terms of [tex]\( y \)[/tex]
We are asked to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] given the equation:
[tex]\[ 2x - 5y = 10 \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 10 + 5y \][/tex]
Divide both sides by 2:
[tex]\[ x = \frac{10 + 5y}{2} \][/tex]
So, [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] is:
[tex]\[ x = \frac{10 + 5y}{2} \][/tex]
### Finding Two Solutions
Now let's find two specific solutions of the equation [tex]\( 2x - 5y = 10 \)[/tex].
#### Solution 1:
Choose [tex]\( y = 1 \)[/tex]:
[tex]\[ x = \frac{10 + 5 \cdot 1}{2} \][/tex]
[tex]\[ x = \frac{10 + 5}{2} \][/tex]
[tex]\[ x = \frac{15}{2} \][/tex]
[tex]\[ x = 7.5 \][/tex]
So, the first solution is:
[tex]\[ (y, x) = (1, 7.5) \][/tex]
#### Solution 2:
Choose [tex]\( y = -2 \)[/tex]:
[tex]\[ x = \frac{10 + 5 \cdot (-2)}{2} \][/tex]
[tex]\[ x = \frac{10 - 10}{2} \][/tex]
[tex]\[ x = \frac{0}{2} \][/tex]
[tex]\[ x = 0 \][/tex]
So, the second solution is:
[tex]\[ (y, x) = (-2, 0) \][/tex]
To summarize, the solutions are:
- [tex]\( (x, y) = (4, 1.5) \)[/tex] from the first part.
- Two specific solutions [tex]\( (y, x) = (1, 7.5) \)[/tex] and [tex]\( (y, x) = (-2, 0) \)[/tex] for the equation [tex]\( 2x - 5y = 10 \)[/tex].
