Let’s start solving the equations one by one.
First Set of Equations
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Equation 1: \[ 9x + 16 = 6x + 1 \] To isolate \( x \), first subtract \( 6x \) from both sides: \[ 9x - 6x + 16 = 1 \] This simplifies to: \[ 3x + 16 = 1 \] Next, subtract 16 from both sides: \[ 3x = 1 - 16 \] Simplifying further: \[ 3x = -15 \] Finally, divide by 3: \[ x = -5 \]
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Equation 2: \[ 4x - 3 = 3x - 9 \] To isolate \( x \), first subtract \( 3x \) from both sides: \[ 4x - 3x - 3 = -9 \] This simplifies to: \[ x - 3 = -9 \] Next, add 3 to both sides: \[ x = -9 + 3 \] Thus, we find: \[ x = -6 \]
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Equation 3: \[ 13x + 26 = 3x - 34 \] First, subtract \( 3x \) from both sides: \[ 13x - 3x + 26 = -34 \] Which simplifies to: \[ 10x + 26 = -34 \] Next, subtract 26 from both sides: \[ 10x = -34 - 26 \] Simplifying gives: \[ 10x = -60 \] Divide by 10: \[ x = -6 \]
Summary of the First Set:
- From the three equations we solved, we got:
- \( x = -5 \)
- \( x = -6 \)
- \( x = -6 \)
Second Set of Equation
Solve for \( ax \): \[ ax - 12 = 9 \] First, add 12 to both sides: \[ ax = 9 + 12 \] Simplifying gives: \[ ax = 21 \]
Real-World Problem
Finding the Length of a Rectangle: Given the area \( A = lw \): \[ A = 52.5 \text{ m}^2 \quad \text{and} \quad w = 5 \text{ m} \] Substituting these into the area formula gives: \[ 52.5 = l \times 5 \] To find \( l \), divide both sides by 5: \[ l = \frac{52.5}{5} \] Calculating this gives: \[ l = 10.5 \text{ m} \]
Explanation:
In this real-world problem, we found that the area of the rectangle is 52.5 square meters and the width is 5 meters. We set up the formula for area \( A = lw \) and rearranged the formula to solve for length \( l \). This process is practical when designing products or spaces, where knowing dimensions is crucial for fitting spaces or ensuring resources meet requirements. Here, the rectangle with a width of 5 m will have a length of 10.5 m to maintain an area of 52.5 m².
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