linear or nonlinear
state whether the given functions are linear or nonlinear:

| function | linear/nonlinear |
| --- | --- |
| 9) $3x + 2y = 7$ | |
| 10) $xy + 4 = 1$ | |
| 11) $\frac{3}{x} = y$ | |
| 12) $x^2 - 3x + 4 = y$ | |
| 13) $y - x = 2$ | |
| 14) $4x^2 + y = 5$ | |
| 15) $9x - 2 = \frac{y}{2}$ | |
| 16) $\frac{y}{2x - 5} = -2$ | |
| 17) $3x - 2 = \frac{1}{y}$ | |
| 18) $3.6 = 2.1x + y$ | |

Question

linear or nonlinear
state whether the given functions are linear or nonlinear:

| function | linear/nonlinear |
| --- | --- |
| 9) $3x + 2y = 7$ |  |
| 10) $xy + 4 = 1$ |  |
| 11) $\frac{3}{x} = y$ |  |
| 12) $x^2 - 3x + 4 = y$ |  |
| 13) $y - x = 2$ |  |
| 14) $4x^2 + y = 5$ |  |
| 15) $9x - 2 = \frac{y}{2}$ |  |
| 16) $\frac{y}{2x - 5} = -2$ |  |
| 17) $3x - 2 = \frac{1}{y}$ |  |
| 18) $3.6 = 2.1x + y$ |  |

Accepted Answer

# Explanation:
## Step1: Recall Linear Function Form
A linear function has the form $ y = mx + b $ (or can be rewritten in this form), where $ m $ and $ b $ are constants, and the exponents of $ x $ and $ y $ are 1, and there are no products of $ x $ and $ y $, or variables in denominators.

## Step2: Analyze Each Function
### 9) $ 3x + 2y = 7 $
Rewrite as $ 2y=-3x + 7 $ or $ y=-\frac{3}{2}x+\frac{7}{2} $. This is in $ y = mx + b $ form. So **Linear**.

### 10) $ xy + 4 = 1 $
Simplify to $ xy=-3 $. There is a product of $ x $ and $ y $, so it cannot be written as $ y = mx + b $. **Nonlinear**.

### 11) $ \frac{3}{x}=y $
Rewrite as $ y = 3x^{-1} $. The variable $ x $ is in the denominator (exponent -1), so not in $ y = mx + b $ form. **Nonlinear**.

### 12) $ x^{2}-3x + 4 = y $
The term $ x^{2} $ has an exponent of 2, so it's a quadratic (not linear). **Nonlinear**.

### 13) $ y - x = 2 $
Rewrite as $ y=x + 2 $, which is in $ y = mx + b $ form ($ m = 1 $, $ b = 2 $). **Linear**.

### 14) $ 4x^{2}+y = 5 $
Rewrite as $ y=-4x^{2}+5 $. The $ x^{2} $ term means it's quadratic (not linear). **Nonlinear**.

### 15) $ 9x - 2=\frac{y}{2} $
Multiply both sides by 2: $ 18x-4 = y $ or $ y = 18x-4 $, which is in $ y = mx + b $ form. **Linear**.

