Raniasibaouih
Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Here are graphs of two equations in a system.

Determine if each of these systems could be represented by the graphs. Be prepared to explain how you know.

[tex]\[
\left\{
\begin{array} { l }
x + 2y = 8 \\
x = -5
\end{array}
\right.
\quad
\left\{
\begin{array} { l }
y = -7x + 13 \\
y = -1
\end{array}
\right.
\quad
\left\{
\begin{array} { l }
3x = 8 \\
3x + y = 15
\end{array}
\right.
\quad
\left\{
\begin{array}{l}
y = 2x - 7 \\
4 + y = 12
\end{array}
\right.
\][/tex]

Sagot :

GinnyAnswer
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.
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.