Wiley Edge Equations Quiz-2

Let the distance between Town X and Town Y be "D" kilometers.
Time taken to go from X to Y = Distance / Speed = D / 20
Time taken to return from Y to X = Distance / Speed = D / 30
Total travel time is given as 5 hours:
D / 20 + D / 30 = 5
To solve for D, find a common denominator:
(3D + 2D) / (20 * 30) = 5
Combine like terms:
5D / 600 = 5
Now, solve for D:
5D = 5 * 600
5D = 3000
Divide both sides by 5:
D = 3000 / 5
D = 600 km

Expand the left side: 6x - 12 + 5x = 20
Combine like terms: 11x - 12 = 20
Add 12 to both sides: 11x = 32
Divide both sides by 11: x = 32 / 11.

Solution:
Solve the equation for y^{2}: 3y^{2} = 12
Add 12 to both sides of the equation: 3y^{2} = 12.
Divide both sides by 3:
y^{2} = 4.<br /> So, the value of y^{2} is 4.

Solution:
Let the speed of the boat in still water be "B" km/h.
When traveling downstream (with the current), the boat's effective speed is (B + 5) km/h, and the time is 2 hours:
Distance = Speed × Time
Distance = (B + 5) km/h × 2 h = 2B + 10 km
When traveling upstream (against the current), the boat's effective speed is (B - 5) km/h, and the time is 3 hours:
Distance = Speed × Time
Distance = (B - 5) km/h × 3 h = 3B - 15 km
Since the distance from A to B is the same as the distance from B to A, we can set the two expressions for distance equal to each other:
2B + 10 = 3B - 15
Now, solve for B:
B - 3B = -15 - 10
-2B = -25
Divide both sides by -2:
B = 25 / 2
B = 12.5 km/h

Solution: To solve this quadratic equation, we can use the quadratic formula:
x= \frac{-b\pm{\sqrt{b^{2}-4ac}}}{2a}
In this case, a = 2, b = 5, and c = -3.
x= \frac{-5±{\sqrt{5^{2}-4.3.(-3)}}}{2.2}<br /> x= \frac{-5±{\sqrt{25+24}}}{4}<br /> x= \frac{-5±{\sqrt{49}}}{4}
Now, we can simplify the square root:
x= \frac{-5\pm7}{4}
Now, calculate both roots:
x= \frac{-5+7}{4}= x= \frac{2}{4}= \frac{1}{2}<br /> x= \frac{-5-7}{4}= x= \frac{-12}{4}=-3<br /> So, the solutions are x = \frac{1}{2} and x= -3

Solution: Start by isolating the square root term: \sqrt{(2x + 5)}= 2x - 4
Square both sides: 2x + 5 = 4x^{2} - 16x + 16
Rearrange and simplify: 4x^{2} - 18x + 11 = 0<br /> Use the quadratic formula: x = \frac{(18 ± \sqrt{(18^{2} - 4 * 4 * 11)})}{(2 * 4)}
Solve for x:
x ≈ 4.63 or x ≈ 0.37.

Solution: Start by isolating the exponential term: 3e^{(2x)} = 12
Divide both sides by 3: e^{(2x)} = 4
Take the natural logarithm of both sides: 2x = ln{4}
Solve for x: x = \frac{\ln4 }{2}.

Answer:
To solve this equation, you can use logarithmic properties. First, combine the logarithms on the left side using the product rule for logarithms:
log_{2}((x+3)(x − 1)) = 3

Now, rewrite the equation in exponential form:
2^{3} = (x+3)(x-1).

Simplify: 8= x^{2}+2x-3.
Rearrange: x^{2}+2x-11-0.
Now, you can solve this quadratic equation using the quadratic formula:
x= \frac{-b\pm{\sqrt{b^{2}-4ac}}}{2a}
In this case, a = 1, b = 2, and c = -11.

Calculate:
x= \frac{-2\pm{\sqrt{2^{2}-4(1)(-11)}}}{2(1)}
x= \frac{-2\pm{\sqrt{4+44}}}{2}
x= \frac{-2\pm4{\sqrt{3}}}{2}

Now, simplify the fraction:
x= -1\pm2{\sqrt{3}}
So, the solutions are x= -1+2{\sqrt{3}} and x= -1-2{\sqrt{3}}

Solution: Combine the logarithms using logarithmic properties: log_{2}((y + 3)y) = 4
Simplify: log_{2}(y^{2} + 3y) = 4
Convert to exponential form: y^{2} + 3y = 2^{4}
Solve the quadratic equation: y^{2} + 3y - 16 = 0
Use the quadratic formula: y = (-3 \pm \sqrt{(3^{2} - 4 * 1 * -16)}) / (2 * 1)
Solve for y: y = 4 or y = -7.

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