### 16) $ \frac{y}{2x - 5}=-2 $
Multiply both sides by $ 2x - 5 $: $ y=-2(2x - 5)=-4x + 10 $? Wait, no—wait, the denominator has $ x $, so when we multiply, we assume $ 2x - 5
eq0 $, but the original equation has $ x $ in the denominator (since $ 2x - 5 $ is in the denominator of $ y $'s fraction). Wait, actually, rewrite: $ y=-2(2x - 5)=-4x + 10 $? No, wait, the original is $ \frac{y}{2x - 5}=-2 $, so $ y=-2(2x - 5) $ only if $ 2x - 5
eq0 $, but the equation as a function would have a restriction, and also, if we rearrange, it's $ y=-4x + 10 $ (but with a domain restriction). Wait, no—wait, the left side is $ \frac{y}{2x - 5} $, so to solve for $ y $, we get $ y=-2(2x - 5)=-4x + 10 $, but the original equation has $ 2x - 5 $ in the denominator, which would imply that $ 2x - 5
eq0 $, but the function $ y=-4x + 10 $ is linear, but the original equation has a variable in the denominator? Wait, no—wait, $ \frac{y}{2x - 5}=-2 $ can be rewritten as $ y=-2(2x - 5) $, which is $ y=-4x + 10 $, but the original equation has a denominator with $ x $, but when we solve for $ y $, the $ x $ is in the numerator. Wait, maybe I made a mistake. Wait, the equation is $ \frac{y}{2x - 5}=-2 $, so cross-multiplying gives $ y=-2(2x - 5) $, which is linear. Wait, but the original equation has $ 2x - 5 $ in the denominator, so the domain is $ x
eq\frac{5}{2} $, but the function $ y=-4x + 10 $ is linear. Wait, but actually, the equation $ \frac{y}{2x - 5}=-2 $ is equivalent to $ y=-4x + 10 $ (with $ x
eq\frac{5}{2} $), so it is linear? Wait, no—wait, the original equation is a rational equation in terms of $ x $ and $ y $, but when solved for $ y $, it's linear. Wait, maybe I was wrong earlier. Wait, let's check again. The key is: can it be written as $ y = mx + b $ with integer exponents 1 on $ x $ and $ y $, no products, no denominators with variables. Wait, in the original equation, $ 2x - 5 $ is in the denominator, but when we solve for $ y $, we get $ y=-4x + 10 $, which is linear. But the original equation has a denominator with $ x $, which would mean that the relation is not a function over all real numbers, but as a function (with domain $ x
eq\frac{5}{2} $), is it linear? Wait, maybe the problem is considering the equation's form. Wait, the left side is $ \frac{y}{2x - 5} $, which is a rational expression, so the equation is nonlinear because it has a variable in the denominator of the fraction containing $ y $. Wait, maybe I messed up. Let's re-express: $ \frac{y}{2x - 5}=-2 $ can be written as $ y=-2(2x - 5) $, but the original equation has a denominator with $ x $, so the equation is not linear because it's a rational equation. So **Nonlinear**.

### 17) $ 3x - 2=\frac{1}{y} $
Rewrite as $ y=\frac{1}{3x - 2} $, which has a variable in the denominator. So cannot be written as $ y = mx + b $. **Nonlinear**.

### 18) $ 3.6 = 2.1x + y $
Rewrite as $ y=-2.1x + 3.6 $, which is in $ y = mx + b $ form. So **Linear**.

# Answer:
| Function       | Linear/Nonlinear |
|----------------|------------------|
| $ 3x + 2y = 7 $ | Linear           |
| $ xy + 4 = 1 $  | Nonlinear        |
| $ \frac{3}{x}=y $ | Nonlinear        |
| $ x^{2}-3x + 4 = y $ | Nonlinear        |
| $ y - x = 2 $   | Linear           |
| $ 4x^{2}+y = 5 $ | Nonlinear        |
| $ 9x - 2=\frac{y}{2} $ | Linear           |
| $ \frac{y}{2x - 5}=-2 $ | Nonlinear        |
| $ 3x - 2=\frac{1}{y} $ | Nonlinear        |
| $ 3.6 = 2.1x + y $ | Linear           |

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Linear Function Form

Step2: Analyze Each Function

9) \( 3x + 2y = 7 \)

10) \( xy + 4 = 1 \)

11) \( \frac{3}{x}=y \)

12) \( x^{2}-3x + 4 = y \)

13) \( y - x = 2 \)

14) \( 4x^{2}+y = 5 \)

15) \( 9x - 2=\frac{y}{2} \)

16) \( \frac{y}{2x - 5}=-2 \)

Answer:

function	linear/nonlinear
10) $xy + 4 = 1$
11) $\frac{3}{x} = y$
12) $x^2 - 3x + 4 = y$
13) $y - x = 2$
14) $4x^2 + y = 5$
15) $9x - 2 = \frac{y}{2}$
16) $\frac{y}{2x - 5} = -2$
17) $3x - 2 = \frac{1}{y}$
18) $3.6 = 2.1x + y$

Function	Linear/Nonlinear
\( xy + 4 = 1 \)	Nonlinear
\( \frac{3}{x}=y \)	Nonlinear
\( x^{2}-3x + 4 = y \)	Nonlinear
\( y - x = 2 \)	Linear
\( 4x^{2}+y = 5 \)	Nonlinear
\( 9x - 2=\frac{y}{2} \)	Linear
\( \frac{y}{2x - 5}=-2 \)	Nonlinear
\( 3x - 2=\frac{1}{y} \)	Nonlinear
\( 3.6 = 2.1x + y \)	Linear