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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.
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