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nikitadnepr [17]
2 years ago
5

How to solve the problem 3.2d-4d=2.3d+3

Mathematics
1 answer:
Liula [17]2 years ago
5 0
3.2d - 4d = 2.3d + 3...simplify by combining like terms
-0.8d = 2.3d + 3....subtract 2.3d from both sides
-0.8d - 2.3d = 3 ...simplify again
-3.1d = 3...divide both sides by -3.1
d = 3/ -3.1
d = - 0.97

or
3.2d - 4d = 2.3d + 3....multiply the equation by 10, gets rid of the decimals
32d - 40d = 23d + 30....subtract 23d from both sides
32d - 40d - 23d = 30....simplify
-31d = 30...divide by -31
d = -30/31
d = - 0.97
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1 step (B): raise both sides of the equation to the power of 2.

(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\   (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4.

2 step (A): simplify to obtain the final radical term on one side of the equation.

-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2.

3 step (F): raise both sides of the equation to the power of 2 again.

(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2.

4 step (E): simplify to get a quadratic equation.

4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0.

5 step (D): use the quadratic formula to find the values of x.

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6 step (C): apply the zero product rule.

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Additional 7 step: check these solutions, substituting into the initial equation.

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The final answer is 500.

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