Answer :
Let's determine which of the given equations represent linear relationships.
### Equation 1: [tex]\( 5 + 2y = 13 \)[/tex]
First, we rearrange it to solve for [tex]\( y \)[/tex]:
[tex]\[ 5 + 2y = 13 \][/tex]
Subtract 5 from both sides:
[tex]\[ 2y = 8 \][/tex]
Divide both sides by 2:
[tex]\[ y = 4 \][/tex]
This equation is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 0 \)[/tex] and [tex]\( b = 4 \)[/tex], which is a linear equation.
### Equation 2: [tex]\( y = \frac{1}{2}x^2 + 7 \)[/tex]
This equation includes [tex]\( x^2 \)[/tex], which makes it a quadratic equation. Since it's not in the form [tex]\( y = mx + b \)[/tex], it is not a linear equation.
### Equation 3: [tex]\( y - 5 = 2(x - 1) \)[/tex]
First, let's simplify and rearrange this equation:
[tex]\[ y - 5 = 2(x - 1) \][/tex]
Distribute the 2 on the right side:
[tex]\[ y - 5 = 2x - 2 \][/tex]
Add 5 to both sides:
[tex]\[ y = 2x + 3 \][/tex]
This equation is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 2 \)[/tex] and [tex]\( b = 3 \)[/tex], which is a linear equation.
### Equation 4: [tex]\( \frac{y}{2} = x + 7 \)[/tex]
First, let's solve for [tex]\( y \)[/tex]:
[tex]\[ \frac{y}{2} = x + 7 \][/tex]
Multiply both sides by 2:
[tex]\[ y = 2(x + 7) \][/tex]
[tex]\[ y = 2x + 14 \][/tex]
This equation is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 2 \)[/tex] and [tex]\( b = 14 \)[/tex], which is a linear equation.
### Equation 5: [tex]\( x = -4 \)[/tex]
This equation does not have a [tex]\( y \)[/tex] variable and thus represents a vertical line. Even though it is a linear equation in one variable, for the purposes of identifying linear relationships typically involving both [tex]\( x \)[/tex] and [tex]\( y \)[/tex], it is considered linear in the sense that it maintains the constant.
### Summary
After inspecting all the equations, the ones that are linear are:
[tex]\[ 5 + 2y = 13 \][/tex]
[tex]\[ y - 5 = 2(x - 1) \][/tex]
[tex]\[ \frac{y}{2} = x + 7 \][/tex]
[tex]\[ x = -4 \][/tex]
Thus, the equations representing linear relationships are:
[tex]\[ 5 + 2y = 13 \][/tex]
[tex]\[ y - 5 = 2(x - 1) \][/tex]
[tex]\[ \frac{y}{2} = x + 7 \][/tex]
[tex]\[ x = -4 \][/tex]
The correct indices are:
[tex]\[ [0, 2, 3, 4] \][/tex]
### Equation 1: [tex]\( 5 + 2y = 13 \)[/tex]
First, we rearrange it to solve for [tex]\( y \)[/tex]:
[tex]\[ 5 + 2y = 13 \][/tex]
Subtract 5 from both sides:
[tex]\[ 2y = 8 \][/tex]
Divide both sides by 2:
[tex]\[ y = 4 \][/tex]
This equation is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 0 \)[/tex] and [tex]\( b = 4 \)[/tex], which is a linear equation.
### Equation 2: [tex]\( y = \frac{1}{2}x^2 + 7 \)[/tex]
This equation includes [tex]\( x^2 \)[/tex], which makes it a quadratic equation. Since it's not in the form [tex]\( y = mx + b \)[/tex], it is not a linear equation.
### Equation 3: [tex]\( y - 5 = 2(x - 1) \)[/tex]
First, let's simplify and rearrange this equation:
[tex]\[ y - 5 = 2(x - 1) \][/tex]
Distribute the 2 on the right side:
[tex]\[ y - 5 = 2x - 2 \][/tex]
Add 5 to both sides:
[tex]\[ y = 2x + 3 \][/tex]
This equation is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 2 \)[/tex] and [tex]\( b = 3 \)[/tex], which is a linear equation.
### Equation 4: [tex]\( \frac{y}{2} = x + 7 \)[/tex]
First, let's solve for [tex]\( y \)[/tex]:
[tex]\[ \frac{y}{2} = x + 7 \][/tex]
Multiply both sides by 2:
[tex]\[ y = 2(x + 7) \][/tex]
[tex]\[ y = 2x + 14 \][/tex]
This equation is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 2 \)[/tex] and [tex]\( b = 14 \)[/tex], which is a linear equation.
### Equation 5: [tex]\( x = -4 \)[/tex]
This equation does not have a [tex]\( y \)[/tex] variable and thus represents a vertical line. Even though it is a linear equation in one variable, for the purposes of identifying linear relationships typically involving both [tex]\( x \)[/tex] and [tex]\( y \)[/tex], it is considered linear in the sense that it maintains the constant.
### Summary
After inspecting all the equations, the ones that are linear are:
[tex]\[ 5 + 2y = 13 \][/tex]
[tex]\[ y - 5 = 2(x - 1) \][/tex]
[tex]\[ \frac{y}{2} = x + 7 \][/tex]
[tex]\[ x = -4 \][/tex]
Thus, the equations representing linear relationships are:
[tex]\[ 5 + 2y = 13 \][/tex]
[tex]\[ y - 5 = 2(x - 1) \][/tex]
[tex]\[ \frac{y}{2} = x + 7 \][/tex]
[tex]\[ x = -4 \][/tex]
The correct indices are:
[tex]\[ [0, 2, 3, 4] \][/tex]
