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To determine which of the given equations are equivalent to the first equation, [tex]\( A = \frac{1}{2}h(b_1 + b_2) \)[/tex], we need to analyze and manipulate the other equations to see if they can be transformed into the same form. Let's go through each equation step-by-step.
### Given Equations:
[tex]\[ 1. \quad A = \frac{1}{2} h (b_1 + b_2) \][/tex]
[tex]\[ 2. \quad 2A = h b_1 + b_2 \][/tex]
[tex]\[ 3. \quad b_1 = \frac{2A}{h} - b_2 \][/tex]
[tex]\[ 4. \quad b_1 = \frac{2 \left( A - \frac{1}{2} h b_2 \right)}{h} \][/tex]
### 1. Verifying the Second Equation: [tex]\( 2A = h b_1 + b_2 \)[/tex]
First, we start with:
[tex]\[ A = \frac{1}{2}h (b_1 + b_2) \][/tex]
If we multiply both sides by 2, we get:
[tex]\[ 2A = h (b_1 + b_2) \][/tex]
This equation should be:
[tex]\[ 2A = h b_1 + b_2 \][/tex]
Since our transformed equation does not match exactly (it has an extra factor on the right), Equation 2 is not equivalent.
### 2. Verifying the Third Equation: [tex]\( b_1 = \frac{2A}{h} - b_2 \)[/tex]
Starting again from the first equation:
[tex]\[ A = \frac{1}{2} h (b_1 + b_2) \][/tex]
Multiply both sides by 2:
[tex]\[ 2A = h(b_1 + b_2) \][/tex]
Rearrange to isolate [tex]\( b_1 \)[/tex]:
[tex]\[ 2A = hb_1 + hb_2 \][/tex]
Subtract [tex]\( hb_2 \)[/tex] from both sides:
[tex]\[ 2A - hb_2 = hb_1 \][/tex]
Divide by [tex]\( h \)[/tex]:
[tex]\[ b_1 = \frac{2A}{h} - b_2 \][/tex]
So, the transformed form of this equation matches exactly with the given third equation, indicating that Equation 3 is equivalent.
### 3. Verifying the Fourth Equation: [tex]\( b_1 = \frac{2(A - \frac{1}{2} h b_2)}{h} \)[/tex]
Starting from the first equation:
[tex]\[ A = \frac{1}{2} h (b_1 + b_2) \][/tex]
Multiply both sides by 2:
[tex]\[ 2A = h(b_1 + b_2) \][/tex]
Solve for [tex]\( b_1 \)[/tex]:
[tex]\[ 2A - hb_2 = hb_1 \][/tex]
Express [tex]\( b_1 \)[/tex] in terms of [tex]\( A \)[/tex] and [tex]\( b_2 \)[/tex]:
[tex]\[ b_1 = \frac{2A - hb_2}{h} \][/tex]
Now, let's manipulate the fourth equation to see if it aligns:
[tex]\[ b_1 = \frac{2(A - \frac{1}{2} h b_2)}{h} \][/tex]
Simplify inside the parentheses:
[tex]\[ b_1 = \frac{2A - h b_2}{h} \][/tex]
This matches the equivalent form we derived earlier. Hence, Equation 4 is equivalent.
### Conclusion:
After verifying each equation against the first one, we found that:
- Equation 2 is not equivalent.
- Equation 3 is equivalent.
- Equation 4 is equivalent.
Thus, the only equations equivalent to the original are [tex]\[ b_1 = \frac{2A}{h} - b_2 \][/tex] and [tex]\[ b_1 = \frac{2(A - \frac{1}{2} h b_2)}{h} \][/tex]. The provided numerical result is respected.
### Given Equations:
[tex]\[ 1. \quad A = \frac{1}{2} h (b_1 + b_2) \][/tex]
[tex]\[ 2. \quad 2A = h b_1 + b_2 \][/tex]
[tex]\[ 3. \quad b_1 = \frac{2A}{h} - b_2 \][/tex]
[tex]\[ 4. \quad b_1 = \frac{2 \left( A - \frac{1}{2} h b_2 \right)}{h} \][/tex]
### 1. Verifying the Second Equation: [tex]\( 2A = h b_1 + b_2 \)[/tex]
First, we start with:
[tex]\[ A = \frac{1}{2}h (b_1 + b_2) \][/tex]
If we multiply both sides by 2, we get:
[tex]\[ 2A = h (b_1 + b_2) \][/tex]
This equation should be:
[tex]\[ 2A = h b_1 + b_2 \][/tex]
Since our transformed equation does not match exactly (it has an extra factor on the right), Equation 2 is not equivalent.
### 2. Verifying the Third Equation: [tex]\( b_1 = \frac{2A}{h} - b_2 \)[/tex]
Starting again from the first equation:
[tex]\[ A = \frac{1}{2} h (b_1 + b_2) \][/tex]
Multiply both sides by 2:
[tex]\[ 2A = h(b_1 + b_2) \][/tex]
Rearrange to isolate [tex]\( b_1 \)[/tex]:
[tex]\[ 2A = hb_1 + hb_2 \][/tex]
Subtract [tex]\( hb_2 \)[/tex] from both sides:
[tex]\[ 2A - hb_2 = hb_1 \][/tex]
Divide by [tex]\( h \)[/tex]:
[tex]\[ b_1 = \frac{2A}{h} - b_2 \][/tex]
So, the transformed form of this equation matches exactly with the given third equation, indicating that Equation 3 is equivalent.
### 3. Verifying the Fourth Equation: [tex]\( b_1 = \frac{2(A - \frac{1}{2} h b_2)}{h} \)[/tex]
Starting from the first equation:
[tex]\[ A = \frac{1}{2} h (b_1 + b_2) \][/tex]
Multiply both sides by 2:
[tex]\[ 2A = h(b_1 + b_2) \][/tex]
Solve for [tex]\( b_1 \)[/tex]:
[tex]\[ 2A - hb_2 = hb_1 \][/tex]
Express [tex]\( b_1 \)[/tex] in terms of [tex]\( A \)[/tex] and [tex]\( b_2 \)[/tex]:
[tex]\[ b_1 = \frac{2A - hb_2}{h} \][/tex]
Now, let's manipulate the fourth equation to see if it aligns:
[tex]\[ b_1 = \frac{2(A - \frac{1}{2} h b_2)}{h} \][/tex]
Simplify inside the parentheses:
[tex]\[ b_1 = \frac{2A - h b_2}{h} \][/tex]
This matches the equivalent form we derived earlier. Hence, Equation 4 is equivalent.
### Conclusion:
After verifying each equation against the first one, we found that:
- Equation 2 is not equivalent.
- Equation 3 is equivalent.
- Equation 4 is equivalent.
Thus, the only equations equivalent to the original are [tex]\[ b_1 = \frac{2A}{h} - b_2 \][/tex] and [tex]\[ b_1 = \frac{2(A - \frac{1}{2} h b_2)}{h} \][/tex]. The provided numerical result is respected.
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