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# 1.5 - Linear Equations, Functions, zeros.

1. Distribute.
2. Gather like terms.
3. Do the same to both sides of the equation.
4. Isolate the variable (often x).

## Example 1. Solve $$7x+2=17$$

| Step | Operation |
|---------------------------------|-------------------------------|
| $7x + 2 = 17$ | Original equation |
| $7x + 2 - 2 = 17 - 2$ | Subtract 2 from both sides |
| $7x = 15$ | Simplify $17-2=15$ |
| $\dfrac{7x}{7} = \dfrac{15}{7}$| Divide both sides by 7 |
| $x = \dfrac{15}{7}$| Simplify $7x/7=x$ |

## Example 2. Solve $$4x+9=2x-7$$

| Step | Operation |
|------------------------------------|------------------------------------|
| $4x + 9 = 2x - 7$ | Original equation |
| $4x + 9 - 2x = 2x - 7 - 2x$ | Subtract $2x$ from both sides |
| $2x + 9 = -7$ | Simplify $4x-2x=2x$ |
| $2x + 9 - 9 = -7 - 9$ | Subtract 9 from both sides |
| $2x = -16$ | Simplify $-7-9=-16$ |
| $\dfrac{2x}{2} = \dfrac{-16}{2}$ | Divide both sides by 2 |
| $x = -8$ | Simplify $-16/2=-8$ |

## Example 3. Solve $$2x+3=2x-7$$

| Step | Operation |
|------------------------------------|------------------------------------|
| $2x + 3 = 2x - 7$ | Original equation |
| $2x + 3 - 2x = 2x - 7 - 2x$ | Subtract $2x$ from both sides |
| $3 = -7$ | Simplify $2x-2x=0$ |
| $3 \neq -7$ | This is a contradiction |

**Conclusion: The equation has no solution.**

## Example 4. Solve $$4(x+2)=2(2x-7)+22$$

| Step | Operation |
|--------------------------------------------|------------------------------------|
| $4(x + 2) = 2(2x - 7) + 22$ | Original equation |
| $4 \cdot x + 4 \cdot 2 = 2 \cdot 2x + 2 \cdot (-7) + 22$ | Apply distribution: $4$ distributes over $(x+2)$ and $2$ over $(2x-7)$ |
| $4x + 8 = 4x - 14 + 22$ | Simplify: $4 \cdot 2 = 8$, $2 \cdot 2x = 4x$, $2 \cdot (-7) = -14$ |
| $4x + 8 = 4x + 8$ | Simplify: $-14 + 22 = 8$ |
| $4x + 8 - 4x = 4x + 8 - 4x$ | Subtract $4x$ from both sides |
| $8 = 8$ | Simplify |

### Conclusion:
This equation is always true, meaning any value of $x$ will satisfy the equation. Therefore, the solution is that **the equation has infinitely many solutions**.

## Example 5. Solve $$9x + 4 = 20$$

| Step | Operation |
|------------------------------------|------------------------------------|
| $9x + 4 = 20$ | Original equation |
| $9x + 4 - 4 = 20 - 4$ | Subtract 4 from both sides |
| $9x = 16$ | Simplify $20-4=16$ |
| $\dfrac{9x}{9} = \dfrac{16}{9}$ | Divide both sides by 9 |
| $x = \dfrac{16}{9}$ | Simplify $9x/9=x$ |

## Example 6. Solve $$2x - 8 = 7x + 9$$

| Step | Operation |
|------------------------------------|------------------------------------|
| $2x - 8 = 7x + 9$ | Original equation |
| $2x - 7x - 8 = 7x + 9 - 7x$ | Subtract $7x$ from both sides |
| $-5x - 8 = 9$ | Simplify |
| $-5x - 8 + 8 = 9 + 8$ | Add 8 to both sides |
| $-5x = 17$ | Simplify |
| $\dfrac{-5x}{-5} = \dfrac{17}{-5}$| Divide both sides by $-5$ |
| $x = -\dfrac{17}{5}$ | Simplify |

## Example 7. Solve $$3(x - 5) = x + 2(x + 4) - 2$$

| Step | Operation |
|---------------------------------------------|------------------------------------|
| $3(x - 5) = x + 2(x + 4) - 2$ | Original equation |
| $3x - 15 = x + 2x + 8 - 2$ | Distribute $3$ and $2$ |
| $3x - 15 = 3x + 6$ | Simplify |
| $3x - 3x - 15 = 3x + 6 - 3x$ | Subtract $3x$ from both sides |
| $-15 = 6$ | Simplify |

### Conclusion:
There is a contradiction ($-15 \neq 6$), so **there is no solution** for this equation.

## Example 8. Solve $$2(x - 5) = 3(x + 2) - x - 16$$

| Step | Operation |
|----------------------------------------------|------------------------------------|
| $2(x - 5) = 3(x + 2) - x - 16$ | Original equation |
| $2x - 10 = 3x + 6 - x - 16$ | Distribute $2$ and $3$ |
| $2x - 10 = 2x - 10$ | Simplify |
| $2x - 10 - 2x = 2x - 10 - 2x$ | Subtract $2x$ from both sides |
| $-10 = -10$ | Simplify |

