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nalin [4]
3 years ago
9

Solve: 3(2x -5) = 6x -15

Mathematics
2 answers:
poizon [28]3 years ago
5 0

Answer:

x=x

Step-by-step explanation:

3 (2x-5) = 6x - 15

6x - 15 = 6x - 15

6x = 6x

x = x

Debora [2.8K]3 years ago
3 0

Answer:

x = all real values

Step-by-step explanation:

3(2x -5) = 6x -15

Distribute

3*2x - 3*5 = 6x-15

6x -15 = 6x -15

Subtract 6x from each side

6x-15-6x = 6x-15-6x

-15 = -15

This is always true so there are infinite solutions

x = all real values

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Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2
inna [77]
(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

6 x^{6}/3x = 2 x^{5} \\  \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\

We repeat previous steps until we run out of numbers:
3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2
\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )
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