coefficient:
if t = 0 then h(0) = 100

if t = 4 then h(4) = 338.4
16a + 4b + 100 = 338.4

if t = 10 then h(10) = 2
100a + 10b + 100 = 2  solve for the system (full solution)

We are given a system of two equations and are asked to solve for \(a\) and \(b\). Let's go step by step:

### Equations:
1. \( 16a + 4b + 100 = 338.4 \)
2. \( 100a + 10b + 100 = 2 \)

We will simplify these two equations first and then solve the system.

### Simplification:
For equation 1:
\[
16a + 4b + 100 = 338.4
\]
Subtract 100 from both sides:
\[
16a + 4b = 238.4 \tag{1}
\]

For equation 2:
\[
100a + 10b + 100 = 2
\]
Subtract 100 from both sides:
\[
100a + 10b = -98 \tag{2}
\]

### Solving the system:
We now have the system:
\[
16a + 4b = 238.4 \tag{1}
\]
\[
100a + 10b = -98 \tag{2}
\]

#### Step 1: Eliminate one variable (b)
To eliminate \(b\), we can multiply equation (1) by 2.5 so that the coefficient of \(b\) in both equations becomes the same.

Multiply equation (1) by 2.5:
\[
2.5(16a + 4b) = 2.5(238.4)
\]
\[
40a + 10b = 596 \tag{3}
\]

Now subtract equation (2) from equation (3):
\[
(40a + 10b) - (100a + 10b) = 596 - (-98)
\]
Simplifying:
\[
-60a = 694
\]
Solving for \(a\):
\[
a = \frac{694}{-60} = -\frac{347}{30} \approx -11.5667
\]

#### Step 2: Solve for \(b\)
Substitute \(a = -\frac{347}{30}\) into one of the original equations. We can use equation (1):
\[
16a + 4b = 238.4
\]
Substitute \(a = -\frac{347}{30}\):
\[
16\left(-\frac{347}{30}\right) + 4b = 238.4
\]
Simplify:
\[
-\frac{5552}{30} + 4b = 238.4
\]
Convert \(-\frac{5552}{30}\) to decimal:
\[
-184.0667 + 4b = 238.4
\]
Add 184.0667 to both sides:
\[
4b = 238.4 + 184.0667 = 422.4667
\]
Solve for \(b\):
\[
b = \frac{422.4667}{4} = 105.6167
\]

### Final Solution:
- \(a = -\frac{347}{30} \approx -11.5667\)
- \(b = 105.6167\)

Would you like a more detailed explanation, or do you have any questions?

#### 5 Related Questions:
1. How can we verify the solution for \(a\) and \(b\)?
2. What is the method to solve a system of linear equations by substitution?
3. How do you interpret the solution \(a\) and \(b\) in a real-world scenario?
4. Can we solve this system graphically?
5. What happens if we change one of the constants in the equations?

#### Tip:
When solving systems of equations, elimination is often faster than substitution when coefficients can be easily scaled.

coefficient: if t = 0 then h(0) = 100

if t = 4 then h(4) = 338.4
16a + 4b + 100 = 338.4

if t = 10 then h(10) = 2
100a + 10b + 100 = 2  solve for the system (full solution)

Learn how to solve a system of linear equations with the example of 16a + 4b + 100 = 338.4 and 100a + 10b + 100 = 2. This problem covers key algebraic concepts and uses the elimination method. Suitable for students in Grades 10-12.

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