Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Sure, let's determine if each of these systems could be represented by the graphs. We'll solve each system step by step:
### 1. System:
[tex]\[ \left\{ \begin{array}{rcl} x + 2y & = & 8 \\ x & = & -5 \end{array} \right. \][/tex]
We'll substitute [tex]\( x = -5 \)[/tex] into the first equation:
[tex]\[ -5 + 2y = 8 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ 2y = 8 + 5 \][/tex]
[tex]\[ 2y = 13 \][/tex]
[tex]\[ y = \frac{13}{2} \][/tex]
Thus, the solution to the first system is [tex]\((x, y) = \left( -5, \frac{13}{2} \right) \)[/tex].
### 2. System:
[tex]\[ \left\{ \begin{array}{rcl} y & = & -7x + 13 \\ y & = & -1 \end{array} \right. \][/tex]
We set the two expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -1 = -7x + 13 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -7x = -1 - 13 \][/tex]
[tex]\[ -7x = -14 \][/tex]
[tex]\[ x = 2 \][/tex]
So, substituting [tex]\( x = 2 \)[/tex] into either equation for [tex]\( y \)[/tex]:
[tex]\[ y = -1 \][/tex]
Therefore, the solution to the second system is [tex]\((x, y) = (2, -1) \)[/tex].
### 3. System:
[tex]\[ \left\{ \begin{array}{rcl} 3x & = & 8 \\ 3x + y & = & 15 \end{array} \right. \][/tex]
First, solve the first equation for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{8}{3} \][/tex]
Next, substitute [tex]\( x = \frac{8}{3} \)[/tex] into the second equation:
[tex]\[ 3 \left(\frac{8}{3}\right) + y = 15 \][/tex]
This simplifies to:
[tex]\[ 8 + y = 15 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = 15 - 8 \][/tex]
[tex]\[ y = 7 \][/tex]
Thus, the solution to the third system is [tex]\((x, y) = \left( \frac{8}{3}, 7 \right) \)[/tex].
### 4. System:
[tex]\[ \left\{ \begin{array}{rcl} y & = & 2x - 7 \\ 4 + y & = & 12 \end{array} \right. \][/tex]
First, solve the second equation for [tex]\( y \)[/tex]:
[tex]\[ 4 + y = 12 \][/tex]
[tex]\[ y = 8 \][/tex]
Now, substitute [tex]\( y = 8 \)[/tex] into the first equation:
[tex]\[ 8 = 2x - 7 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 2x = 8 + 7 \][/tex]
[tex]\[ 2x = 15 \][/tex]
[tex]\[ x = \frac{15}{2} \][/tex]
Therefore, the solution to the fourth system is [tex]\((x, y) = \left( \frac{15}{2}, 8 \right) \)[/tex].
### Summary of Solutions:
1. [tex]\(\left( -5, \frac{13}{2} \right)\)[/tex]
2. [tex]\((2, -1)\)[/tex]
3. [tex]\(\left( \frac{8}{3}, 7 \right)\)[/tex]
4. [tex]\(\left( \frac{15}{2}, 8 \right)\)[/tex]
These solutions can indeed be represented by the graphs of these systems of equations.
### 1. System:
[tex]\[ \left\{ \begin{array}{rcl} x + 2y & = & 8 \\ x & = & -5 \end{array} \right. \][/tex]
We'll substitute [tex]\( x = -5 \)[/tex] into the first equation:
[tex]\[ -5 + 2y = 8 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ 2y = 8 + 5 \][/tex]
[tex]\[ 2y = 13 \][/tex]
[tex]\[ y = \frac{13}{2} \][/tex]
Thus, the solution to the first system is [tex]\((x, y) = \left( -5, \frac{13}{2} \right) \)[/tex].
### 2. System:
[tex]\[ \left\{ \begin{array}{rcl} y & = & -7x + 13 \\ y & = & -1 \end{array} \right. \][/tex]
We set the two expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -1 = -7x + 13 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -7x = -1 - 13 \][/tex]
[tex]\[ -7x = -14 \][/tex]
[tex]\[ x = 2 \][/tex]
So, substituting [tex]\( x = 2 \)[/tex] into either equation for [tex]\( y \)[/tex]:
[tex]\[ y = -1 \][/tex]
Therefore, the solution to the second system is [tex]\((x, y) = (2, -1) \)[/tex].
### 3. System:
[tex]\[ \left\{ \begin{array}{rcl} 3x & = & 8 \\ 3x + y & = & 15 \end{array} \right. \][/tex]
First, solve the first equation for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{8}{3} \][/tex]
Next, substitute [tex]\( x = \frac{8}{3} \)[/tex] into the second equation:
[tex]\[ 3 \left(\frac{8}{3}\right) + y = 15 \][/tex]
This simplifies to:
[tex]\[ 8 + y = 15 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = 15 - 8 \][/tex]
[tex]\[ y = 7 \][/tex]
Thus, the solution to the third system is [tex]\((x, y) = \left( \frac{8}{3}, 7 \right) \)[/tex].
### 4. System:
[tex]\[ \left\{ \begin{array}{rcl} y & = & 2x - 7 \\ 4 + y & = & 12 \end{array} \right. \][/tex]
First, solve the second equation for [tex]\( y \)[/tex]:
[tex]\[ 4 + y = 12 \][/tex]
[tex]\[ y = 8 \][/tex]
Now, substitute [tex]\( y = 8 \)[/tex] into the first equation:
[tex]\[ 8 = 2x - 7 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 2x = 8 + 7 \][/tex]
[tex]\[ 2x = 15 \][/tex]
[tex]\[ x = \frac{15}{2} \][/tex]
Therefore, the solution to the fourth system is [tex]\((x, y) = \left( \frac{15}{2}, 8 \right) \)[/tex].
### Summary of Solutions:
1. [tex]\(\left( -5, \frac{13}{2} \right)\)[/tex]
2. [tex]\((2, -1)\)[/tex]
3. [tex]\(\left( \frac{8}{3}, 7 \right)\)[/tex]
4. [tex]\(\left( \frac{15}{2}, 8 \right)\)[/tex]
These solutions can indeed be represented by the graphs of these systems of equations.
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.