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# Solving Multi-Step Linear Equations and Compound Inequalities

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What is the solution to the equation 15(x3+x2)=-5x24(x3+2)?

May 13, 2015

#6
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You are right up until the very last step shaniab29544.  On the next to last step you have x3 = 9.  Because the 3 comes after the x we assume this means x to the power 3 (at least that is what heureka and I have assumed), so x-cubed = 9.  This means that x is the cube root of 9 or 3√9.

If you prefer to interpret this as just x times 3 then we would have x times 3 = 9 so that x would be 9 divided by 3, or x = 3.

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May 16, 2015

#1
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What is the solution to the equation 1−5(x3+x2)=-5x2−4(x3+2) ?

$$\small{\text{ \begin{array}{rcl} 1-5(x^3+x^2) &=& -5x^2-4(x^3+2) \\\\ 1-5(x^3+x^2) + 5x^2 +4(x^3+2) &=& 0\\\\ 1 - 5x^3 - 5x^2 + 5x^2 + 4x^3 + 8 &=& 0 \\\\ -5x^3 - 5x^2 + 5x^2 + 4x^3 + 9 &=& 0 \\\\ -5x^3 \underbrace{- 5x^2 + 5x^2}_{=0} + 4x^3 + 9 &=& 0 \\\\ -5x^3 + 4x^3 + 9 &=& 0 \\\\ -x^3 + 9 &=& 0 \\\\ 9 &=& x^3 \\\\ x^3 &=& 9 \qquad | \qquad \sqrt{}\\\\ x &=& \sqrt{9}\\\\ x &=& 2.0800838231 \end{array} }}\\\\ \mathbf{ x = 2.0800838231 }$$ .
May 13, 2015
#2
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Another beautifully presented answer.  Thank you Heureka May 14, 2015
#3
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it wrong

May 15, 2015
#4
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What makes you think it's wrong shaniab29544?

heureka has given the correct real number root (there are two other roots, but they are complex).

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May 15, 2015
#5
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it like this  i think

1−5(x3+x2)=-5x2−4(x3+2)1−5x3−5x2=-5x2−4x3−81−5x2=-5x2+x3−89−5x2=-5x2+x39=x3       27=x

May 15, 2015
#6
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