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abruzzese [7]
3 years ago
9

3(x + 4) + 2 = 2 + 5(x – 4) what is the fifth step in solving this equation.

Mathematics
1 answer:
kotegsom [21]3 years ago
8 0
3(x+4)+2= 2+5(x-4)

3x+12+2= 2+ 5x-20

3x+14= 5x-18

3x+14-14= 5x-18-14

3x= 5x-32

3x-5x = 5x-32-5x

-2x= -32

-2x/ -2= -32/-2

x= 16

So answer A combine
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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.

D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}.

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

x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}.

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

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