Answer :
To determine which of the given equations represents a nonlinear function, we must analyze each equation in terms of its degree. The degree of an equation is defined as the highest power of the variable(s) in the equation.
Let's examine these equations one by one:
### Equation 1: [tex]\( x(y-5) = 2 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ x(y-5) = 2 \][/tex]
[tex]\[ xy - 5x = 2 \][/tex]
The degree of a term is the sum of the exponents of the variables in that term:
- The term [tex]\( xy \)[/tex] has a degree of [tex]\( 1 + 1 = 2 \)[/tex]
- The term [tex]\( -5x \)[/tex] has a degree of 1
The highest degree term is [tex]\( xy \)[/tex], which is of degree 2. Therefore, this equation is nonlinear.
### Equation 2: [tex]\( y - 2(x+9) = 0 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ y - 2(x+9) = 0 \][/tex]
[tex]\[ y - 2x - 18 = 0 \][/tex]
- The term [tex]\( y \)[/tex] has a degree of 1
- The term [tex]\( -2x \)[/tex] has a degree of 1
- The term [tex]\( -18 \)[/tex] is a constant with degree 0
The highest degree term is of degree 1. Therefore, this equation is linear.
### Equation 3: [tex]\( 3y + 6(2-x) = 5 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ 3y + 6(2-x) = 5 \][/tex]
[tex]\[ 3y + 12 - 6x = 5 \][/tex]
[tex]\[ 3y - 6x + 12 = 5 \][/tex]
[tex]\[ 3y - 6x + 7 = 0 \][/tex]
- The term [tex]\( 3y \)[/tex] has a degree of 1
- The term [tex]\( -6x \)[/tex] has a degree of 1
- The term [tex]\( 7 \)[/tex] is a constant with degree 0
The highest degree term is of degree 1. Therefore, this equation is linear.
### Equation 4: [tex]\( 2(y + x) = 0 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ 2(y + x) = 0 \][/tex]
[tex]\[ 2y + 2x = 0 \][/tex]
- The term [tex]\( 2y \)[/tex] has a degree of 1
- The term [tex]\( 2x \)[/tex] has a degree of 1
The highest degree term is of degree 1. Therefore, this equation is linear.
After evaluating all the equations:
- The first equation [tex]\( x(y-5)=2 \)[/tex] is nonlinear.
- The second equation [tex]\( y - 2(x+9)=0 \)[/tex] is linear.
- The third equation [tex]\( 3y + 6(2-x)=5 \)[/tex] is linear.
- The fourth equation [tex]\( 2(y+x)=0 \)[/tex] is linear.
Thus, the equation that represents a nonlinear function is:
[tex]\[ x(y-5) = 2 \][/tex]
Let's examine these equations one by one:
### Equation 1: [tex]\( x(y-5) = 2 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ x(y-5) = 2 \][/tex]
[tex]\[ xy - 5x = 2 \][/tex]
The degree of a term is the sum of the exponents of the variables in that term:
- The term [tex]\( xy \)[/tex] has a degree of [tex]\( 1 + 1 = 2 \)[/tex]
- The term [tex]\( -5x \)[/tex] has a degree of 1
The highest degree term is [tex]\( xy \)[/tex], which is of degree 2. Therefore, this equation is nonlinear.
### Equation 2: [tex]\( y - 2(x+9) = 0 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ y - 2(x+9) = 0 \][/tex]
[tex]\[ y - 2x - 18 = 0 \][/tex]
- The term [tex]\( y \)[/tex] has a degree of 1
- The term [tex]\( -2x \)[/tex] has a degree of 1
- The term [tex]\( -18 \)[/tex] is a constant with degree 0
The highest degree term is of degree 1. Therefore, this equation is linear.
### Equation 3: [tex]\( 3y + 6(2-x) = 5 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ 3y + 6(2-x) = 5 \][/tex]
[tex]\[ 3y + 12 - 6x = 5 \][/tex]
[tex]\[ 3y - 6x + 12 = 5 \][/tex]
[tex]\[ 3y - 6x + 7 = 0 \][/tex]
- The term [tex]\( 3y \)[/tex] has a degree of 1
- The term [tex]\( -6x \)[/tex] has a degree of 1
- The term [tex]\( 7 \)[/tex] is a constant with degree 0
The highest degree term is of degree 1. Therefore, this equation is linear.
### Equation 4: [tex]\( 2(y + x) = 0 \)[/tex]
Rewrite this equation in standard form:
[tex]\[ 2(y + x) = 0 \][/tex]
[tex]\[ 2y + 2x = 0 \][/tex]
- The term [tex]\( 2y \)[/tex] has a degree of 1
- The term [tex]\( 2x \)[/tex] has a degree of 1
The highest degree term is of degree 1. Therefore, this equation is linear.
After evaluating all the equations:
- The first equation [tex]\( x(y-5)=2 \)[/tex] is nonlinear.
- The second equation [tex]\( y - 2(x+9)=0 \)[/tex] is linear.
- The third equation [tex]\( 3y + 6(2-x)=5 \)[/tex] is linear.
- The fourth equation [tex]\( 2(y+x)=0 \)[/tex] is linear.
Thus, the equation that represents a nonlinear function is:
[tex]\[ x(y-5) = 2 \][/tex]