### Conclusion:
This equation is always true, so **there are infinitely many solutions**.

## Example 9. Solve. $$2x+3=\dfrac{1}{2}x-\dfrac{1}{3}$$

| Step | Operation |
|------------------------------------------------------|------------------------------------|
| $2x + 3 = \dfrac{1}{2}x - \dfrac{1}{3}$ | Original equation |
| $2x - \dfrac{1}{2}x + 3 = \dfrac{1}{2}x - \dfrac{1}{2}x - \dfrac{1}{3}$ | Subtract $\dfrac{1}{2}x$ from both sides |
| $\dfrac{4}{2}x - \dfrac{1}{2}x + 3 = -\dfrac{1}{3}$ | Express $2x$ as $\dfrac{4}{2}x$ to simplify |
| $\dfrac{3}{2}x + 3 = -\dfrac{1}{3}$ | Combine $\dfrac{4}{2}x - \dfrac{1}{2}x = \dfrac{3}{2}x$ |
| $\dfrac{3}{2}x = -\dfrac{1}{3} - 3$ | Subtract 3 from both sides |
| $\dfrac{3}{2}x = -\dfrac{1}{3} - \dfrac{9}{3}$ | Rewrite 3 as $\dfrac{9}{3}$ |
| $\dfrac{3}{2}x = -\dfrac{10}{3}$ | Combine fractions |
| $x = \dfrac{-10}{3} \times \dfrac{2}{3}$ | Multiply both sides by $\dfrac{2}{3}$ |
| $x = -\dfrac{20}{9}$ | Simplify |

## Example 10. Solve. $$3x+\dfrac{1}{2}=\dfrac{1}{2}x-5$$

| Step | Operation |
|------------------------------------------------------|------------------------------------|
| $3x + \dfrac{1}{2} = \dfrac{1}{2}x - 5$ | Original equation |
| $3x - \dfrac{1}{2}x + \dfrac{1}{2} = \dfrac{1}{2}x - \dfrac{1}{2}x - 5$ | Subtract $\dfrac{1}{2}x$ from both sides |
| $\dfrac{5}{2}x + \dfrac{1}{2} = -5$ | Simplify $3x - \dfrac{1}{2}x$ as $\dfrac{5}{2}x$ |
| $\dfrac{5}{2}x = -5 - \dfrac{1}{2}$ | Subtract $\dfrac{1}{2}$ from both sides |
| $\dfrac{5}{2}x = -\dfrac{10}{2} - \dfrac{1}{2}$ | Rewrite $-5$ as $-\dfrac{10}{2}$ |
| $\dfrac{5}{2}x = -\dfrac{11}{2}$ | Combine fractions |
| $x = \dfrac{-11}{2} \times \dfrac{2}{5}$ | Divide both sides by $\dfrac{5}{2}$ |
| $x = \dfrac{-11}{5}$ | Simplify |

# Zeroes of a Function

- If $f(c)=0$, we say: "c is a zero of $f(x)$"
- On the graph of $f(x)$, $(c,0)$ is an x-intercept.

## Example 11. Find the zeroes of $$f(x) = 5x + 7$$

| Step | Operation |
|------------------------------------|------------------------------------|
| $5x + 7 = 0$ | Set $f(x) = 0$ to find zeroes |
| $5x = -7$ | Subtract 7 from both sides |
| $\dfrac{5x}{5} = \dfrac{-7}{5}$ | Divide both sides by 5 |
| $x = -\dfrac{7}{5}$ | Simplify |

## Example 12. Find the zeroes of $$f(x) = -3x + 8$$

| Step | Operation |
|------------------------------------|------------------------------------|
| $-3x + 8 = 0$ | Set $f(x) = 0$ to find zeroes |
| $-3x = -8$ | Subtract 8 from both sides |
| $\dfrac{-3x}{-3} = \dfrac{-8}{-3}$ | Divide both sides by $-3$ |
| $x = \dfrac{8}{3}$ | Simplify |

## Example 13. Find the zeroes of $$f(x) = -\dfrac{2}{3}x + \dfrac{5}{4}$$

| Step | Operation |
|---------------------------------------------|----------------------------------------|
| $-\dfrac{2}{3}x + \dfrac{5}{4} = 0$ | Set $f(x) = 0$ to find zeroes |
| $-\dfrac{2}{3}x = -\dfrac{5}{4}$ | Subtract $\dfrac{5}{4}$ from both sides |
| $x = \dfrac{-5/4}{-2/3}$ | Divide both sides by $-\dfrac{2}{3}$ |
| $x = \dfrac{-5}{4} \times \dfrac{-3}{2}$ | Multiply by the reciprocal of $-\dfrac{2}{3}$ |
| $x = \dfrac{15}{8}$ | Simplify |

## Example 14. Find the zeroes of $$f(x) = 6$$

| Step | Operation |
|-------------------------|-----------------------------------|
| $6 = 0$ | Set $f(x) = 0$ to find zeroes |
| $6 = 0$ | Contradiction. |

### Conclusion:
There is **no solution** because the constant function \(f(x) = 6\) never equals zero. Thus, the equation has **no zeroes**.

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## Simple Interest Formula

- Simple interest. If an account begins with an amount $P$ (called the principal) and adds money at an annual rate of $r$ (given as a decimal) for a period of $t$ years, then the money earned (called interest) is given by $I=Prt$.

### **Example 15:**
If you begin with \$400 and invest it at an annual rate of 4% for 5 years, what is the final amount?

**Solution:**
The interest rate is found by shifting the decimal on 4% two units to the left: 0.04.

The interest earned is:
$I = (400)(0.04)(5) = 80$

The interest is \$80, and so the final amount is \$480.

### **Example 16:**
If you begin with \$700 and invest it at an annual rate of 3% for 7 years, what is the final amount?

To find the final amount, we can calculate the interest using the simple interest formula:

$$
I = P \cdot r \cdot t
$$

Where:
- $P = 700$ (the principal)
- $r = 0.03$ (the annual rate of 3% as a decimal)
- $t = 7$ years

The interest earned will be:

$$
I = 700 \cdot 0.03 \cdot 7 = 147
$$

The total final amount is the principal plus the interest:

$$
\text{Final amount} = 700 + 147 = 847
$$

Thus, the final amount is **\$847**.