Solve for
. Round to the nearest 100th if needed.

20x+3=21               x=

-16x+9=40            x=

Let's solve each equation step by step.

### 1. Solve for \( x \) in the equation \( 20x + 3 = 21 \):

**Step 1:** Subtract 3 from both sides to isolate the term with \( x \).

\[
20x + 3 - 3 = 21 - 3
\]

\[
20x = 18
\]

**Step 2:** Divide both sides by 20 to solve for \( x \).

\[
x = \frac{18}{20} = 0.9
\]

So, \( x = 0.9 \).

### 2. Solve for \( x \) in the equation \( -16x + 9 = 40 \):

**Step 1:** Subtract 9 from both sides to isolate the term with \( x \).

\[
-16x + 9 - 9 = 40 - 9
\]

\[
-16x = 31
\]

**Step 2:** Divide both sides by -16 to solve for \( x \).

\[
x = \frac{31}{-16} \approx -1.94
\]

So, \( x \approx -1.94 \) when rounded to the nearest hundredth.

### Final Answers:
1. \( x = 0.9 \)
2. \( x \approx -1.94 \)

Would you like further details or have any questions?

Here are some related questions:
1. What happens if you substitute \( x = 0.9 \) back into the first equation?
2. What is the solution if \( 20x + 3 = 0 \)?
3. How would the solution change if the second equation was \( -16x + 9 = 0 \)?
4. What is the significance of rounding \( x \) to the nearest hundredth?
5. How do you solve a quadratic equation, such as \( ax^2 + bx + c = 0 \)?

**Tip:** Always double-check your solutions by substituting them back into the original equation.

Learn how to solve linear equations 20x + 3 = 21 and -16x + 9 = 40 step-by-step. Understand the process of isolating variables and performing operations to find solutions. Suitable for students in Grades 7-9.

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