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Nataly_w [17]
3 years ago
9

(1/(x-1))-(3/(x^2+2x-3))

%7B%7D%5E%7B2%7D%20%2B%202x%20-%203%20%7D%20" id="TexFormula1" title=" \frac{1}{x - 1} - \frac{3}{x {}^{2} + 2x - 3 } " alt=" \frac{1}{x - 1} - \frac{3}{x {}^{2} + 2x - 3 } " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
ycow [4]3 years ago
4 0

\dfrac{1}{x-1}-\dfrac{3}{x^2+2x-3} \\\dfrac{1}{x-1}-\dfrac{3}{(x+3)(x-1)} \\\dfrac{1\cdot(x+3)-3}{(x+3)(x-1)} \\\dfrac{x+3-3}{(x+3)(x-1)} \\\dfrac{x}{(x+3)(x-1)}

Hope this helps.

r3t40

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This problem is an example of solving equations with variables on both sides. To solve, we must first set up an equation for both the red balloon and the blue balloon. 

Since the red balloon rises at 2.6 meters per second, we can represent this part of the equation as 2.6s. The balloon is already 7.3 meters off of the ground, so we just add the 7.3 to the 2.6s: 

2.6s + 7.3

Since the blue balloon rises at 1.5 meters per second, we can represent this part of the equation as 1.5s. The balloon is already 12.4 meters off of the ground, so we just add the 12.4 to the 1.5:

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To determine when both balloons are at the same height, we set the two equations equal to each other: 

2.6s + 7.3 = 1.5s + 12.4

Then, we solve for s. First, the variables must be on the same side of the equation. We can do this by subtracting 1.5s from both sides of the equation: 

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This means that it will take 4.63 seconds for both balloons to reach the same height. If we want to know what height that is, we simply plug the 4.63 back into each equation: 

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